Question

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

Answer 1

The correct option is B

$\frac{10}{\sqrt{83}}$ units

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 $n_1=\hat{i}-2\hat{j}+3\hat{k}and{n}_2=2\hat{i}-\hat{j}-\hat{k}$
$\therefore$ Any vector n prependicular to both $n_1,n_2$ is given by $n=n_1\times n_2$
$\left|\begin{array}{ccc}\hat{i}\hat{j}\hat{k}& & \\ 1-23& & \\ 2-1-1& & \end{array}\right|=5\hat{i}+7\hat{j}+3\hat{k}$
$\therefore$ Equation of a plane passing through (1, -1, -1) and perpendicular to n is given by
$5(x-1)+7(y+1)+3(z+1)=0\Rightarrow 5x+7y+3z+5=0$
$\therefore$ Required distance =$\left|\frac{5+21-21+5}{\sqrt{5^2+7^2+3^2}}\right|=\frac{10}{\sqrt{83}}$ units