vector math problem

Question

Write the vector equation of a line through the point of intersection of 𝑙1 and 𝑙2 and perpendicular to both 𝑙1 and 𝑙2. 𝑙1:[𝑥,𝑦,𝑧] = [2,1,1] + 𝑠[4,0,−1]

𝑙2:[𝑥,𝑦,𝑧] = [3,−1,1] + 𝑡[9,−2,−2]

Answer 1

To get intersaction point we write parametric equations of lines.

Line l_1: x = 2 + 4s, y = 1, z = 1 - s. Line l₂: x = 3 + 9t, y = - 1 - 2t, z = 1 - 2t.

To get point of intersection we have system of 3 equations in 2 variables s and t:

2 + 4s = 3 + 9t, 1 = -1 - 2t, 1 - s = 1 - 2t.

From second equation t = - 1. Then from 3rd equation 1 - s = 1 + 2, so s = - 2.

Let check 1st equation: 2 + 4(- 2) = 3 + 9(- 1), - 6 = - 6, OK!

So we get intersaction point: x = - 6, y = 1, z = 3 or (- 6, 1, 3).

Let now symmetric equation of line (x + 6)/m = (y - 1)/n = (z - 3)/p.

Now because this line perpendicular to l₁ and l₂ we have: 4m - p = 0 and 9m - 2n - 2p = 0.

So, p = 4m, 9m - 2n - 8m = 0, m = 2n, p = 8n.

Symmetric equation of line:(x + 6)/2n = (y-1)/n = (z - 3)/8n. After reducing all denominators by n we get

(x + 6)/2 = (y - 1)/1 = (z - 3)/8 = u, x = - 6 + 2u, y = 1 + u, z = 3 + 8u.

So vector equation of this line: r = [x,y,z] = [- 6, 1, 3] + u[2, 1, 8]