Find the vector equation of the plane which passes through the points
2 i + 4 j + 2k ̧ 2 i + 3 j + 5k and parallel to the vector 3 i − 2 j + k . Also find the point where this plane meets the line joining the points 2 i + j + 3k and 4 i − 2 j + 3k ?

r = (1 − s)a + sb + tc
a = 2 i + 4 j + 2k, b = 2 i + 3 j + 5k and
c = 3i − 2 j + k
Equation of the plane is
r = (1 − s)(2 i + 4 j + 2k) + s(2 i + 3 j + 5k) + t(3t − 2 j + k)
= [2 − 2s + 2s + 3t]i + [4 − 4s + 3s − 2t] j + [2 − 2s + 5s + t]t
r = [3t + 2] i + [4 − s − 2t] j + [2 + 3 j + t]k ...(1)
Equation of the line is
r = (1 − p)a + pb where p is a scalar.
r = (1 − p)(2 i + j + 3k) + p(4 i − 2 j + 3k)
= (2 − 2p + 4p) i + (1 − p − 2p) j + (3 − 3p + 3p)k
r = [2 + 2p] i + [1 − 3p] j + [3]k
...(2)
At the point of intersection (1) = (2)
(3t + 2) i + (4 − s − 2t) j + (2 + 3s + t)k = (2 + 2p) i + (1 − 3p) j + 3k
By comparing the like term, we get
3t + 2 = 2p + 2
3t − 2p = 0
...(i)
4 − s − 2t = 1 − 3p
2t + s − 3p = 3
...(ii)
2 + 3s + t = 3
3s + t = 1
...(iii)
Solving (ii) and (iii)
(ii) × 3 ⇒ 6t + 3s – 9p = 9
(iii) ⇒ t + 3s = 1
5t – 9p = 8 ...(iv)
Solving (i) – (iv) :
15t – 10 p = 0
15t – 27p = 24
17p = –24 ⇒ p = −
24
17
To find point of intersection,
Put p = −
24
in (2)
17
48 ⎞
⎛
⎛ 72 ⎞
r = ⎜ 2 − ⎟ i + ⎜ 1 + ⎟ j + 3k
17 ⎠
⎝
⎝ 17 ⎠
⎛ 14 ⎞
⎛ 89 ⎞
= ⎜ − ⎟ i + ⎜ ⎟ j + 3k
⎝ 17 ⎠
⎝ 17 ⎠
14 89
∴ Point of intersection is ⎛ − , ,3 ⎞ .
⎜
⎟