Question

Find the vector and cartesian equation of the plane passing through the point (2, 5, - 3), (-2, -3, 5) and (5, 3, -3). Also, find the point of intersection of this plane with the line passing through points (3, 1, 5) and (-1, -3, -1).

Answer 1

Let `veca = 2hati + 5hatj - 3hatk, vecb = -2hati - 3hatj + 5hatk, vecc = 5hati + 3hatj - 3hatk`

The vector equation of the plane passing through `veca, vecb "and" vecc` is given by

`(vecr - veca). [ (vecb - veca) xx (vecc - veca) ] = 0`

⇒ `[ vecr - (2hati + 5hatj - 3hatk)] . [(-4hati - 8hatj + 8hatk) xx (3hati - 2hatj)] = 0`

⇒ `[ vecr - (2hati + 5hatj - 3hatk)] . (2hati + 3hatj + 4hatk)= 0`

Cartesian equation of the plane passing through the points (2, 5, - 3), (-2, -3, 5) and (5, 3, -3) is given by

`|(x - 2, y - 5, z + 3), (-2 - 2, -3-5, 5 + 3), (5 - 2, 3 - 5, - 3 + 3)| = 0`

`|(x - 2, y - 5, z + 3), (-4, -8, 8), (3, -2, 0)| = 0`

⇒ ( x - 2)(16) - (y - 5)(- 24) + (z + 3) (32) = 0

⇒ 2x + 3y + 4z = 7               ....(i)

Equation of the line passing through points (3,1,5) and (−1,−3,−1) is given by

`( x - 3)/2 = (y - 1)/2 = (z - 5)/3 = λ`          ....(ii)

Any point on line (ii) is given by (2λ + 3, 2λ + 1, 3λ + 5)
If plane (i) intersects with the line (ii), then

2(2λ + 3) + 3(2λ + 1) + 4(3λ + 5) = 7

⇒ 22λ = - 22
⇒ λ = -1

Therefore, point of intersection is (1, -1, 2).