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

The Correct Option is B

\(\left(-2, \frac{7}{2}, \frac{3}{2}\right)\) 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): \(\sqrt{26}\)