3D Geometry 3 Question 29
29. If $L _1$ is the line of intersection of the planes $2 x-2 y+3 z-2=0, x-y+z+1=0$ and $L _2$ is the line of intersection of the planes $x+2 y-z-3=0$, $3 x-y+2 z-1=0$, then the distance of the origin from the plane, containing the lines $L _1$ and $L _2$ is (2018 Main)
(a) $\frac{1}{4 \sqrt{2}}$
(b) $\frac{1}{3 \sqrt{2}}$
(c) $\frac{1}{2 \sqrt{2}}$
(d) $\frac{1}{\sqrt{2}}$
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Answer:
Correct Answer: 29. (b)
Solution:
- $L _1$ is the line of intersection of the plane $2 x-2 y+3 z-2=0$ and $x-y+z+1=0$ and $L _2$ is the line of intersection of the plane $x+2 y-z-3=0$ and $3 x-y+2 z-1=0$
Since $L_ {i}$ is parallel to
$\hat i \quad\quad \hat j \quad \hat k$
$2 \quad -2 \quad 3 \quad \quad \quad = \hat i + \hat j$
$1 \quad -1 \quad 1$
Since $L_ {2}$ is parallel to
$\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix}$ $= 3 \hat i - 5 \hat j - 7 \hat k$
Also, $L _2$ passes through $\frac{5}{7}, \frac{8}{7}, 0$.
[put $z=0$ in last two planes]
So, equation of plane is
$ \begin{array}{ccc} x-\frac{5}{7} & y-\frac{8}{7} & z \\ 1 & 1 & 0 \\ 3 & -5 & -7 \end{array}=0 \Rightarrow 7 x-7 y+8 z+3=0 $
Now, perpendicular distance from origin is
$ \frac{3}{\sqrt{7^{2}+7^{2}+8^{2}}}=\frac{3}{\sqrt{162}}=\frac{1}{3 \sqrt{2}} $
