NCERT Solutions
Class 12 Maths
Three Dimensional Geometry

Ex.Misc.Q.18
Find the distance of the point (−1, −5, −10) from the point of intersection of the line
r = 2i – j + 2k + λ(3i + 4j + 2k) and the plane r.(i – j + k) = 5
The equation of the given line is
r = 2i – j + 2k + λ(3i + 4j + 2k) ……….1
The equation of the given plane is
r.(i – j + k) = 5 ……….2
Substituting the value of from equation 1 in equation 2, we get
[2i – j + 2k + λ(3i + 4j + 2k)].(i – j + k) = 5
=> [(3λ + 2)i + (4λ - 1)j + (2λ + 2)k].(i – j + k) = 5
=> (3λ + 2) - (4λ - 1) + (2λ + 2) = 5
=> λ = 0
Substituting this value in equation 1, we get the equation of the line as
r = 2i – j + 2k
This means that the position vector of the point of intersection of the line and the plane Is r = 2i – j + 2k
This shows that the point of intersection of the given line and plane is given by the coordinates, (2, −1, 2). The point is (−1, −5, −10).
The distance d between the points, (2, −1, 2) and (−1, −5, −10), is
d = √[(-1 - 2)2 + (-5 + 1)2 + (-10 - 2)2] = √(9 + 16 + 144) = √(169) = 13