Solveeit Logo
Login

Question

Question: The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line (1/2...

The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line (1/2) x = (1/3)y = (–1/6) z is –

A

1

B

2

C

½

D

4

Answer

2

Explanation

Solution

In this example, we are not required to find the perpendicular distance of the point (1, –2, 3) from the given plane, but the distance of this point from the given plane measured parallel to the given line whose direction ratios are 2, 3, –6.

Now the equation of the line through (1, –2, 3) and parallel to the line whose d.c.’s are 2, 3, –6 are x−12\frac{x - 1}{2} = y+23\frac{y + 2}{3} = z−3−6\frac{z - 3}{- 6}

Any point on this line is (1 + 2r, –2 + 3r, 3 – 6r) … (i)

If it is lie on the given plane x – y + z = 5, then we have

(1 + 2r) – (–2 + 3r) + (3 – 6r) = 5

Or 1 – 7r = 0 or r = (1/7)

Substituting this value of r in (i), the point is [(9/7), (–11/7), (15/7)] and therefore the required distance of this point from the given point (1, –2, 3)

=[(97−1)2+(−117+2)2+(157−3)2]\sqrt{\left\lbrack \left( \frac{9}{7} - 1 \right)^{2} + \left( - \frac{11}{7} + 2 \right)^{2} + \left( \frac{15}{7} - 3 \right)^{2} \right\rbrack}

=[449+949+3649]\sqrt{\left\lbrack \frac{4}{49} + \frac{9}{49} + \frac{36}{49} \right\rbrack} = 1.