Question

Find the distance of the plane 3x+4y-5z-7=0.

(a) \(\frac{7}{\sqrt{40}}\)

(b) \(\frac{6}{\sqrt{34}}\)

(c) \(\frac{8}{\sqrt{50}}\)

(d) \(\frac{7}{\sqrt{50}}\)

I have been asked this question during an interview for a job.

Origin of the question is Three Dimensional Geometry topic in section Three Dimensional Geometry of Mathematics – Class 12

Answer 1

Correct answer is (d) \(\frac{7}{\sqrt{50}}\)

To explain: From the given equation, the direction ratios of the normal to the plane are 3, 4, -5; the direction cosines are

\(\frac{3}{\sqrt{3^2+4^2+(-5)^2}},\frac{4}{\sqrt{3^2+4^2+(-5)^2}},\frac{-5}{\sqrt{3^2+4^2+(-5)^2}}\),i.e. \(\frac{3}{\sqrt{50}},\frac{4}{\sqrt{50}},\frac{-5}{\sqrt{50}}\)

Dividing the equation throughout by √50, we get

\(\frac{3}{\sqrt{50}} x+\frac{4}{\sqrt{50}} y-\frac{5}{\sqrt{50}} z=\frac{7}{\sqrt{50}}\)

The above equation is in the form of lx+my+nz=d, where d is the distance of the plane from the origin. So, the distance of the plane from the origin is \(\frac{7}{\sqrt{50}}\).