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Question: The point of intersection of the line \(\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{- 2}\) and ...

The point of intersection of the line x−13=y+24=z−3−2\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{- 2} and plane 2x−y+3z−1=02x - y + 3z - 1 = 0 is

A

(10,−10,3)(10, - 10,3)

B

(10,10,−3)(10,10, - 3)

C

(−10,10,3)( - 10,10,3)

D

None of these

Answer

(10,10,−3)(10,10, - 3)

Explanation

Solution

Given line is, x−13=y+24=z−3−2=k\frac { x - 1 } { 3 } = \frac { y + 2 } { 4 } = \frac { z - 3 } { - 2 } = k, (say)

∴ Point on the line is

x=3k+1,y=4k−2x = 3 k + 1 , y = 4 k - 2 z=−2k+3z = - 2 k + 3 .....(i)

This point must satisfies the equation of plane

∴ 2(3k+1)−(4k−2)+3(−2k+3)−1=0⇒k=32 ( 3 k + 1 ) - ( 4 k - 2 ) + 3 ( - 2 k + 3 ) - 1 = 0 \Rightarrow k = 3

From (i),(x,y,z)=(10,10,−3)( x , y , z ) = ( 10,10 , - 3 ).