Question

The perpendicular distance from the origin to the plane containing the two lines,
$\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}$ and $\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}$ , is:

• A. $11$
• B. $11\sqrt{6}$
• C. $\frac{11}{\sqrt{6}}$
• D. $6\sqrt{11}$

Answer 1

The correct option is C $\frac{11}{\sqrt{6}}$

The plane containing is the lines
$L_1:\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}$ and
$L_2:\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}$
$\therefore$ normal vector of plane is -
$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 5& 7\\ 1& 4& 7\end{array}\right|$
$=\hat{i}(35-28)-\hat{j}(21-7)+\hat{k}(12-5)$
$=7\hat{i}-14\hat{j}+7\hat{k}$
$\therefore$ Equation of plane is
$7(x+2)-14(y-2)+7(z+5)=0$
$\Rightarrow 7x-14y+7z+77=0$
$\Rightarrow x-2y+z+11=0$
Now, perpendicular distance from $(0,0,0)$ to plane is given by -
$d=\left|\frac{0-0+0+11}{\sqrt{1+4+1}}\right|$
$\Rightarrow d=\frac{11}{\sqrt{6}}$