Question

determine whether the line of parametric equations, x = 4+2t, y = −5t, z = 4+t, t is in ℝ, intersects the plane with equation 3x+4y+6z − 52 = 0.

Answer 1

After substituting the parametric equations for x, y, and z into the plane's equation and solving for t, we find that the line intersects the plane at t = -2.

Step-by-step explanation:

To determine if the line given by the parametric equations x = 4+2t, y = −5t, z = 4+t, where t is in ℝ, intersects with the plane represented by the equation 3x+4y+6z − 52 = 0, we substitute the parametric equations into the plane equation and solve for t.

Substituting x, y, and z into the plane equation gives:

3(4+2t) + 4(-5t) + 6(4+t) − 52 = 0
Expanding and simplifying:

12 + 6t - 20t + 24 + 6t - 52 = 0
− 8t - 16 = 0
− 8t = 16
t = − 2

Since we found a real value for t, that means the line does intersect with the plane at the point corresponding to t = − 2.