Question

If the foot of the perpendicular from the point $A(-1,4,3)$ on the plane $P:2x+my+nz=$ 4 , is $\left(-2,\frac{7}{2},\frac{3}{2}\right)$, then the distance of the point A from the plane $P$, measured parallel to a line with direction ratios $3,-1,-4$, is equal to :

• A. 1
• B. $\sqrt{26}$
• C. $2\sqrt{2}$
• D. $\sqrt{14}$

Answer 1

Answer: (B)

Let $B$ be foot of $\perp$ coordinates of $B=\left(-2,\frac{7}{2},\frac{3}{2}\right)$
Direction ratio of line $AB$ is $⟨2,1,3⟩$ so $m=1,n=3$
So equation of $AC$ is $\frac{x+1}{3}=\frac{y-4}{-1}=\frac{z-3}{-4}=\lambda$
So point $C$ is $(3\lambda -1,-\lambda +4,-4\lambda +3)$. But
$C$ lies on the plane, so
$6\lambda -2-\lambda +4-12\lambda +9=4$
$\Rightarrow \lambda =1\Rightarrow C(2,3,-1)$
$\Rightarrow AC=\sqrt{26}$