3D Geometry 3 Question 61
####61. 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 $P_{1}: 7 x+y+2 z=3, \quad P_{2}: 3 x+5 y-6 z=4 . \quad$ 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.
(2013 Adv.)
| List I | List II | ||
|---|---|---|---|
| P. | $a=$ | 1. | 13 |
| Q. | $b=$ | 2. | -3 |
| R. | $c=$ | 3. | 1 |
| S. | $d=$ | 4. | -2 |
Codes
| $\mathrm{P}$ | $\mathrm{Q}$ | $\mathrm{R}$ | $\mathrm{S}$ | $\mathrm{P}$ | $\mathrm{Q}$ | $\mathrm{R}$ | $\mathrm{S}$ | ||
|---|---|---|---|---|---|---|---|---|---|
| (a) | 3 | 2 | 4 | 1 | (b) 1 | 3 | 4 | 2 | |
| (c) | 3 | 2 | 1 | 4 | (d) 2 | 4 | 1 | 3 |
Show Answer
Answer:
Correct Answer: 61. (a)
Solution:
- $L_{1}: \frac{x-1}{2}=\frac{y-0}{-1}=\frac{z-(-3)}{1}$
$ \begin{aligned} \text { Normal of plane } P: \mathbf{n} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 7 & 1 & 2 \\ 3 & 5 & -6 \end{array}\right| \\ & =\hat{\mathbf{i}}(-16)-\hat{\mathbf{j}}(-42-6)+\hat{\mathbf{k}}(32) \\ & =-16 \hat{\mathbf{i}}+48 \hat{\mathbf{j}}+32 \hat{\mathbf{k}} \end{aligned} $
DR’s of normal $\mathbf{n}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
Point of intersection of $L_{1}$ and $L_{2}$.
$ \begin{aligned} \Rightarrow & & 2 K_{1}+1 & =K_{2}+4 \\ \text { and } & & -k_{1} & =k_{2}-3 \\ \Rightarrow & & k_{1} & =2 \text { and } k_{2}=1 \end{aligned} $
$\therefore$ Point of intersection $(5,-2,-1)$
Now equation of plane,
$ \begin{aligned} & 1 \cdot(x-5)-3(y+2)-2(z+1)=0 \\ & \Rightarrow \quad x-3 y-2 z-13=0 \\ & \Rightarrow \quad x-3 y-2 z=13 \\ & \therefore \quad a \rightarrow 1, b \rightarrow-3, c \rightarrow-2, d \rightarrow 13 \end{aligned} $
