First slide

Planes in 3D

Question

$T h e e q u a t i o n o f t h e p l a n e s p a s \sin g t h r o u g h t h e l i n e o f i n t e r s e c t i o n o f$ $the planes 3 x - y - 4 z = 0 and$ $x + 3 y + 6 = 0 whose distance$ $from the origin is 1, are$

Moderate

A

$x - 2 y - 2 z - 3 = 0,$ $2 x + y - 2 z + 3 = 0$
B

$x - 2 y + 2 z - 3 = 0$ $2 x + y + 2 z + 3 = 0$
C

$x + 2 y - 2 z - 3 = 0,$ $2 x + y - 2 z + 3 = 0$
D

None of these

Solution

$\begin{array}{l} Equation of planes passing through line of intersection of two planes is \\ (3 x - y - 4 z) + λ (x + 3 y + 6) = 0 \\ (3 + λ) x + (3 λ - 1) y - 4 z + 6 λ = 0 \\ Given the distance of plane from orgin is 1 \\ ∴ |\frac{6 λ}{\sqrt{{(3 + λ)}^{2} + {(3 λ - 1)}^{2} + {(- 4)}^{2}}}| = 1 \\ \Rightarrow λ = \pm 1 \\ Therefore, the equation of the planes are 2 x - 4 y - 4 z - 6 = 0 i e; x - 2 y - 2 z - 3 = 0 \\ a n d 4 x + 2 y - 4 z + 6 = 0 i e; 2 x + y - 2 z + 3 = 0 \end{array}$

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