Find the point of intersection of the line x = 2t, y = t - 1, z = 3t + 4 and the plane x - 5y + 3z = 47.
A)(10, 4, 19)
B)(-10, 4, 19)
C)(0, -4, -1)
D)(10, -4, 19)
E)(0, -1, 4)
F)(0, 1, -4)
G)(0, -1, -4)
H)(10, -4, -19)

Question

Quizplus · Accepted Answer

To find the point of intersection, substitute the parametric equations of the line (x = 2t, y = t - 1, z = 3t + 4) into the equation of the plane (x - 5y + 3z = 47). This gives:2t - 5(t - 1) + 3(3t + 4) = 47Solving for t gives t = 5. Substituting t = 5 back into the parametric equations of the line gives the point of intersection as (10, 4, 19).

Find the Point of Intersection of the Line X =