Consider the lines L 1: ( x -1/2)=( y -3/2)=( z -2/-1) and L 2: ( x -2/1)=( y -6/-1)=( z +2/3). The distance of the point P (√10, √10, √10) from the plane passing through origin and whose normal is perpendicular to both the lines L 1 and L 2, is

Question

Consider the lines L 1: ( x -1/2)=( y -3/2)=( z -2/-1) and L 2: ( x -2/1)=( y -6/-1)=( z +2/3). The distance of the point P (√10, √10, √10) from the plane passing through origin and whose normal is perpendicular to both the lines L 1 and L 2, is (A) √10 (B) 4 (C) 2 (D) 2 √10

Tardigrade Admin · Accepted Answer

The normal vector of plane is parallel to vector

= ∣ ∣ \hat{i} 21 \hat{j} 2 - 1 \hat{k} - 1 3 ∣ ∣ = 5 \hat{i} - 7 \hat{j} - 4 \hat{k}

∴

Equation of plane is

5 x - 7 y - 4 z = 0

.....(1)
So, distance of plane in equation (1) from

P (10, 10, 10)

= \frac{∣5 ( 10 ) - 7 ( 10 ) - 4 ( 10 ) ∣}{( 5 ) ^{2} + ( - 7 ) ^{2} + ( - 4 ) ^{2}} = \frac{6 10}{3 10} = 2.

$10$

4

2

$210$

Q. Consider the lines L1​:2x−1​=2y−3​=−1z−2​ and L2​:1x−2​=−1y−6​=3z+2​. The distance of the point P(10​,10​,10​) from the plane passing through origin and whose normal is perpendicular to both the lines L1​ and L2​, is

10​

4

2

210​

$10$

$210$