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Question: The reflection of the point (2, –1, 3) in the plan\(3 x - 2 y - z = 9\) is...

The reflection of the point (2, –1, 3) in the plan3x−2y−z=93 x - 2 y - z = 9 is

A

(267,157,177)\left( \frac { 26 } { 7 } , \frac { 15 } { 7 } , \frac { 17 } { 7 } \right)

B

(267,−157,177)\left( \frac { 26 } { 7 } , \frac { - 15 } { 7 } , \frac { 17 } { 7 } \right)

C

(157,267,−177)\left( \frac { 15 } { 7 } , \frac { 26 } { 7 } , \frac { - 17 } { 7 } \right)

D

(267,177,−157)\left( \frac { 26 } { 7 } , \frac { 17 } { 7 } , \frac { - 15 } { 7 } \right)

Answer

(267,−157,177)\left( \frac { 26 } { 7 } , \frac { - 15 } { 7 } , \frac { 17 } { 7 } \right)

Explanation

Solution

Let P be the point (2, –1, 3) and Q be its reflection in the given plane.

Then, PQ is perpendicular to the given plane

Hence, d.r.’s of PQ are 3, –2, 1 and consequently, equations of PQ are x−23=y+1−2=z−3−1\frac { x - 2 } { 3 } = \frac { y + 1 } { - 2 } = \frac { z - 3 } { - 1 }

Any point on this line is (3r+2,−2r−1,−r+3)( 3 r + 2 , - 2 r - 1 , - r + 3 )

Let this point be Q.

Then midpoint of PQ

=(3r+2+22,−2r−1−12,−r+3+32)=(3r+42,−r−1,−r+62)= \left( \frac { 3 r + 2 + 2 } { 2 } , \frac { - 2 r - 1 - 1 } { 2 } , \frac { - r + 3 + 3 } { 2 } \right) = \left( \frac { 3 r + 4 } { 2 } , - r - 1 , \frac { - r + 6 } { 2 } \right) This point lies in given plane i.e.

3(3r+42)−2(−r−1)−(−r+62)=93 \left( \frac { 3 r + 4 } { 2 } \right) - 2 ( - r - 1 ) - \left( \frac { - r + 6 } { 2 } \right) = 9

⇒ 9r+12+4r+4+r−6=99 r + 12 + 4 r + 4 + r - 6 = 9 ⇒ 14r=814 r = 8 ⇒ r=47r = \frac { 4 } { 7 }

Hence, the required point Q is

(3(47)+2,−2(47)−1,−47+3)=(267,−157,177)\left( 3 \left( \frac { 4 } { 7 } \right) + 2 , - 2 \left( \frac { 4 } { 7 } \right) - 1 , \frac { - 4 } { 7 } + 3 \right) = \left( \frac { 26 } { 7 } , \frac { - 15 } { 7 } , \frac { 17 } { 7 } \right)