Line intersection and plane calculus.?

If (x, y, z) is on both the line and the plane, then there is a real number t such that

........(1) 1 + t = x and 6 - 2t = y and z = 11 - 2t

........(2) x - y = 4

Therefore, x - y = (1 + t) - (6 - 2t) = -5 + 3t = 4 which implies that t = 3. Plugging in 3 in place of t in (1) we find x = 4, y = 0, and z = 5.

Answer:

The line L(t) = (1+t, 6-2t, 11-2t) intersects the plane x - y = 4 at the point (4, 0, 5) when t = 3.

The directional vector v, of a line perpendicular to the airplane is the comparable with the aid of fact the conventional vector of the airplane. v = <3, -2, 7> With the directionla vector of the line and a element on the line O(0,0,0) we can write the equation of the airplane. r(t) = O + television r(t) = <0, 0, 0> + t<3, -2, 7> with the aid of fact the element on the line is the beginning place we can simplify the equation of the line further. r(t) = t<3, -2, 7> the place parameter t stages over the real numbers