Question

A plane passes through $(2,3,-1)$ and is perpendicular to the line having direction ratios $3,-4,7$. The perpendicular distance from the origin to this plane is

• A. $\frac{3}{\sqrt{74}}$
• B. $\frac{5}{\sqrt{74}}$
• C. $\frac{6}{\sqrt{74}}$
• D. $\frac{13}{\sqrt{74}}$

Answer 1

Answer: (D)

The equation of the plane passes through the point $(2,3,-1)$ is
$a(x-2)+b(y-3)+c(z+1)=0\dots (i)$
where $a,b,c$ are the direction ratio of the normal to the plane.
Also, given the plane is perpendicular to the line whose direction ratio is $(3,-4,7)$. So,
that line and the normal of the plane are parallel.
$\Rightarrow \frac{a}{3}=\frac{b}{-4}=\frac{c}{7}=k$
$\Rightarrow a=3k,b=-4k,c=7k$
From Eq. (i)
$3k(x-2)-4k(y-3)+7k(z+1)=0$
$\Rightarrow 3x-4y+7z+13=0\dots (ii)$
Now, perpendicular distance from the origin to this plane
$=\frac{\mid 3\times 0-4\times 0+7\times 0+13|}{\sqrt{9+16+49}}$
$=\frac{13}{\sqrt{74}}$