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. $\frac{10}{\sqrt{74}}$
• B. $\frac{20}{\sqrt{74}}$
• C. $\frac{10}{\sqrt{83}}$
• D. $\frac{5}{\sqrt{83}}$

Answer 1

The correct option is C $\frac{10}{\sqrt{83}}$

The two lines: $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$

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

Let DR of the plane be $<a,b,c>$

$\therefore a+(-2b)+3c=0$

$2a-b-c=0$

On solving:

$\frac{a}{2+3}=\frac{b}{7}=\frac{c}{-1+4}$

$\Rightarrow a=5,b=7,c=3$

$\therefore$ Equation of the plane with DR $<5,7,3>$ and passing through $(1,-1,-1)$

$\Rightarrow 5(x-1)+7(y+1)+3(z+1)=0$

$5x+7y+3z+5=0$

Distance of the point $(1,3,7)$ from the plane $5x+7y+3z+5=0$

$\Rightarrow \frac{|5+21-21+5|}{\sqrt{25+49+9}}$

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

Option C is correct