First slide

Planes in 3D

Question

$The equation of the plane which is equally inclined to the lines \frac{x - 1}{2} = \frac{y}{- 2} = \frac{z + 2}{- 1} and$ $\frac{x + 3}{8} = \frac{y - 4}{1} = \frac{z}{- 4} and passing through the origin is/are$

Difficult

A

$14 x - 5 y - 7 z = 0$
B

$2 x + 7 y - z = 0$
C

$3 x - 4 y - z = 0$
D

$x + 2 y - 5 z = 0$

Solution

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e r e q u i r e d p l a n e b e a x + b y + c z = 0 . \\ S i n c e t h e g i v e n l i n e s a r e e q u a l l y i n c l i n e d t o t h e p l a n e, t h e a n g l e s \\ b e t w e e n t h e l i n e s a n d t h e p l a n e m u s t b e e q u a l . \\ \Rightarrow \frac{| 2 a - 2 b - c |}{3} = \frac{| 8 a + b - 4 c |}{9} \\ Only options (1) and (2) satisfy the above condition. \end{array}$

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