Find the distance of the point (–1, –5, –10), from the point of intersection of the line and the plane [CBSE 2011]

Given, the equation of the line is –

The cartesian form of the equation is –

∴ x = 3λ + 2 …(1)

y = 4λ – 1 …(2)

and z = 2λ + 2 …(3)

Given equation of plane is –

The cartesian form of above equation is –

x – y + z = 5 …(4)

At the point of intersection of line and the plane-

We have –

3λ +2 – (4λ – 1) + (2λ + 2) = 5

⇒ λ + 5 = 5

∴ λ = 0

Hence, x = 2 ; y = -1 and z = 2

∴ position vector of the intersection point is or coordinate is (2,-1,2)

By distance formula, we know that distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by –

∴ distance between (-1, -5, -10) and (2,-1, 2) is given by –

⇒ D = = 13 units.

Hence, a distance of the point (–1, –5, –10), from the point of intersection of the line and the plane is 13 units