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#### Question 8:

Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are $\frac{x-3}{2}=\frac{y+1}{7}=\frac{z-2}{-3}.$

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is .

Here,

Vector equation of the required line is

#### Question 9:

The cartesian equations of a line are $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}.$ Find a vector equation for the line.

The cartesian equation of the given line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$.
It can be re-written as

$\frac{x-5}{3}=\frac{y-\left(-4\right)}{7}=\frac{z-6}{2}$

Thus, the given line passes through the point having position vector $\stackrel{\to }{a}=5\stackrel{^}{i}-4\stackrel{^}{j}+6\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=3\stackrel{^}{i}+7\stackrel{^}{j}+2\stackrel{^}{k}$.

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is .

Vector equation of the required line is

#### Question 10:

Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2}$. Also, reduce the equation obtained in vector form.

We know that the cartesian equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{m}$ is $\frac{x-{x}_{1}}{a}=\frac{y-{y}_{2}}{b}=\frac{z-{z}_{3}}{c}$.

Here,

Here,

Cartesian equation of the required line is

$\frac{x-1}{1}=\frac{y-\left(-1\right)}{2}=\frac{z-2}{-2}\phantom{\rule{0ex}{0ex}}⇒\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2}\phantom{\rule{0ex}{0ex}}$

We know that the cartesian equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{m}$ is $\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{m}$.
Here, the line is passing through the point and its direction ratios are proportional to 1, 2, $-$2.

Vector equation of the required line is
$\stackrel{\to }{r}=\left(\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}-2\stackrel{^}{k}\right)$

#### Question 11:

Find the direction cosines of the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}.$ Also, reduce it to vector form.

The cartesian equation of the given line is

$\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$

It can be re-written as

$\frac{x-4}{-2}=\frac{y-0}{6}=\frac{z-1}{-3}$

This shows that the given line passes through the point and its direction ratios are proportional to .

So, its direction cosines are

Thus, the given line passes through the point having position vector $\stackrel{\to }{a}=4\stackrel{^}{i}+\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=-2\stackrel{^}{i}+6\stackrel{^}{j}-3\stackrel{^}{k}$.

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is .

Here,

Vector equation of the required line is

#### Question 12:

The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.

The cartesian equation of the given line is

It can be re-written as

$\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c}$

Thus, the given line passes through the point  and its direction ratios are proportional to a, 1, c. It is also parallel to the vector $\stackrel{\to }{b}=a\stackrel{^}{i}+\stackrel{^}{j}+c\stackrel{^}{k}$.

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is .

Vector equation of the required line is

#### Question 13:

Find the vector equation of a line passing through the point with position vector $\stackrel{^}{i}-2\stackrel{^}{j}-3\stackrel{^}{k}$and parallel to the line joining the points with position vectors Also, find the cartesian equivalent of this equation.

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is .

Here,

Vector equation of the required line is

Reducing (1) to cartesian form, we get

#### Question 14:

Find the points on the line $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$ at a distance of 5 units from the point P (1, 3, 3).

The coordinates of any point on the line $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$ are given by

Let the coordinates of the desired point be .

The distance between this point and (1, 3, 3) is 5 units.

Substituting the values of $\lambda$ in (1), we get the coordinates of the desired point as ($-$2, $-$1, 3) and (4, 3, 7).

#### Question 15:

Show that the points whose position vectors are are collinear.

Let the given points be P, Q and R and let their position vectors be

Vector equation of line passing through P and Q is

If points P, Q and R are collinear, then point R must satisfy (1).

$7\stackrel{^}{i}+9\stackrel{^}{k}=\left(-2\stackrel{^}{i}+3\stackrel{^}{j}\right)+\lambda \left(3\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}$

Comparing the coefficients of , we get

$\lambda$=3

These three equations are consistent, i.e. they give the same value of $\lambda$.
Hence, the given three points are collinear.

Disclaimer: The question given in the book has a minor error. The third vectors should be $7\stackrel{^}{i}+9\stackrel{^}{k}$. The solution here is created accordingly.

#### Question 16:

Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line $\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}.$

We have

$\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}$

It can be re-written as

$\frac{x+2}{-1}=\frac{y+3}{7}=\frac{z-3}{\left(\frac{3}{2}\right)}\phantom{\rule{0ex}{0ex}}=\frac{x+2}{-2}=\frac{y+3}{14}=\frac{z-3}{3}$

This shows that the given line passes through the point and its direction ratios are proportional to $-$2, 14, 3.

Thus, the parallel vector is $\stackrel{\to }{b}=-2\stackrel{^}{i}+14\stackrel{^}{j}+3\stackrel{^}{k}$.

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is .

Here,
.

Vector equation of the required line is

Reducing (1) to cartesian form, we get

#### Question 17:

The cartesian equation of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios and also its vector equation.

The cartesian equation of the given line is

3x + 1 = 6y − 2 = 1 − z

It can be re-written as

$\frac{x+\frac{1}{3}}{\frac{1}{3}}=\frac{y-\frac{1}{3}}{\frac{1}{6}}=\frac{z-1}{-1}\phantom{\rule{0ex}{0ex}}=\frac{x-\left(-\frac{1}{3}\right)}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}$

Thus, the given line passes through the point $\left(-\frac{1}{3},\frac{1}{3},1\right)$ and its direction ratios are proportional to 2, 1, $-$6. It is parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}+\stackrel{^}{j}-6\stackrel{^}{k}$.

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is .

Vector equation of the required line is

#### Question 18:

Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.

The equation of the line 5x − 25 = 14 − 7y = 35z can be re-written as

Since the required line is parallel to the given line, so the direction ratios of the required line are proportional to .

The vector equation of the required line passing through the point (1, 2, $-$1) and having direction ratios proportional to is

$\stackrel{\to }{r}=\left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)+\lambda \left(7\stackrel{^}{i}-5\stackrel{^}{j}+\stackrel{^}{k}\right)$.

#### Question 1:

Show that the three lines with direction cosines are mutually perpendicular.

The direction cosines of the three lines are

Hence, the given lines are perpendicular to each other.

#### Question 2:

Show that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).

Suppose vector $\stackrel{\to }{a}$ is passing through the points (1, $-$1, 2) and (3, 4,$-$2) and $\stackrel{\to }{b}$ is passing through the points (0, 3, 2) and (3, 5, 6).

Then,

Now,
$\stackrel{\to }{a}.\stackrel{\to }{b}=\left(2\stackrel{^}{i}+5\stackrel{^}{j}-4\stackrel{⏜}{k}\right).\left(3\stackrel{^}{i}+2\stackrel{^}{j}+4\stackrel{⏜}{k}\right)=0$

Hence, the given lines are perpendicular to each other.

#### Question 3:

Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and, (1, 2, 5).

Equations of lines passing through the points are given by

$\frac{x-{x}_{1}}{{x}_{2}-{x}_{1}}=\frac{y-{y}_{1}}{{y}_{2}-{y}_{1}}=\frac{z-{z}_{1}}{{z}_{2}-{z}_{1}}$

So, the equation of a line passing through (4, 7, 8) and (2, 3, 4) is

$\frac{x-4}{2-4}=\frac{y-7}{3-7}=\frac{z-8}{4-8}\phantom{\rule{0ex}{0ex}}⇒\frac{x-4}{-2}=\frac{y-7}{-4}=\frac{z-8}{-4}$

Also, the equation of the line passing through the points ($-$1, $-$2, 1) and (1, 2, 5) is

$\frac{x+1}{1+1}=\frac{y+2}{2+2}=\frac{z-1}{5-1}\phantom{\rule{0ex}{0ex}}⇒\frac{x+1}{2}=\frac{y+2}{4}=\frac{z-1}{4}$

We know that two lines are parallel if

We observe

$\frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4}=-1\phantom{\rule{0ex}{0ex}}$

Hence, the given lines are parallel to each other.

#### Question 4:

Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}.$

We know that the cartesian equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is $\frac{x-{x}_{1}}{a}=\frac{y-{y}_{2}}{b}=\frac{z-{z}_{3}}{c}$.

Here,

The cartesian equation of the required line is

$\frac{x-\left(-2\right)}{3}=\frac{y-4}{5}=\frac{z-\left(-5\right)}{6}\phantom{\rule{0ex}{0ex}}=\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\phantom{\rule{0ex}{0ex}}$

#### Question 5:

Show that the lines are perpendicular to each other.

We have

These equations can be re-written as

Now,

Hence, the given two lines are perpendicular to each other.

#### Question 6:

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1).

The direction ratios of the line joining the origin to the point (2, 1, 1) are 2, 1, 1.

Let $\stackrel{\to }{{b}_{1}}=2\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}$

The direction ratios of the line joining the points (3, 5, $-$1) and (4, 3, $-$1) are 1, $-$2, 0.

Let $\stackrel{\to }{{b}_{2}}=\stackrel{^}{i}-2\stackrel{^}{j}+0\stackrel{^}{k}$

Now,

Hence, the two lines joining the given points are perpendicular to each other.

#### Question 7:

Find the equation of a line parallel to x-axis and passing through the origin.

The direction ratios of the line parallel to x-axis are proportional to 1, 0, 0.

Equation of the line passing through the origin and parallel to x-axis is

$\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}\phantom{\rule{0ex}{0ex}}=\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

#### Question 8:

Find the angle between the following pairs of lines:
(i)

(ii)

(iii)

(i)

Let be vectors parallel to the given lines.

Now,

If $\theta$ is the angle between the given lines, then

(ii)

Let be vectors parallel to the given lines.

Now,

If $\theta$ is the angle between the given lines, then

(iii)

Let be vectors parallel to the given lines.

Now,

If $\theta$ is the angle between the given lines, then

#### Question 9:

Find the angle between the following pairs of lines:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(i)

Let be vectors parallel to the given lines.

If $\theta$ is the angle between the given lines, then

(ii)

Let be vectors parallel to the given lines.

Now,

If $\theta$ is the angle between the given lines, then

(iii)

The equations of the given lines can be re-written as

Let be vectors parallel to the given lines.

Now,

If $\theta$ is the angle between the given lines, then

(iv)

The equations of the given lines can be re-written as

Let be vectors parallel to the given lines.

Now,

If $\theta$ is the angle between the given lines, then

(v )

The equations of the given lines can be re-written as

Let be vectors parallel to the given lines.

Now,
$\stackrel{\to }{{b}_{1}}=\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{{b}_{2}}=3\stackrel{^}{i}+4\stackrel{^}{j}+5\stackrel{^}{k}$

If $\theta$ is the angle between the given lines, then

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

(vi)

The equations of the given lines can be re-written as

Let be vectors parallel to the given lines.

Now,

If $\theta$ is the angle between the given lines, then

#### Question 10:

Find the angle between the pairs of lines with direction ratios proportional to
(i) 5, −12, 13 and −3, 4, 5
(ii) 2, 2, 1 and 4, 1, 8
(iii) 1, 2, −2 and −2, 2, 1
(iv) a, b, c and bc, ca, ab.

(i) 5, −12, 13 and −3, 4, 5

(ii) 2, 2, 1 and 4, 1, 8

(iii) 1, 2, −2 and −2, 2, 1

(iv) a, b, c and bc, ca, ab.

#### Question 11:

Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the  other one is obtained by joining the points (3, 1, 4) and (7, 2, 12).

The direction ratios of the line joining the points (3, 1, 4) and (7, 2, 12) are proportional to 4, 1, 8.

Let be vectors parallel to the lines having direction ratios proportional to 2, 2, 1 and 4, 1, 8.

Now,

If $\theta$ is the angle between the given lines, then

#### Question 12:

Find the equation of the line passing through the point (1, 2, −4) and parallel to the line $\frac{x-3}{4}=\frac{y-5}{2}=\frac{z+1}{3}.$

The direction ratios of the line parallel to line $\frac{x-3}{4}=\frac{y-5}{2}=\frac{z+1}{3}$ are proportional to 4, 2, 3.

Equation of the required line passing through the point (1, 2, $-$4) having direction ratios proportional to 4, 2, 3 is

$\frac{x-1}{4}=\frac{y-2}{2}=\frac{z-\left(-4\right)}{3}\phantom{\rule{0ex}{0ex}}=\frac{x-1}{4}=\frac{y-2}{2}=\frac{z+4}{3}$

#### Question 13:

Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line $\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}.$

The equation of line $\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}$ can be re-written as

$\frac{x-\frac{1}{2}}{2}=\frac{y+\frac{5}{3}}{\frac{2}{3}}=\frac{z-2}{-3}$

The direction ratios of the line parallel to line $\frac{2x-1}{4}=\frac{3y+5}{2}=\frac{2-z}{3}$ are proportional to 2, $\frac{2}{3}$, $-$3.

Equation of the required line passing through the point ($-$1, 2, 1) having direction ratios proportional to 2, $\frac{2}{3}$, $-$3 is

$\frac{x-\left(-1\right)}{2}=\frac{y-2}{\frac{2}{3}}=\frac{z-1}{-3}\phantom{\rule{0ex}{0ex}}=\frac{x+1}{2}=\frac{y-2}{\frac{2}{3}}=\frac{z-1}{-3}$

#### Question 14:

Find the equation of the line passing through the point (2, −1, 3) and parallel to the line $\stackrel{\to }{r}=\left(\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\right)+\mathrm{\lambda }\left(2\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}\right).$

The given line is parallel to the vector $2\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}$ and the required line is parallel to the given line.
So, the required line is parallel to the vector $2\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}$.

Hence, the equation of the required line passing through the point (2,$-$1, 3) and parallel to the vector $2\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}$ is $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}\right)+\mathrm{\lambda }\left(2\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}\right)$.

#### Question 15:

Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines

Let:

Since the required line is perpendicular to the lines parallel to the vectors , it is parallel to the vector $\stackrel{\to }{b}=\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}$.

Now,

Thus, the direction ratios of the required line are proportional to 2, $-$7, 4.

The equation of the required line passing through the point (2, 1, 3) and having direction ratios proportional to 2, $-$7, 4 is $\frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-3}{4}$.

#### Question 16:

Find the equation of the line passing through the point $\stackrel{^}{i}+\stackrel{^}{j}-3\stackrel{^}{k}$ and perpendicular to the lines

The required line is perpendicular to the lines parallel to the vectors .

So, the required line is parallel to the vector $\stackrel{\to }{b}=\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}$.

Now,

Equation of the required line passing through the point $\left(\stackrel{^}{i}+\stackrel{^}{j}-3\stackrel{^}{k}\right)$ and parallel to $\left(4\stackrel{^}{i}-5\stackrel{^}{j}+\stackrel{^}{k}\right)$ is $\stackrel{\to }{r}=\left(\stackrel{^}{i}+\stackrel{^}{j}-3\stackrel{^}{k}\right)+\lambda \left(4\stackrel{^}{i}-5\stackrel{^}{j}+\stackrel{^}{k}\right)$.

#### Question 17:

Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).

The direction ratios of the line joining the points (4, 3, 2), (1,$-$1, 0) and (1, 2, $-$1), (2, 1, 1) are $-$3, $-$4, $-$2 and 1, $-$1, 2, respectively.

Let:

Since the required line is perpendicular to the lines parallel to the vectors , it is parallel to the vector $\stackrel{\to }{b}=\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}$.

Now,

So, the direction ratios of the required line are proportional to $-$10, 4, 7.

The equation of the required line passing through the point (1, $-$1, 1) and having direction ratios proportional to $-$10, 4, 7 is $\frac{x-1}{-10}=\frac{y+1}{4}=\frac{z-1}{7}$.

#### Question 18:

Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines .

We have

$\frac{x-8}{8}=\frac{y+9}{-16}=\frac{z-10}{7}\phantom{\rule{0ex}{0ex}}\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$

Let:

Since the required line is perpendicular to the lines parallel to the vectors , it is parallel to the vector $\stackrel{\to }{b}=\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}$.

Now,

The direction ratios of the required line are proportional to 24, 61, 112.

The equation of the required line passing through the point (1, 2, $-$4) and having direction ratios proportional to 24, 61, 112 is $\frac{x-1}{24}=\frac{y-2}{61}=\frac{z+4}{112}$.

#### Question 19:

Show that the lines are perpendicular to each other.

The direction ratios of the lines are proportional to 7, $-$5, 1 and 1, 2, 3, respectively.

Let:

Now,

Hence, the given lines are perpendicular to each other.

#### Question 20:

Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2.

The equation of the line 6x − 2 = 3y + 1 = 2z − 2 can be re-written as

Since the required line is parallel to the given line, the direction ratios of the required line are proportional to .

The vector equation of the required line passing through the point (2, $-$1, $-$1) and having direction ratios proportional to is $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{j}-\stackrel{^}{k}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)$.

#### Question 21:

If the lines are perpendicular, find the value of λ.

The equations of the given lines are

Since the given lines are perpendicular to each other, we have

$-3\left(3\lambda \right)+2\lambda \left(1\right)+2\left(-5\right)=0\phantom{\rule{0ex}{0ex}}⇒-9\lambda +2\lambda -10=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{10}{7}$

#### Question 22:

If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

The direction ratios of AB and CD are proportional to 3, 3, 4 and 6, 6, 8, respectively.

Let $\theta$ be the angle between AB and CD. Then,

#### Question 23:

Find the value of λ so that the following lines are perpendicular to each other.

The equations of the given lines  can be re-written as

Since the given lines are perpendicular to each other, we have

$\left(5\lambda +2\right)1-5\left(2\lambda \right)+1\left(3\right)=0\phantom{\rule{0ex}{0ex}}⇒5\lambda =5\phantom{\rule{0ex}{0ex}}⇒\lambda =1$

#### Question 24:

Find the direction cosines of the line $\frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}$. Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.                                                              [CBSE 2014]

The equation of the given line is $\frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}$.
The given equation can be re-written as $\frac{x+2}{2}=\frac{y-\frac{7}{2}}{3}=\frac{z-5}{-6}$.

This line passes through the point $\left(-2,\frac{7}{2},5\right)$ and has direction ratios proportional to 2, 3, −6.

So, its direction cosines are

$\frac{2}{\sqrt{{2}^{2}+{3}^{2}+{\left(-6\right)}^{2}}},\frac{3}{\sqrt{{2}^{2}+{3}^{2}+{\left(-6\right)}^{2}}},\frac{-6}{\sqrt{{2}^{2}+{3}^{2}+{\left(-6\right)}^{2}}}$

Or $\frac{2}{7},\frac{3}{7},\frac{-6}{7}$

The required line passes through the point having position vector $\stackrel{\to }{a}=-\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}+3\stackrel{^}{j}-6\stackrel{^}{k}$.

So, its vector equation is

$\stackrel{\to }{r}=\left(-\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+3\stackrel{^}{j}-6\stackrel{^}{k}\right)$

#### Question 1:

Show that the lines intersect and find their point of intersection.

The coordinates of any point on the first line are given by

The coordinates of a general point on the first line are .

Also, the coordinates of any point on the second line are given by

The coordinates of a general point on the second line are

If the lines intersect, then they have a common point. So, for some values of , we must have

#### Question 2:

Show that the lines do not intersect.

The coordinates of any point on the first line are given by

The coordinates of a general point on the first line are .

The coordinates of any point on the second line are given by

The coordinates of a general point on the second line are .

If the lines intersect, then they have a common point. So, for some values of , we must have

#### Question 3:

Show that the lines intersect. Find their point of intersection.

The coordinates of any point on the first line are given by

The coordinates of a general point on the first line are .

The coordinates of any point on the second line are given by

The coordinates of a general point on the second line are .

If the lines intersect, then they have a common point. So, for some values of , we must have

#### Question 4:

Prove that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection.

The coordinates of any point on the line AB are given by

The coordinates of a general point on AB are .

The coordinates of any point on the line CD are given by

The coordinates of a general point on CD are .
If the lines AB and CD intersect, then they have a common point. So, for some values of , we must have

#### Question 5:

Prove that the line intersect and find their point of intersection.

The position vectors of two arbitrary points on the given lines are

If the lines intersect, then they have a common point. So, for some values of , we must have

Equating the coefficients of , we get

Solving (2) and (3), we get

Substituting the values in (1), we get

Substituting $\mu =0$ in the second line, we get $\stackrel{\to }{r}=4\stackrel{^}{i}+0\stackrel{^}{j}-\stackrel{^}{k}$ as the position vector of the point of intersection.

Thus, the coordinates of the point of intersection are (4, 0, $-$1).

#### Question 6:

Determine whether the following pair of lines intersect or not:
(i)

(ii)

(iii)

(iv)

(i)

The position vectors of two arbitrary points on the given lines are

If the lines intersect, then they have a common point. So, for some values of , we must have

Equating the coefficients of , we get

Solving (2) and (3), we get
.

Substituting the values in (1), we get

(ii)

The coordinates of any point on the first line are given by

The coordinates of a general point on the first line are .

Also, the coordinates of any point on the second line are given by

The coordinates of a general point on the second line are .

If the lines intersect, then they have a common point. So, for some values of , we must have

(iii)

The coordinates of any point on the first line are given by

The coordinates of a general point on the first line are .

Also, the coordinates of any point on the second line are given by

The coordinates of a general point on the second line are .

If the lines intersect, then they have a common point. So, for some values of , we must have

(iv)

The coordinates of any point on the first line are given by

The coordinates of a general point on the first line are .

The coordinates of any point on the second line are given by

The coordinates of a general point on the second line are .

If the lines intersect, then they have a common point. So, for some values of , we must have

Disclaimer: The question printed in the book is incorrect. Instead of z, 3 is printed.

#### Question 7:

Show that the lines are intersecting. Hence, find their point of intersection.

The position vectors of two arbitrary points on the given lines are

If the lines intersect, then they have a common point. So, for some values of , we must have

Equating the coefficients of , we get

Solving (1) and (2), we get
$\lambda =-4\phantom{\rule{0ex}{0ex}}\mu =-2$.

Substituting the values in (3), we get

Substituting $\mu =-2$ in the second line, we get $\stackrel{\to }{r}=5\stackrel{^}{i}-2\stackrel{^}{j}-6\stackrel{^}{i}-4\stackrel{^}{j}-12\stackrel{^}{k}=-\stackrel{^}{i}-6\stackrel{^}{j}-12\stackrel{^}{k}$ as the position vector of the point of intersection.

Thus, the coordinates of the point of intersection are ($-$1, $-$6, $-$12).

#### Question 1:

Find the perpendicular distance of the point (3, −1, 11) from the line $\frac{x}{2}=\frac{y-2}{-3}=\frac{z-3}{4}.$

Let the point (3, $-$1, 11) be P and the point through which the line passes be Q (0, 2, 3).
The line is parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}-3\stackrel{^}{j}+4\stackrel{^}{k}$.

Now,

$\stackrel{\to }{PQ}=-3\stackrel{^}{i}+3\stackrel{^}{j}-8\stackrel{^}{k}$

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

#### Question 2:

Find the perpendicular distance of the point (1, 0, 0) from the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}.$ Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

Let the point (1, 0, 0) be P and the point through which the line passes be Q (1, $-$1, $-$10).
The line is parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}-3\stackrel{^}{j}+8\stackrel{^}{k}$.

Now,

$\stackrel{\to }{PQ}=0\stackrel{^}{i}-\stackrel{^}{j}-10\stackrel{^}{k}$

Let L be the foot of the perpendicular drawn from the point P (1, 0, 0) to the given line.

The coordinates of a general point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ are given by

Let the coordinates of L be .

The direction ratios of PL are proportional to .

The direction ratios of the given line are proportional to 2 ,$-$3, 8, but PL is perpendicular to the given line.

Substituting $\lambda =1$ in , we get the coordinates of L as (3, $-$4, $-$2).

Equation of the line PL is given by

$\frac{x-1}{3-1}=\frac{y-0}{-4-0}=\frac{z-0}{-2-0}\phantom{\rule{0ex}{0ex}}=\frac{x-1}{1}=\frac{y}{-2}=\frac{z}{-1}$

$⇒\stackrel{\to }{r}=\stackrel{^}{i}+\lambda \left(\stackrel{^}{i}-2\stackrel{^}{j}-\stackrel{^}{k}\right)$

#### Question 3:

Find the foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3).

Let D be the foot of the perpendicular drawn from the point A (1, 0, 3) to the line BC.

The coordinates of a general point on the line BC are given by

Let the coordinates of D be .

The direction ratios of AD are proportional to .

The direction ratios of the line BC are proportional to 1, 2, $-$2, but AD is perpendicular to the line BC.

$\therefore 1\left(\lambda +3\right)+2\left(2\lambda +7\right)-2\left(-2\lambda -2\right)=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{7}{3}$
Substituting $\lambda =-\frac{7}{3}$ in , we get the coordinates of D as .

#### Question 4:

A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D.

Point D is the foot of the perpendicular drawn from the point A (1, 0, 4) to the line BC.

The coordinates of a general point on the line BC are given by

Let the coordinates of D be .

The direction ratios of AD are proportional to .

The direction ratios of the line BC are proportional to 2, 8,$-$2, but AD is perpendicular to the line BC.

Substituting $\lambda =\frac{11}{9}$ in , we get the coordinates of D as .

#### Question 5:

Find the foot of perpendicular from the point (2, 3, 4) to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}.$ Also, find the perpendicular distance from the given point to the line.

Let L be the foot of the perpendicular drawn from the point P (2, 3, 4) to the given line.

The coordinates of a general point on the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ are given by

Let the coordinates of L be .

The direction ratios of PL are proportional to .

The direction ratios of the given line are proportional to $-$2, 6, $-$3, but PL is perpendicular to the given line.

$\therefore -2\left(-2\lambda +2\right)+6\left(6\lambda -3\right)-3\left(-3\lambda -3\right)=0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{13}{49}$

Substituting $\lambda =\frac{13}{49}$ in , we get the coordinates of L as  .

.

Hence, the length of the perpendicular from P on PL is .

#### Question 6:

Find the equation of the perpendicular drawn from the point P (2, 4, −1) to the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}.$ Also, write down the coordinates of the foot of the perpendicular from P.

Let L be the foot of the perpendicular drawn from the point P (2, 4, $-$1) to the given line.

The coordinates of a general point on the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$ are given by

Let the coordinates of L be .

The direction ratios of PL are proportional to .

The direction ratios of the given line are proportional to 1, 4, $-$9, but PL is perpendicular to the given line.

$\therefore 1\left(\lambda -7\right)+4\left(4\lambda -7\right)-9\left(-9\lambda +7\right)=0\phantom{\rule{0ex}{0ex}}⇒\lambda =1$

Substituting $\lambda =1$ in , we get the coordinates of L as .

Equation of the line PL is

$\frac{x-2}{-4-2}=\frac{y-4}{1-4}=\frac{z+1}{-3+1}\phantom{\rule{0ex}{0ex}}=\frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z+1}{-2}$

#### Question 7:

Find the length of the perpendicular drawn from the point (5, 4, −1) to the line $\stackrel{\to }{r}=\stackrel{^}{i}+\lambda \left(2\stackrel{^}{i}+9\stackrel{^}{j}+5\stackrel{^}{k}\right).$

Let the point (5, 4, $-$1) be P and the the point through which the line passes be Q (1, 0, 0).
The line is parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}+9\stackrel{^}{j}+5\stackrel{^}{k}$.

Now,

$\stackrel{\to }{PQ}=-4\stackrel{^}{i}-4\stackrel{^}{j}+\stackrel{^}{k}$

#### Question 8:

Find the foot of the perpendicular drawn from the point $\stackrel{^}{i}+6\stackrel{^}{j}+3\stackrel{^}{k}$ to the line $\stackrel{\to }{r}=\stackrel{^}{j}+2\stackrel{^}{k}+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right).$ Also, find the length of the perpendicular

Let L be the foot of the perpendicular drawn from the point P ($\stackrel{^}{i}+6\stackrel{^}{j}+3\stackrel{^}{k}$) to the line $\stackrel{\to }{r}=\stackrel{^}{j}+2\stackrel{^}{k}+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right).$

Let the position vector L be
$\stackrel{\to }{r}=\stackrel{^}{j}+2\stackrel{^}{k}+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)=\lambda \stackrel{^}{i}+\left(1+2\lambda \right)\stackrel{^}{j}+\left(2+3\lambda \right)\stackrel{^}{k}$        ...(1)

Now,

Since $\stackrel{\to }{PL}$ is perpendicular to the given line, which is parallel to $\stackrel{\to }{b}=\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$, we have

Substituting $\lambda =1$ in (1), we get the position vector of L as $\stackrel{^}{i}+3\stackrel{^}{j}+5\stackrel{^}{k}$.

Substituting $\lambda =1$ in (2), we get
$\stackrel{\to }{PL}=-3\stackrel{^}{j}+2\stackrel{^}{k}$

Hence, the length of the perpendicular from point P on PL is .

#### Question 9:

Find the equation of the perpendicular drawn from the point P (−1, 3, 2) to the line $\stackrel{\to }{r}=\left(2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}\right).$ Also, find the coordinates of the foot of the perpendicular from P.

Let L be the foot of the perpendicular drawn from the point P ($-$1, 3, 2) to the line $\stackrel{\to }{r}=\left(2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}\right).$

Let the position vector L be
$\stackrel{\to }{r}=\left(2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}\right)=2\lambda \stackrel{^}{i}+\left(2+\lambda \right)\stackrel{^}{j}+\left(3+3\lambda \right)\stackrel{^}{k}$        ...(1)

Now,

Since $\stackrel{\to }{PL}$ is perpendicular to the given line, which is parallel to $\stackrel{\to }{b}=2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}$, we have

Substituting $\lambda =-\frac{2}{7}$ in (1), we get the position vector of L as $-\frac{4}{7}\stackrel{^}{i}+\frac{12}{7}\stackrel{^}{j}+\frac{15}{7}\stackrel{^}{k}$.

So, the coordinates of the foot of the perpendicular from P to the given line is L .
Substituting $\lambda =-\frac{2}{7}$ in (2), we get
$\stackrel{\to }{PL}=\frac{3}{7}\stackrel{^}{i}-\frac{9}{7}\stackrel{^}{j}+\frac{1}{7}\stackrel{^}{k}$

Equation of the perpendicular drawn from P to the given line is

#### Question 10:

Find the foot of the perpendicular from (0, 2, 7) on the line $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}.$

Let L be the foot of the perpendicular drawn from the point P (0, 2, 7) to the given line.

The coordinates of a general point on the line $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}$ are given by

Let the coordinates of L be .

The direction ratios of PL are proportional to .

The direction ratios of the given line are proportional to $-$1, 3, $-$2, but PL is perpendicular to the given line.

$\therefore -1\left(-\lambda -2\right)+3\left(3\lambda -1\right)-2\left(-2\lambda -4\right)=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{1}{2}$
Substituting $\lambda =-\frac{1}{2}$ in , we get the coordinates of L as .

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

#### Question 11:

Find the foot of the perpendicular from (1, 2, −3) to the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}.$

Let L be the foot of the perpendicular drawn from the point P (1, 2, $-$3) to the given line.

The coordinates of a general point on the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}$ are given by

Let the coordinates of L be .

The direction ratios of PL are proportional to .

The direction ratios of the given line are proportional to 2, $-$2, $-$1, but PL is perpendicular to the given line.

Substituting $\lambda =1$ in , we get the coordinates of L as .

#### Question 12:

Find the equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD.

Equation of line AB passing through the points A(0, 6,$-$9) and B($-$3, $-$6, 3) is

$\frac{x-0}{-3-0}=\frac{y-6}{-6-6}=\frac{z+9}{3+9}\phantom{\rule{0ex}{0ex}}=\frac{x}{1}=\frac{y-6}{4}=\frac{z+9}{-4}$

Here, D is the foot of the perpendicular drawn from C (7, 4, $-$1) on AB.

The coordinates of a general point on AB are given by

Let the coordinates of D be .
The direction ratios of CD are proportional to .

The direction ratios of AB are proportional to 1, 4, $-$4, but CD is perpendicular to AB.

Substituting $\lambda =-1$ in , we get the coordinates of D as .

Equation of CD is
$\frac{x-7}{-1-7}=\frac{y-4}{2-4}=\frac{z+1}{-5+1}\phantom{\rule{0ex}{0ex}}=\frac{x-7}{4}=\frac{y-4}{1}=\frac{z+1}{2}$

#### Question 13:

Find the distance of the point (2, 4, −1) from the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$.                  [NCERT EXEMPLAR]

We know that the distance d from point P to the line l having equation $\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b}$ is given by $d=\frac{\left|\stackrel{\to }{b}×\stackrel{\to }{\mathrm{PQ}}\right|}{\left|\stackrel{\to }{b}\right|}$, where Q is any point on the line l.

The equation of the given line is $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$.

Let P(2, 4, −1) be the given point. Now, in order to find the required distance we need to find a point Q on the given line.

The given line passes through the point (−5, −3, 6). So, take this point as the required point Q(−5, −3, 6).

Also, the line is parallel to the vector $\stackrel{\to }{b}=\stackrel{^}{i}+4\stackrel{^}{j}-9\stackrel{^}{k}$.

Now, $\stackrel{\to }{\mathrm{PQ}}=\left(-5\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)-\left(2\stackrel{^}{i}+4\stackrel{^}{j}-\stackrel{^}{k}\right)=-7\stackrel{^}{i}-7\stackrel{^}{j}+7\stackrel{^}{k}$

$\therefore \stackrel{\to }{b}×\stackrel{\to }{\mathrm{PQ}}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 1& 4& -9\\ -7& -7& 7\end{array}\right|=-35\stackrel{^}{i}+56\stackrel{^}{j}+21\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{b}×\stackrel{\to }{\mathrm{PQ}}\right|=\sqrt{{\left(-35\right)}^{2}+{56}^{2}+{21}^{2}}=\sqrt{1225+3136+441}=\sqrt{4802}=49\sqrt{2}$

Let d be the required distance.

$\therefore d=\frac{\left|\stackrel{\to }{b}×\stackrel{\to }{\mathrm{PQ}}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{49\sqrt{2}}{\sqrt{1+16+81}}=\frac{49\sqrt{2}}{\sqrt{98}}=\frac{49\sqrt{2}}{7\sqrt{2}}=7$

Thus, the distance of the given point from the given line is 7 units.

#### Question 14:

Find the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, −1, 3) and C(2, −3, −1).                 [NCERT EXEMPLAR]

The Cartesian equation of the line joining points B(0, −1, 3) and C(2, −3, −1) is

Let L be the foot of the perpendicular drawn from the point A(1, 8, 4) to the line $\frac{x}{2}=\frac{y+1}{-2}=\frac{\mathrm{z}-3}{-4}$.

The coordinates of general point on the line $\frac{x}{2}=\frac{y+1}{-2}=\frac{\mathrm{z}-3}{-4}$ are given by

Let the coordinates of L be $\left(2\lambda ,-2\lambda -1,-4\lambda +3\right)$. Therefore, the direction ratios of AL are proportional to

$2\lambda -1,-2\lambda -1-8,-4\lambda +3-4$ or $2\lambda -1,-2\lambda -9,-4\lambda -1$

Direction ratios of the given line are proportional to 2, −2, −4.

But, AL is perpendicular to the given line.

$\therefore 2×\left(2\lambda -1\right)+\left(-2\right)×\left(-2\lambda -9\right)+\left(-4\right)×\left(-4\lambda -1\right)=0\phantom{\rule{0ex}{0ex}}⇒4\lambda -2+4\lambda +18+16\lambda +4=0\phantom{\rule{0ex}{0ex}}⇒24\lambda +20=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{5}{6}$

Putting $\lambda =-\frac{5}{6}$ in $\left(2\lambda ,-2\lambda -1,-4\lambda +3\right)$, we get

$\left(2×\left(-\frac{5}{6}\right),-2×\left(-\frac{5}{6}\right)-1,-4×\left(-\frac{5}{6}\right)+3\right)=\left(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\right)$

Thus, the required coordinates of the foot of the perpendicular are $\left(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\right)$.

#### Question 1:

Find the shortest distance between the following pairs of lines whose vector equations are:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii) $\stackrel{\to }{r}=\left(8+3\lambda \right)\stackrel{^}{i}-\left(9+16\lambda \right)\stackrel{^}{j}+\left(10+7\lambda \right)\stackrel{^}{k}$ and $\stackrel{\to }{r}=15\stackrel{^}{i}+29\stackrel{^}{j}+5\stackrel{^}{k}+\mu \left(3\stackrel{^}{i}+8\stackrel{^}{j}-5\stackrel{^}{k}\right)$              [NCERT EXEMPLAR]

(i)

Comparing the given equations with the equations , we get

The shortest distance between the lines is given by

(ii)

Comparing the given equations with the equations , we get

The shortest distance between the lines is given by

(iii)

Comparing the given equations with the equations, we get

The shortest distance between the lines is given by

(iv)

The vector equations of the given lines can be re-written as

Comparing the given equations with the equations , we get

The shortest distance between the line is given by

(v)

The vector equations of the given lines can be re-written as

Comparing the given equations with the equations , we get

The shortest distance between the lines is given by

(vi)

Comparing the given equations with the equations , we get

The shortest distance between the lines is given by

(vii)

Comparing the given equations with the equations, we get

The shortest distance between the lines is given by

(viii) The vector equations of the given lines can be re-written as

$\stackrel{\to }{r}=8\stackrel{^}{i}-9\stackrel{^}{j}+10\stackrel{^}{k}+\lambda \left(3\stackrel{^}{i}-16\stackrel{^}{j}+7\stackrel{^}{k}\right)$ and $\stackrel{\to }{r}=15\stackrel{^}{i}+29\stackrel{^}{j}+5\stackrel{^}{k}+\mu \left(3\stackrel{^}{i}+8\stackrel{^}{j}-5\stackrel{^}{k}\right)$

Comparing the given equations with the equations , we get

$\stackrel{\to }{{a}_{1}}=8\stackrel{^}{i}-9\stackrel{^}{j}+10\stackrel{^}{k}$

$\stackrel{\to }{{b}_{1}}=3\stackrel{^}{i}-16\stackrel{^}{j}+7\stackrel{^}{k}$

$\stackrel{\to }{{a}_{2}}=15\stackrel{^}{i}+29\stackrel{^}{j}+5\stackrel{^}{k}$

$\stackrel{\to }{{b}_{2}}=3\stackrel{^}{i}+8\stackrel{^}{j}-5\stackrel{^}{k}$

$\therefore \stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=\left(15\stackrel{^}{i}+29\stackrel{^}{j}+5\stackrel{^}{k}\right)-\left(8\stackrel{^}{i}-9\stackrel{^}{j}+10\stackrel{^}{k}\right)=7\stackrel{^}{i}+38\stackrel{^}{j}-5\stackrel{^}{k}$

$\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 3& -16& 7\\ 3& 8& -5\end{array}\right|=24\stackrel{^}{i}+36\stackrel{^}{j}+72\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}\right|=\sqrt{{24}^{2}+{36}^{2}+{72}^{2}}=\sqrt{576+1296+5184}=\sqrt{7056}=84$

Also,

$\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right).\left(\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}\right)\phantom{\rule{0ex}{0ex}}=\left(7\stackrel{^}{i}+38\stackrel{^}{j}-5\stackrel{^}{k}\right).\left(24\stackrel{^}{i}+36\stackrel{^}{j}+72\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}=7×24+38×36+\left(-5\right)×72\phantom{\rule{0ex}{0ex}}=168+1368-360\phantom{\rule{0ex}{0ex}}=1176$

We know that the shortest distance between the lines is given by .

∴ Required shortest distance between the given pairs of lines,

#### Question 2:

Find the shortest distance between the following pairs of lines whose cartesian equations are:
(i)

(ii)

(iii)

(iv)

(i) The equations of the given lines are

Since line (1) passes through the point (1, 2, 3) and has direction ratios proportional to 2, 3, 4, its vector equation is

Also, line (2) passes through the point (2, 3, 5) and has direction ratios proportional to 3, 4, 5.
Its vector equation is

Now,

The shortest distance between the lines is given by

(ii) The equations of the given lines are

Since line (1) passes through the point (1, $-$1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is

Also, line (2) passes through the point ($-$1, 2, 2) and has direction ratios proportional to 3, 1, 0.
Its vector equation is

Now,

The shortest distance between the lines is given by

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

(iii)

Since line (1) passes through the point (1, $-$2, 3) and has direction ratios proportional to $-$1, 1, $-$2, its vector equation is

Also, line (2) passes through the point (1, $-$1, $-$1) and has direction ratios proportional to 1, 2, $-$2.
Its vector equation is

Now,

The shortest distance between the lines is given by

(iv)

Since line (1) passes through the point (3, 5, 7) and has direction ratios proportional to 1, $-$2, 1, its vector equation is

Also, line (2) passes through the point ($-$1, $-$1, $-$1) and has direction ratios proportional to 7, $-$6, 1.
Its vector equation is

Now,

The shortest distance between the lines is given by

#### Question 3:

By computing the shortest distance determine whether the following pairs of lines intersect or not:
(i)

(ii)

(iii)

(iv)

(i)

Comparing the given equations with the equations , we get

(ii)

Comparing the given equations with the equations , we get

(iii)

Since the first line passes through the point (1, $-$1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is

Also, the second line passes through the point ($-$1, 2, 2) and has direction ratios proportional to 5, 1, 0.
Its vector equation is

Now,

(iv)

Since the first line passes through the point (5, 7, $-$3) and has direction ratios proportional to 4, $-$5, $-$5, its vector equation is

Also, the second line passes through the point (8, 7, 5) and has direction ratios proportional to 7, 1, 3.
Its vector equation is

Now,

#### Question 4:

Find the shortest distance between the following pairs of parallel lines whose equations are:
(i)

(ii)

(i) The vector equations of the given lines are

These two lines pass through the points having position vectors and are parallel to the vector $\stackrel{\to }{b}=\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$.

Now,
$\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=\stackrel{^}{i}-3\stackrel{^}{j}-4\stackrel{^}{k}$
and

The shortest distance between the two lines is given by

$\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{\sqrt{78}}{\sqrt{3}}=\sqrt{26}$

(ii)

These two lines pass through the points having position vectors and are parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$.

Now,
$\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=\stackrel{^}{i}-\stackrel{^}{k}$
and

The shortest distance between the two lines is given by

$\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{\sqrt{11}}{\sqrt{6}}$

#### Question 5:

Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(i) (0, 0, 0) and (1, 0, 2)
(ii) (1, 3, 0) and (0, 3, 0)

(i) The equation of the line passing through the points (0, 0, 0) and (1, 0, 2) is

$\frac{x-0}{1-0}=\frac{y-0}{0-0}=\frac{z-0}{2-0}\phantom{\rule{0ex}{0ex}}=\frac{x}{1}=\frac{y}{0}=\frac{z}{2}$

(ii) The equation of the line passing through the points (1, 3, 0) and (0, 3, 0) is

$\frac{x-1}{0-1}=\frac{y-3}{3-3}=\frac{z-0}{0-0}\phantom{\rule{0ex}{0ex}}=\frac{x-1}{-1}=\frac{y-3}{0}=\frac{z}{0}$

Since the first line passes through the point (0, 0, 0) and has direction ratios proportional to 1, 0, 2, its vector equation is

Also, the second line passes through the point (1, 3, 0) and has direction ratios proportional to $-$1, 0, 0.
Its vector equation is

Now,

The shortest distance between the lines is given by

#### Question 6:

Write the vector equations of the following lines and hence determine the distance between them

We have
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\phantom{\rule{0ex}{0ex}}\frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}$

Since the first line passes through the point (1, 2, $-$4) and has direction ratios proportional to 2, 3, 6, its vector equation is

Also, the second line passes through the point (3, 3, $-$5) and has direction ratios proportional to 4, 6, 12.
Its vector equation is

These two lines pass through the points having position vectors and are parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}$.

Now,

$\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}$

and

The shortest distance between the two lines is given by

$\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{\sqrt{293}}{7}\mathrm{units}$

#### Question 7:

Find the shortest distance between the lines
(i)

(ii)

(iii)

(iv)

(i)

Comparing the given equations with the equations , we get

The shortest distance between the lines is given by

(ii)

Since the first line passes through the point ($-$1, $-$1, $-$1) and has direction ratios proportional to 7, $-$6, 1, its vector equation is

Also, the second line passing through the point (3, 5, 7) has direction ratios proportional to 1,$-$2, 1.
Its vector equation is

Now,

The shortest distance between the lines is given by

(iii)

Comparing the given equations with the equations , we get

The shortest distance between the lines is given by

(iv)

Comparing the given equations with the equations , we get

The shortest distance between the lines is given by

#### Question 8:

Find the distance between the lines l1 and l2 given by

Given:

These two lines pass through the points having position vectors and are parallel to the vector $\stackrel{\to }{b}=2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}$.

Now,
$\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}=2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}$
and

The shortest distance between the two lines is given by

$\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}=\frac{\sqrt{293}}{7}$

#### Question 9:

Find the vector equation of a line passing through the point (2, 3, 2) and parallel to the line  Also find the distance between these lines.

Vector equation of a line passing through (2, 3, 2) and parallel to the line $\stackrel{\to }{r}=\left(-2\stackrel{^}{i}+3\stackrel{^}{j}\right)+\lambda \left(2\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)$ has direction ratios (2, −3, 6)
Hence, its equation is

$\stackrel{\to }{r}=\left(2\stackrel{^}{i}+3\stackrel{^}{j}+2\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)$
Hence, distance between

#### Question 1:

The angle between the straight lines
is
(a) 45°
(b) 30°
(c) 60°
(d) 90°

(d) 90°

We have

The direction ratios of the given lines are proportional to 2, 5, 4 and 1, 2, $-$3.

The given lines are parallel to the vectors .

Let $\theta$ be the angle between the given lines.

Now,

#### Question 2:

The lines $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and $\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}$ are
(a) coincident
(b) skew
(c) intersecting
(d) parallel

(a) coincident

The equation of the given lines are

Thus, the two lines are parallel to the vector $\stackrel{\to }{b}=\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$ and pass through the points (0, 0, 0) and (1, 2, 3).

Now,

Since the distance between the two parallel line is 0, the given lines are coincident.

#### Question 3:

The direction ratios of the line perpendicular to the lines

are proportional to
(a) 4, 5, 7
(b) 4, −5, 7
(c) 4, −5, −7
(d) −4, 5, 7

(a) 4, 5, 7

We have

The direction ratios of the given lines are proportional to 2, $-$3, 1 and 1, 2, $-$2.

The vectors parallel to the given vectors are .

Vector perpendicular to the given two lines is

Hence, the direction ratios of the line perpendicular to the given two lines are proportional to 4, 5, 7.

#### Question 4:

The angle between the lines is
(a) ${\mathrm{cos}}^{-1}\left(\frac{1}{65}\right)$

(b) $\frac{\mathrm{\pi }}{6}$

(c) $\frac{\mathrm{\pi }}{3}$

(d) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{\mathrm{\pi }}{3}$

We have

The direction ratios of the given lines are proportional to 1, 1, 2 and .

The given lines are parallel to vectors .

Let $\theta$ be the angle between the given lines.

Now,

#### Question 5:

The direction ratios of the line xy + z − 5 = 0 = x − 3y − 6 are proportional to
(a) 3, 1, −2

(b) 2, −4, 1

(c)

(d)

(a) 3, 1, −2

We have
xy + z − 5 = 0 = x − 3y − 6

From (1) and (2), we get

So, the given equation can be re-written as

$\frac{x-6}{3}=\frac{y}{1}=\frac{z+1}{-2}$

Hence, the direction ratios of the given line are proportional to 3, 1,$-$2.

#### Question 6:

The perpendicular distance of the point P (1, 2, 3) from the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ is
(a) 7
(b) 5
(c) 0
(d) none of these

(a) 7

We have

$\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$

Let point (1, 2, 3) be P and the point through which the line passes be Q (6, 7, 7). Also, the line is parallel to the vector $\stackrel{\to }{b}=3\stackrel{^}{i}+2\stackrel{^}{j}-2\stackrel{^}{k}$.

Now,

$\stackrel{\to }{PQ}=5\stackrel{^}{i}+5\stackrel{^}{j}+4\stackrel{^}{k}$

#### Question 7:

The equation of the line passing through the points is
(a)

(b)

(c)

(d) none of these

(c)

Equation of the line passing through the points having position vectors is

#### Question 8:

If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =
(a) −2
(b) −1
(c) 1
(d) 2

(b) −1

If a line makes angles α, β and γ with the axes, then
${\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{2}\beta +{\mathrm{cos}}^{2}\gamma =1$                           ...(1)

We have

#### Question 9:

If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
(a)

(b)

(c)

(d)

(a)

The direction ratios of the line are proportional to 1, $-$3, 2.

$\therefore$ The direction cosines of the line are

#### Question 10:

If a line makes angle with x-axis and y-axis respectively, then the angle made by the line with z-axis is
(a) π/2
(b) π/3
(c) π/4
(d) 5π/12

(b) π/3

If a line makes angles α, β and γ with the axes, then ${\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{2}\beta +{\mathrm{cos}}^{2}\gamma =1$.

Here,

$\alpha =\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}\beta =\frac{\mathrm{\pi }}{4}$

Now,

#### Question 11:

The projections of a line segment on X, Y and Z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are
(a)

(b)

(c)

(d) none of these

(a)

If a line makes angles α, β and γ with the axes, then

Let r be the length of the line segment. Then,

Substituting r = 13 in (2), we get

Thus, the direction cosines of the line are .

#### Question 12:

The lines are
(a) parallel
(b) intersecting
(c) skew
(d) coincident

(d) coincident

The equations of the given lines are

Thus, the two lines are parallel to the vector $\stackrel{\to }{b}=\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$ and pass through the points (0, 0, 0) and (1, 2, 3).

Now,

Since, the distance between the two parallel lines is 0, the given two lines are coincident lines.

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

#### Question 13:

The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is
(a) parallel to x-axis
(b) parallel to y-axis
(c) parallel to z-axis
(d) perpendicular to z-axis

(d) perpendicular to z-axis

We have

$\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$

Also, the given line is parallel to the vector $\stackrel{\to }{b}=3\stackrel{^}{i}+\stackrel{^}{j}+0\stackrel{^}{k}$.

Let $x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}$ be perpendicular to the given line.

Now,
$3x+4y+0z=0$
It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).

Hence, the given line is perpendicular to z-axis.

#### Question 14:

The shortest distance between the lines
is
(a) $\sqrt{30}$
(b) $2\sqrt{30}$
(c) $5\sqrt{30}$
(d) $3\sqrt{30}$

(d) $3\sqrt{30}$

We have

We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3, $-$1, 1.

Its vector equation is

Also, line (2) passes through the point ($-$3, $-$7, 6) and has direction ratios proportional to $-$3, 2, 4.

Its vector equation is

Now,

The shortest distance between the lines is given by

#### Question 15:

The equation of y-axis in space is
​(a) x = 0, z = 0
(b) x = 0, y = 0
(c) y = 0, z = 0
(d) y = 0

Equation of y-axis is also given by x = 0, = 0
Hence, the correct answer is option A.

#### Question 1:

The vector equations of OX, OY and OZ are ______________.

Vector equation for OX is of the form $\lambda \stackrel{^}{i}$ since y and z coordinates are zero i.e. $\stackrel{\to }{r}=\lambda \stackrel{^}{i}$

Similarly, vector equation for OY is of the form i.e. $\stackrel{\to }{r}=\mathrm{\mu }\stackrel{^}{j}$ and vector equation for OZ is of the form $\stackrel{\to }{r}=v\stackrel{^}{k}$

i.e.  represents vector equation of OX, OY and OZ respectively.

#### Question 2:

The vector equation of the line $\frac{5-x}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is _________________.

For given line $\frac{5-x}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is parametric form

Direction ratios are a = −3, b = 7, and c = 2
fixed point is x1 = 5, x2 = −4, x3 = 6

#### Question 3:

The vector equation of the line passing through the points (3, 4, –7) and (1, –1, 6) is _______________.

Since vector equation of a line passing through is

#### Question 4:

If l, l, l are direction cosines of a line, then l = ___________________.

Since direction cosines of a line are l, l, l
i.e. l = l, m = l and n = l is given
Since l2 + m2 + n2 = 1
i.e. l2 + l2 + l2 = 1
i.e. 3l= 1
i.e. l$\frac{1}{3}$
i.e. $l=±\frac{1}{\sqrt{3}}$

#### Question 5:

If the line $\frac{x-1}{l}=\frac{y-2}{m}=\frac{z+1}{n}$ passes through the point (–1, 0, 1), then its direction cosines l, m, n are _______________.

#### Question 6:

The equations of a line passing through point (–2, 3, 4) and equally inclined with the coordinate axes OX, OY and OZ are ______________.

Since for equally inclined line with the coordinate axes OX, OY, and OZ direction ratios is of the form (k, k, k) or (1, 1, 1)

Since line passes through (−2, 3, 4)

i.e. x1 = −2, y1 = 3, z1 = 4 and a = 1, b = 1, c = 1

#### Question 7:

The angle between the lines whose direction ratios are proportional to a, b, c and  is _________________.

Since, the angle between the lines whose direction ratios are proportional to (a1, b1, c1) and (a2, b2, c2)
is

∴ angle between the lines whose direction ratios are proportional to

#### Question 8:

The equation of the straight line passing through (a, b, c) and (a – b. b – c, c – a) are _______________ .

Equation of a straight line passing through

#### Question 9:

The equations of straight line passing through (a, b, c) and parallel to z-axis are _______________ .

Direction ratios of a line parallel to z-axis is (0, 0, 1)

Hence, equation of straight line passing through (a, b, c) and parallel to z-axis is

#### Question 10:

If the lines  are at right angle, then k = _______________ .

#### Question 11:

The Cartesian equations of the straight line passing through the point (2, 1, –1) and making angles with the positive directions of the coordinate axes are _____________.

Since straight line makes angle  with positive direction of co-ordinate axes
∴ direction cosines are given by

∴ Equation of line passing through (x1, y1, z1) and direction cosines l, m, n is given by

#### Question 12:

The equations of x-axis in unsymmetrical form are: ___________.

Since y = 0 and z = 0 represent x-axis
∴ The equation x-axis in unsymmetrical form are y = 0, z = 0

#### Question 13:

The equation of y-axis in symmetrical form is $\frac{x}{0}=\frac{y}{1}=\frac{z}{0}$. ________________.

check question

#### Question 14:

If the lines are coplanar, then  = _____________.

#### Question 15:

If the lines  are coplanar and , then k =​__________.

#### Question 16:

The equation of x-axis in symmetrical form is ___________.

#### Question 17:

​The equation of x-axis in unsymmetrical form is ___________.

Repeated

Since y = 0 and = 0 represent x-axis
∴ The equation x-axis in unsymmetrical form are y = 0, z = 0

#### Question 18:

The vector equation of the line though the points (3, 4, –7) and (1, –1, 6) is ___________.

Since vector equation of a line passing through

#### Question 19:

​The vector equation of the line $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is ____________.

Equation in cartesian form is

#### Question 1:

Write the cartesian and vector equations of X-axis.

Since x-axis passes through the the point (0, 0, 0) having position vector $\stackrel{\to }{a}=0\stackrel{^}{i}+0\stackrel{^}{j}+0\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=1\stackrel{^}{i}+0\stackrel{^}{j}+0\stackrel{^}{k}$ having direction ratios proportional to 1, 0, 0, the cartesian equation of x-axis is

$\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}\phantom{\rule{0ex}{0ex}}=\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

Also, its vector equation is

#### Question 2:

Write the cartesian and vector equations of Y-axis.

Since y-axis passes through the the point (0, 0, 0) having position vector $\stackrel{\to }{a}=0\stackrel{^}{i}+0\stackrel{^}{j}+0\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=0\stackrel{^}{i}+1\stackrel{^}{j}+0\stackrel{^}{k}$ having direction ratios proportional to 0, 1, 0, the cartesian equation of y-axis is

$\frac{x-0}{0}=\frac{y-0}{1}=\frac{z-0}{0}\phantom{\rule{0ex}{0ex}}=\frac{x}{0}=\frac{y}{1}=\frac{z}{0}$

Also, its vector equation is

#### Question 3:

Write the cartesian and vector equations of Z-axis.

Since z-axis passes through the the point (0, 0, 0) having position vector $\stackrel{\to }{a}=0\stackrel{^}{i}+0\stackrel{^}{j}+0\stackrel{^}{k}$ and is parallel to the vector $\stackrel{\to }{b}=0\stackrel{^}{i}+0\stackrel{^}{j}+\stackrel{^}{k}$ having direction ratios proportional to 0, 0, 1, the cartesian equation of z-axis is

$\frac{x-0}{0}=\frac{y-0}{0}=\frac{z-0}{1}\phantom{\rule{0ex}{0ex}}=\frac{x}{0}=\frac{y}{0}=\frac{z}{1}$

Also, its vector equation is

#### Question 4:

Write the vector equation of a line passing through a point having position vector and parallel to vector .

The vector equation of the line passing through the point having position vector $\stackrel{\to }{\alpha }$ and parallel to vector $\stackrel{\to }{\beta }$ is $\stackrel{\to }{r}=\stackrel{\to }{\alpha }+\lambda \stackrel{\to }{\beta }$.

#### Question 5:

Cartesian equations of a line AB are $\frac{2x-1}{2}=\frac{4-y}{7}=\frac{z+1}{2}.$ Write the direction ratios of a line parallel to AB.

We have

$\frac{2x-1}{2}=\frac{4-y}{7}=\frac{z+1}{2}$

The equation of the line AB can be re-written as

$\frac{x-\frac{1}{2}}{1}=\frac{y-4}{-7}=\frac{z+1}{2}$

The direction ratios of the line parallel to AB are proportional to 1, $-$7, 2.

Also, the direction cosines of the line parallel to AB are proportional to

#### Question 6:

Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.

We have
6x − 2 = 3y + 1 = 2z − 4

The equation of given line can be re-written as

$\frac{x-\frac{1}{3}}{\frac{1}{6}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-2}{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{x-\frac{1}{3}}{1}=\frac{y+\frac{1}{3}}{2}=\frac{z-2}{3}$

The direction ratios of the line parallel to AB are proportional to 1, 2, 3.

Hence, the direction cosines of the line parallel to AB are proportional to

.

#### Question 7:

Write the direction cosines of the line

We have

The equation of the given line can be re-written as

The direction ratios of the given line are proportional to 4, $-$3, 0.

Hence, the direction cosines of the given line are proportional to

#### Question 8:

Write the coordinate axis to which the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-1}{0}$is perpendicular.

We have

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-1}{0}$

The given line is parallel to the vector $\stackrel{\to }{b}=3\stackrel{^}{i}+4\stackrel{^}{j}+0\stackrel{^}{k}$.

Let $x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}$ be perpendicular to the given line.

Now,
$3x+4y+0z=0$
It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).

Hence, the given line is perpendicular to z-axis.

#### Question 9:

Write the angle between the lines

We have

The given lines are parallel to the vectors .

Let $\theta$ be the angle between the given lines.

Now,

#### Question 10:

Write the direction cosines of the line whose cartesian equations are 2x = 3y = −z.

We have
2x = 3y = −z

The equation of the given line can be re-written as

$\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{-1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}$

The direction ratios of the line parallel to AB are proportional to 3, 2, $-$6.

Hence, the direction cosines of the line parallel to AB are proportional to

#### Question 11:

Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z.

We have
2x = 3y = −z
6x = −y = −4z

The given lines can be re-written as

These lines are parallel to vectors .

Let $\theta$ be the angle between these lines.

Now,

#### Question 12:

Write the value of λ for which the lines are perpendicular to each other.

We have

The given lines are parallel to vectors .

For $\stackrel{\to }{{b}_{1}}\perp \stackrel{\to }{{b}_{2}}$, we must have

$\stackrel{\to }{{b}_{1}}.\stackrel{\to }{{b}_{2}}=0\phantom{\rule{0ex}{0ex}}⇒\left(-3\stackrel{^}{i}+2\lambda \stackrel{^}{j}+2\stackrel{^}{k}\right).\left(3\lambda \stackrel{^}{i}+\stackrel{^}{j}-5\stackrel{^}{k}\right)=0\phantom{\rule{0ex}{0ex}}⇒-7\lambda -10=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{10}{7}$

#### Question 13:

Write the formula for the shortest distance between the lines

The shortest distance d between the parallel lines is given by

$d=\frac{\left|\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right)×\stackrel{\to }{b}\right|}{\left|\stackrel{\to }{b}\right|}$

#### Question 14:

Write the condition for the lines to be intersecting.

The shortest distance between the lines is given by

For the lines to be intersecting, $d=0$.

#### Question 15:

The cartesian equations of a line AB are $\frac{2x-1}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}.$ Find the direction cosines of a line parallel to AB.

We have

$\frac{2x-1}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}$

The equation of the line AB can be re-written as

$\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{y+2}{2}=\frac{z-3}{3}\phantom{\rule{0ex}{0ex}}=\frac{x-\frac{1}{2}}{\sqrt{3}}=\frac{y+2}{4}=\frac{z-3}{6}$

Thus, the direction ratios of the line parallel to AB are proportional to $\sqrt{3}$, 4, 6.

Hence, the direction cosines of the line parallel to AB are proportional to

#### Question 16:

If the equations of a line AB are $\frac{3-x}{1}=\frac{y+2}{-2}=\frac{z-5}{4},$ write the direction ratios of a line parallel to AB.

We have

$\frac{3-x}{1}=\frac{y+2}{-2}=\frac{z-5}{4}$

The equation of the line AB can be re-written as

$\frac{x-3}{-1}=\frac{y+2}{-2}=\frac{z-5}{4}$

Thus, the direction ratios of the line parallel to AB are proportional to $-$1, $-$2, 4.

#### Question 17:

Write the vector equation of a line given by $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}.$

We have

$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$

The given line passes through the point (5, $-$4, 6) and has direction ratios proportional to 3, 7, 2.

Vector equation of the given line passing through the point having position vector $\stackrel{\to }{a}=5\stackrel{^}{i}-4\stackrel{^}{j}+6\stackrel{^}{k}$ and parallel to a vector $\stackrel{\to }{b}=3\stackrel{^}{i}+7\stackrel{^}{j}+2\stackrel{^}{k}$ is

$\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}=5\stackrel{^}{i}-4\stackrel{^}{j}+6\stackrel{^}{k}+\lambda \left(3\stackrel{^}{i}+7\stackrel{^}{j}+2\stackrel{^}{k}\right)$

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

#### Question 18:

The equations of a line are given by $\frac{4-x}{3}=\frac{y+3}{3}=\frac{z+2}{6}.$ Write the direction cosines of a line parallel to this line.

We have

$\frac{4-x}{3}=\frac{y+3}{3}=\frac{z+2}{6}$

The equation of the given line can be re-written as

$\frac{x-4}{-3}=\frac{y+3}{3}=\frac{z+2}{6}$

The direction ratios of the line parallel to the given line are proportional to $-$3, 3, 6.

Hence, the direction cosines of the line parallel to the given line are proportional to

#### Question 19:

Find the Cartesian equations of the line which passes through the point (−2, 4 , −5) and is parallel to the line $\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}.$

The equation of the given line is $\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}$

It can be re-written as

$\frac{x+3}{3}=\frac{y-4}{-5}=\frac{z+8}{6}$

Since the required line is parallel to the given line, the direction ratios of the required line are proportional to 3, $-$5, 6.

Hence, the cartesian equations of the line passing through the point ($-$2, 4, $-$5) and parallel to a vector having direction ratios proportional to 3, $-$5, 6 is $\frac{x+2}{3}=\frac{y-4}{-5}=\frac{z+5}{6}$.

#### Question 20:

Find the angle between the lines $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-5\stackrel{^}{j}+\stackrel{^}{k}\right)+\lambda \left(3\stackrel{^}{i}+2\stackrel{^}{j}+6\stackrel{^}{k}\right)$ and $\stackrel{\to }{r}=7\stackrel{^}{i}-6\stackrel{^}{k}+\mu \left(\stackrel{^}{i}+2\stackrel{^}{j}+2\stackrel{^}{k}\right)$.                  [CBSE 2014]

Let $\theta$ be the angle between the given lines. The given lines are parallel to the vectors $\stackrel{\to }{{b}_{1}}=3\stackrel{^}{i}+2\stackrel{^}{j}+6\stackrel{^}{k}$ and $\stackrel{\to }{{b}_{2}}=\stackrel{^}{i}+2\stackrel{^}{j}+2\stackrel{^}{k}$, respectively.

So, the angle $\theta$ between the given lines is given by

Thus, the angle between the given lines is ${\mathrm{cos}}^{-1}\left(\frac{19}{21}\right)$.

#### Question 21:

Find the angle between the lines $2x=3y=-z$ and $6x=-y=-4z$.           [CBSE 2015]

The equations of the given lines can be re-written as

$\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}$ and $\frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}$

We know that angle between the lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$ is given by $\mathrm{cos}\theta =\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}+{c}_{1}^{2}}\sqrt{{a}_{2}^{2}+{b}_{2}^{2}+{c}_{2}^{2}}}$.

Let $\theta$ be the angle between the given lines.

Thus, the angle between the given lines is $\frac{\mathrm{\pi }}{2}$.

#### Question 22:

Find the vector equation of a line passing through the point (3, 4, 5) and is parallel to the vector $2\stackrel{^}{i}+2\stackrel{^}{j}-3\stackrel{^}{k}.$

Vector equation of a line passing through (3, 4, 5) and parallel to vector $\stackrel{\to }{a}=2\stackrel{^}{i}+2\stackrel{^}{j}-3\stackrel{^}{k}$ has directions ratios (2, 2, −3)
Hence, equation of line is

#### Question 1:

Find the vector and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector $3\stackrel{^}{i}+2\stackrel{^}{j}-8\stackrel{^}{k}.$

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to vector $\stackrel{\to }{b}$ is .

Here,

Vector equation of the required line is given by

Reducing (1) to cartesian form, we get

#### Question 2:

Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).

We know that the vector equation of a line passing through the points with position vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is , where $\lambda$ is a scalar.

Here,

Vector equation of the required line is

#### Question 3:

Find the vector equation of a line which is parallel to the vector $2\stackrel{^}{i}-\stackrel{^}{j}+3\stackrel{^}{k}$ and which passes through the point (5, −2, 4). Also, reduce it to cartesian form.

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is .

Here,

So, the vector equation of the required line is

Reducing (1) to cartesian form, we get

#### Question 4:

A line passes through the point with position vector $2\stackrel{^}{i}-3\stackrel{^}{j}+4\stackrel{^}{k}$ and is in the direction of $3\stackrel{^}{i}+4\stackrel{^}{j}-5\stackrel{^}{k}.$ Find equations of the line in vector and cartesian form.

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is .

Here,
.

So, the vector equation of the required line is

Reducing (1) to cartesian form, we get

#### Question 5:

ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, Find the vector equation of the line BD. Also, reduce it to cartesian form.

We know that the position vector of the mid-point of $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is $\frac{\stackrel{\to }{a}+\stackrel{\to }{b}}{2}$.

Let the position vector of point D be $x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}$.

Position vector of mid-point of A and C = Position vector of mid-point of B and D

The vector equation of line BD passing through the points with position vectors $\stackrel{\to }{a}$(B) and $\stackrel{\to }{b}$(D) is .

Here,

Vector equation of the required line is

Reducing (1) to cartesian form, we get

#### Question 6:

Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).

We know that the vector equation of a line passing through the points with position vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is , where $\lambda$ is a scalar.

Here,

Vector equation of the required line is

Reducing (1) to cartesian form, we get

#### Question 7:

Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector $\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k}.$ Reduce the corresponding equation in cartesian from.

We know that the vector equation of a line passing through a point with position vector $\stackrel{\to }{a}$ and parallel to the vector $\stackrel{\to }{b}$ is .

Here,

Vector equation of the required line is

Reducing (1) to cartesian form, we get

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