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
| $P$ | $Q$ | $R$ | $S$ | $P$ | $Q$ | $R$ | $S$ | ||
|---|---|---|---|---|---|---|---|---|---|
| (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) 62. $A \rightarrow r, B \rightarrow q ; C \rightarrow p ; D \rightarrow s$
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} $
