Login

find the distance of the point(6,5,9) from the plane determined by the points A(3,-1,2) B(5,2,4) C(-1,-1,6)

$T h e g e n e r a l e q u a t i o n o f a p l a n e p a s \sin g t h r o u g h A (3, - 1, 2) i s, a (x - 3) + b (y + 1) + c (z - 2) = 0 ...... (1) I t w i l l p a s s e s t h r o u g h B (5, 2, 4) a n d C (- 1, - 1, 6) i f, a (5 - 3) + b (2 + 1) + c (4 - 2) = 0 \Rightarrow 2 a + 3 b + 2 c = 0 ..... (2) A n d, a (- 1 - 3) + b (- 1 + 1) + c (6 - 2) = 0 \Rightarrow - 4 a + 4 c = 0 \Rightarrow a - c = 0 ..... (3) S o l v i n g (2) a n d (3) b y c r o s s m u l t i p l i c a t i o n, w e h a v e, \frac{a}{- 3} = \frac{b}{4} = \frac{c}{- 3} = k \Rightarrow a = - 3 k, b = 4 k a n d c = - 3 k S u b s t i t u t i n g a l l t h e v a l u e s o f a, b, a n d c i n e q u a t i o n (1), w e g e t, - 3 k (x - 3) + 4 k (y + 1) - 3 k (z - 2) = 0 \Rightarrow 3 (x - 3) - 4 (y + 1) + 3 (z - 2) = 0 \Rightarrow 3 x - 9 - 4 y - 4 + 3 z - 6 = 0 \Rightarrow 3 x - 4 y + 3 z - 19 = 0 ..... (4) N o w, d i s \tan c e o f t h e p o int (6, 5, 9) f r o m t h e p l a n e 3 x - 4 y + 3 z - 19 = 0 i s, D = \frac{| 3 \times 6 - 4 \times 5 + 3 \times 9 - 19 |}{\sqrt{{(3)}^{2} + {(- 4)}^{2} + {(3)}^{2}}} \Rightarrow D = \frac{| 18 - 20 + 27 - 19 |}{\sqrt{9 + 16 + 9}} \Rightarrow D = \frac{6}{\sqrt{34}} H e n c e, d i s \tan c e o f t h e p o int (6, 5, 9) f r o m t h e p l a n e 3 x - 4 y + 3 z - 19 = 0 i s \frac{6}{\sqrt{34}} .$

15

View Full Answer

Copyright © 2024 Aakash EduTech Pvt. Ltd. All rights reserved.

Aakash EduTech Pvt. Ltd.

What are you looking for?