Question

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

Answer 1

Vector equation of the plane which passes through the points 2i+4j+2k, 2i+3j+5k and parallel to the vector 3i-2j+k is given as follow:

1. r = (1 − s)a + sb + tc

1. a = 2 i + 4 j + 2k, b = 2 i + 3 j + 5k and c = 3i − 2 j + k.
2. 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]k.
3. r =[3t + 2] i + [4 − s − 2t] j + [2 + 3 j + t]k ...(1)
4. Equation of the line is r = (1 − p)a + pb where p is a scalar.
5. 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.
6. r = [2 + 2p] i + [1 − 3p] j + [3]k ...(2).
7. 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.
8. 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

10. To find point of intersection,

put p = -24/17 in (2)

11. r = ( 2+2×(-24/17))i + ( 1-3×(-24/17))j + 3k

r = (-14/17)i + (89/17)j + 3k

12. Hence Point of intersection is:

( -14/17, 89/17, 3)