Find a vector equation (with integral values and parametric parts) of the line that is the intersection of?

Question

9x+11y-5z=50 and 10x+7y-16z=19

Answer 1

Normal vectors to the two planes are (9, 11, -5) and (10, 7, -16), so a vector in the direction of the line is given by the cross product of these two, i.e. (-141, 94, -47). We can take out a common factor of -47 to give a direction vector of (3, -2, 1).

Let x = 0, then 11y - 5z = 50 and 7y - 16z = 19, so we get
7(11y-5z) - 11(7y-16z) = 350 - 209 <=> 141z = 141, so z = 1 and y = 5. So (0, 5, 1) is a point on the line.

So the equation of the line can be written as
(x, y, z) = (0, 5, 1) + t(3, -2, 1)
= (3t, 5 - 2t, 1 + t)
for t ∈ R.