# find unit vector perpendicular to A =2 i +j + k  and B = i-j+2k.

The vector perpendicular to the two vectors would be given by the cross product of the two

now, let

A = 2i + j + k

B = i - j + 2k

and

C = A X B

thus,

C = (2i + j + k) X (i - j + 2k)

or

C = i(2+1) - j(4-1) + k(-2-1)

or

C = 3i - 3j - 3k

now, the unit vector would be

C = C / |C| = (3i - 3j - 3k) / [32 + 32 + 32

or

C =  (3i - 3j - 3k) / √[27]

--

thus, the unit vector will be

C = (i - j - k) / √3

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take the cross product of the two vectors you are given. This will yield a vector which is perpendicular to both. To find the unit vector, proceed to normalize the vector by dividing it by its magnitude.

hi knack.. here is the method....

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thanxxxxxxxxxxxxxxxxxxx

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ur welcome... dear... hope this helps u...

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a vector of magnityude 100 N inclined at an angle of 30 degree to another vector of magnitude 50 N.calculate the magnitude of dot product and cross product of two vectors.

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First qn (1)

we have to find cross product of  A x B orthogonal to A and B

A =2 i +j + k   B = i-j+2k.

A x B = ( 2 i +j + k ) (  i-j+2k )

= 3i - 3j  - 3k

Length of an unit vector  1

therefore the unit vector = 3i - 3j  - 3k / root of ( 9 + 9 + 9)

= 3i - 3j - 3k / 3 root 3 = i/root3 - j/ root 3 - k/root3

so the unit vector orthogonal to both  A =2 i +j + k  and B = i-j+2k is  C = i/root3 , - j/ root 3  , - k/root3

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For second question ( 2 )

magnitude of forces =A =  100N  and B=  50N...angle = x = 30*

Dot product = AB cos x = 100 x 50 x cos 30 = 4330.127

Cross product = AB sin x = 100 x 50 x sin 30 = 2500

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thanxxxxxxxxxxxxxxxx raghu.

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hi

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