Question

The distance of the point $\left(1,3,-7\right)$ from the plane passing through the point $\left(1,-1,-1\right)$, 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. $\frac{10}{\sqrt{83}}$
• B. $\frac{5}{\sqrt{83}}$
• C. $\frac{10}{\sqrt{74}}$
• D. $\frac{20}{\sqrt{74}}$

Answer 1

The correct option is A

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

Explanation for the correct answer:

Step 1: Finding equation of the plane.

Given ,equations of the lines are:

$\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$

$\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$

Direction of vector of the lines can be assumed as

$n_1=1\hat{i}-2\hat{j}+3\hat{k}$ and

$n_2=2\hat{i}-\hat{j}-\hat{k}$

Any vector perpendicular to both the vectors is given by the cross product of both the vectors

$$\begin{gathered} n=n_1\times n_2 \\ n=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 3\\ 2& -1& -1\end{array}\right| \\ =5\hat{i}+7\hat{j}+3\hat{k} \end{gathered}$$

Equation of plane passing through $\left(1,-1,-1\right)$ and perpendicular to $n$is given by

$5(x-x_1)+7(y-y_1)+3(z-z_1)=0$

Substituting $x_1=1,y_1=-1and{z}_1=-1$ in the equation:

$$\begin{gathered} \Rightarrow 5\left(x-1\right)+7\left(y+1\right)+3\left(z+1\right)=0 \\ \Rightarrow 5x+7y+3z+5=0 \end{gathered}$$

This is equation of required plane.

Step 2: Finding the distance of point from plane.

So the distance of a point $\left(x_1,y_1,z_1\right)$ from a plane $ax+by+cz+d=0$ is given by:

$\frac{\left|{\mathrm{ax}}_1+{\mathrm{by}}_1+{\mathrm{cz}}_1+d\right|}{\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}^{\mathbf{2}}}}$ where $a=5,b=7,c=3,d=0$

Distance of $\left(1,3,-7\right)$ from this plane is $=\frac{\left|5\left(1\right)+7\left(3\right)+3\left(-7\right)+5\right|}{\sqrt{25+49+9}}$

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

Hence, option (A) is the correct answer