Consider the lines
$L_{1} : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}, L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2},$ and the planes $P_{1} : 7 x + y + 2 z = 3, P_{2} : 3 x + 5 y - 6 z = 4.$ Let $a x + b y + c z = d$ the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ and perpendicular to planes $P_{1}$ and $P_{2}$ .
Match List I with List II and select the correct answer using the code given below the lists.

$\begin{matrix} L i s t I & L i s t I I P & a = & 1 & 13 Q & b = & 2 & - 3 R & c = & 3 & 1 S & d = & 4 & - 2 \end{matrix}$

P-3, Q-2, R-4, S-1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

P-1, Q-3, R-4, S-2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P-3, Q-2, R-1, S-4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P-2, Q-4, R-1, S-3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
P-3, Q-2, R-4, S-1

$L_{1} : \frac{x - 1}{2} = \frac{y - 0}{- 1} = \frac{z - (- 3)}{1}$
Normat of plane $P : n = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i^j^k 712 35 - 6 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i (- 16) -^j (- 42 - 6) +^k (32) = - 16^i + 48^j + 32^k D R^{'} s o f n o r m a l n =^i - 3^j - 2^k P o i n t o f i n t e r s e c t i o n o f L_{1} a n d L_{2} . \Rightarrow 2 K_{1} + 1 = K_{2} + 4 a n d - k_{1} = k_{2} - 3 \Rightarrow k_{1} = 2 a n d k_{2} = 1 ∴ P o i n t o f i n t e r s e c t i o n (5, - 2, - 1) N o w e q u a t i o n o f p l a n e, 1. (x - 5) - 3 (y + 2) - 2 (z + 1) = 0 \Rightarrow x - 3 y - 2 z - 13 = 0 \Rightarrow x - 3 y - 2 z = 13 ∴ a \to 1, b \to - 3, c \to - 2, d \to 13$

Suggest Corrections

Similar questions

Consider the lines
$L_{1} : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}, L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2},$ and the planes $P_{1} : 7 x + y + 2 z = 3, P_{2} : 3 x + 5 y - 6 z = 4.$ Let $a x + b y + c z = d$ the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ and perpendicular to planes $P_{1}$ and $P_{2}$ .
Match List I with List II and select the correct answer using the code given below the lists.

$\begin{matrix} L i s t I & L i s t I I P & a = & 1 & 13 Q & b = & 2 & - 3 R & c = & 3 & 1 S & d = & 4 & - 2 \end{matrix}$

Consider the lines $L_{1}$ : $\frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}, L_{2}$ : $\frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}$ and the planes $P_{1}$ : $7 x + y + 2 z = 3, P_{2}$ : $3 x + 5 y - 6 z = 4.$ Let $a x + b y + c z = d$ be the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ , and perpendicular to planes $P_{1}$ and $P_{2}$ .