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Question: The point at which the line joining the points (2, –3, 1) and (3, –4, –5) intersects the plane \(2x ...

The point at which the line joining the points (2, –3, 1) and (3, –4, –5) intersects the plane 2x+y+z=72x + y + z = 7is

A

(1, 2, 7)

B

(1, –2, 7)

C

(–1, 2, 7)

D

(1, –2, –7)

Answer

(1, –2, 7)

Explanation

Solution

Ratio −[2(2)+(−3)(1)+(1)(1)−72(3)+(−4)(1)+(−5)(1)−7]=−[−5−10]=−(12)- \left[ \frac { 2 ( 2 ) + ( - 3 ) ( 1 ) + ( 1 ) ( 1 ) - 7 } { 2 ( 3 ) + ( - 4 ) ( 1 ) + ( - 5 ) ( 1 ) - 7 } \right] = - \left[ \frac { - 5 } { - 10 } \right] = - \left( \frac { 1 } { 2 } \right)

∴x=2(2)−3(1)1=1,y=−3(2)−(−4)1=−2\therefore x = \frac { 2 ( 2 ) - 3 ( 1 ) } { 1 } = 1 , y = \frac { - 3 ( 2 ) - ( - 4 ) } { 1 } = - 2

and z=1(2)−(−5)1=7z = \frac { 1 ( 2 ) - ( - 5 ) } { 1 } = 7 . Therefore, P(1,−2,7)P ( 1 , - 2,7 ).

Trick : As (1, – 2, 7) and (– 1, 2, 7) satisfy the equation 2x+y+z=72 x + y + z = 7 but the point (1, – 2, 7) is collinear with
(2, – 3, 1) and (3, – 4, – 5).

Note : If a point dividing the join of two points in some particular ratio, then this point must be collinear with the given points.