When the planes are perpendicular, what is k?

[DespicableMe]

Homework Statement

Determine value of k when thehse are perpendicular:
[x,y,z] = [1+4s+kt, 2+2s+t, 7+2t]
and
[x,y,z] = [4,1,-1] + s[1,0,5] + t[0,-3,3]

The Attempt at a Solution

I learned that in 2 and 3D, two lines are perpendicular only if, after solving for the t in front of the direction vector, x=x , y=y and z=z

My attempt was to Cross product the direction vecotrs in each of the equations, and then dot product those values to find k, since if they are perpend. then dot product=0

I'm not sure how they ended up with k = 31/3

 

[HallsofIvy]

I very much doubt that you learned that. The condition you give, that the "direction vector" of the lines is parallel, tells you that two lines are parallel, not perpendicular.

However, what you are saying here is correct. For the first plane, <4, 2, 0> and <k, 1, 0> are vectors in the plane and so their cross product is normal to the plane. Similarly, for the second plane, <1, 0, 5> and <0, -3, 3> are in the plane and so their cross product is normal to the plan. The two planes will be perpendicular if and only if the dot product of those two normals is 0.

You are correct that 31/3 is incorrect. What did you get?