Question

The distance of the point (1, –2, 3) from the plane

x – y + z = 5, measured parallel to the line is-

• A. 1/7
• B. 1 `
• C. 7
• D. None of these

Answer 1

Answer: (B)

1 `

Answer 2

Equation of the line through (1, –2, 3) and parallel to the given line are

$\frac { x - 1 } { 2 / 7 }$= $\frac { y + 2 } { 3 / 7 }$= = r. Any point on it at a distance r from (1, –2, 3) is

$\left( 1 + \frac { 2 r } { 7 } , - 2 + \frac { 3 r } { 7 } , 3 - \frac { 6 r } { 7 } \right)$Since the point lies in

x – y + z = 5.

1 + $\frac { 2 r } { 7 } + 2 - \frac { 3 r } { 7 } + 3 - \frac { 6 r } { 7 }$ = 5

Ž r = 1b