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:

  1. $L_{1}$ 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}} $



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