Question

The distance of the point (–1, –5, –10) from the point of intersection of the line vector r = 2i - j + 2k + λ(3i + 4j + 12k) and the plane r.(i - j + k) = 5 is 

A. 9

B. 13

C. 17

D. none of these

Answer 1

Correct option is B.13

Let, the point of intersection of the line

and the plane

As (x₀, y₀, z₀) is the point of intersection of the line and the plane, so the position vector of this point i.e.

must satisfy both of the equation of line and the equation of plane.

Substituting, \(\vec r_0\) in place of  \(\vec r\) in both the equations, we

get,

i.e.x₀ =2 + 3λ

y₀ = - 1 + 4λ

z₀ =2 + 12λ

Substituting, these values in equation (2) we get,

((2 + 3λ) × 1) - (1 × ( - 1 + 4λ)) + (1 × (2 + 12λ)) = 5

2 + 3λ + 1 - 4λ + 2 + 12λ = 5

11λ = 0

λ = 0

∴ x₀ = 2 + 3λ

= 2

y₀ = - 1 + 4λ

= -1

z₀ = 2 + 12λ

= 2

Hence, the point of intersection is, (2, - 2, 2).

Now, the distance between the point ( - 1, - 5, - 10) and (2, - 1, 2) is,

= 13

Hence, the required distance between the point ( - 1, - 5, - 10) the point where the line

the plane 

is 13 units.