First slide

Planes in 3D

Question

In $R^{3},$ let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes $P_{1} : x + 2 y - z + 1 = 0$ and . $P_{2} : 2 x - y + z - 1 = 0$ .Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane $P_{1}$ . Which of the following points lie(s) on M?

Moderate

A

$(0, - \frac{5}{6}, - \frac{2}{3})$
B

$(- \frac{1}{6}, - \frac{1}{3}, \frac{1}{6})$
C

$(- \frac{5}{6}, 0, \frac{1}{6})$
D

$(- \frac{1}{3}, 0, \frac{2}{3})$

Solution

$\begin{array}{l} A n y l i n e p a r a l l e l t o t h e l i n e o f i n t e r s e c t i o n o f p l a n e s P_{1} \equiv x + 2 y - z + 1 = 0 \\ a n d P_{2} \equiv 2 x - y + z - 1 = 0 i s p a r a l l e l t o t h e v e c t o r \\ |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 1 \\ 2 & - 1 & 1 \end{matrix}| = \hat{i} - 3 \hat{k} - 5 \hat{k} \\ ∴ A n y p o i n t o n t h e l i n e L i s i n t h e f o r m A (λ, - 3 λ, - 5 λ) w h e r e λ \in R \\ L e t M (h, k, l) b e t h e f o o t o f t h e ⊥ d r a w n f r o m A o n t o t h e p l a n e P_{1} . \\ T h e n \frac{h - λ}{1} = \frac{k + 3 λ}{2} = \frac{l + 5 λ}{- 1} = - \frac{(λ - 6 λ + 5 λ + 1)}{1 + 4 + 1} = - \frac{1}{6} \\ \Rightarrow M (h, k, l) = (λ - \frac{1}{6}, - 3 λ - \frac{1}{3}, - 5 λ + \frac{1}{6}) \\ C l e a r l y f o r λ = \frac{1}{6}, 0, t h e p o i n t s i n o p t i o n s (1), (2) l i e o n M . \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App