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)

$\sqrt{26}$

Answer 2

$\left(-2, \frac{7}{2}, \frac{3}{2}\right)$ satisfies the plane $P : 2x + my + nz = 4$
$-4 + \frac{7m}{2} + \frac{3n}{2} = 4$
$⇒7m+3n=16⋯(i)$

Line joining $A(–1, 4, 3)$ and $\left(-2, \frac{7}{2}, \frac{3}{2}\right)$ is perpendicular to $P : 2x + my + nz = 4$
$\frac{1}{2} = \frac{\frac{1}{2}}{m} = \frac{\frac{3}{2}}{n}$
$⇒m=1\& n=3$

Plane $P : 2x + y \+ 3z = 4$

Distance of P from $A(–1, 4, 3)$ parallel to the line
$\frac{x+1}{3}=\frac{y−4}{−1}=\frac{z−3}{−4}:L$
for point of intersection of P&L
$2(3r – 1) + (–r \+ 4) + 3(–4r \+ 3) = 4 ⇒r = 1$
Point of intersection :$(2, 3, –1)$
Required distance
$\sqrt{3^2 + 1^2 + 4^2}$
$=\sqrt{26}$
So, the correct option is (B): $\sqrt{26}$