Question

Where does the line through (1, 0, 1) and (4, −3, 3) intersect the plane x + y + z = 6?

Answer 1

Parametrically for real t the line is

$(x,y,z) = (1-t)(1,0,1) + t(4,-3,3)$

$(x,y,z) = (1,0,1) + t(4-1,-3-0,3-1)$

$(x,y,z) = (1,0,1) + t(3,-3,2)$

That's three equations; we substitute into

$x+y+z=6$

$(1+3t)+(0-3t) + ( 1+2t)=6$

$2t=4$

$t=2$

$(x,y,z) = (1,0,1) + 2(3,-3,2) = (7,-6,5)$

The intersection is a single point (7,-6,5).