Q. 20

# 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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Questions Based on 3D Geometry26 mins
Revision of straight lines important formulas in one shot59 mins
Reminding 11th - Revision of Circles important formulas in 50 Minutes56 mins
Concept Builder Questions of straight Lines (Quiz Session)55 mins
Conic Section Part 164 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses