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\,\,\, \ldots (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 $
$\Rightarrow a=3 k, b=-4 k, c=7 k$
From Eq. (i)
$3 k(x-2)-4 k(y-3)+7 k(z+1)=0 $
$\Rightarrow 3 x-4 y+7 z+13=0 \,\,\, \ldots(ii)$
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}} $
$=\frac{13}{\sqrt{74}}$