3D Geometry 3 Question 33
####33. The distance of the point $(1,3,-7)$ from the plane passing through the point $(1,-1,-1)$ having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$, is
(a) $\frac{20}{\sqrt{74}}$ units
(b) $\frac{10}{\sqrt{83}}$ units
(c) $\frac{5}{\sqrt{83}}$ units
(d) $\frac{10}{\sqrt{74}}$ units
(2017 Main)
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Answer:
Correct Answer: 33. (b)
Solution:
- Given, equations of lines are
$\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$
and
$ \frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1} $
Let $\mathbf{n}_ {1}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\mathbf{n}_ {2}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$ $\therefore$ Any vector $\mathbf{n}$ perpendicular to both $\mathbf{n}_ {1}, \mathbf{n}_ {2}$ is given by
$ \mathbf{n}=\mathbf{n}_ {1} \times \mathbf{n}_ {2} $
$\Rightarrow \quad \mathbf{n}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & 3 \\ 2 & -1 & -1\end{array}\right|=5 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
$\therefore$ Equation of a plane passing through $(1,-1,-1)$ and perpendicular to $\mathbf{n}$ is given by
$ \begin{aligned} & 5(x-1)+7(y+1)+3(z+1)=0 \\ & \Rightarrow \quad 5 x+7 y+3 z+5=0 \\ & \therefore \text { Required distance }=\left|\frac{5+21-21+5}{\sqrt{5^{2}+7^{2}+3^{2}}}\right|=\frac{10}{\sqrt{83}} \text { units } \end{aligned} $
