Question

The distance of the point (3, – 4 , 5) from the plane

2x + 5y – 6z = 16 measured parallel to the line

$\frac{x}{2}$ = $\frac{y}{1}$ = $\frac{z}{–2}$ must be equal to –

• A. $\frac{30}{7}$
• B. $\frac{60}{7}$
• C. $\frac{60}{11}$
• D. $\frac{30}{11}$

Answer 1

Answer: (B)

$\frac{60}{7}$

Answer 2

2x + 5y – 6z = 16 ..... (i)

$\frac{x}{2} = \frac{y}{1} = \frac{z}{–2}$ ..... (ii)

P(3, – 4, 5)

DR's of line (ii) = DR's of line PQ

Eq. of line PQ

Ž $\frac{x–3}{2} = \frac{y + 4}{1} = \frac{z–5}{–2}$ = l

\ Coordinate of the point Q (3 + 2l, – 4 + l, 5 – 2l)

Put the coordinates of Q in eqn. (i)

Ž l = ?

distance between P & Q