Question

1. Determine the scalar equation of the plane with vector equation Vector r= (3,-1,4) +s(2,-1,5) + t(-3,2,-2).

2. Determine the value of k so that the line with parametric equations x = 2 + 3t, y = -2 + 5t, z = kt is parallel to the plane with equation 4x + 3y – 3z -12 = 0.

3. The planes A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0 are perpendicular. Find the value of A1A2 + B1B2 + C1C2.

Answer 1

Have you done any of the work??

Answer 2

1. notice that you have two direction vectors , find a vector [a,b,c] which is perpendicular to both of those. Such a vector is called the normal and is obtained by taking the cross-product of the 2 direction vectors.
So your equation of the plane will be ax + by + cz = k
plug in the point (3,-1,4) to find k

2. As I said in #1, the normal to 4x + 3y – 3z -12 = 0 is [4,3,-3]
to be parallel with the line, the dot-product of [4,3,-3] with [3,5,k] mus
be zero. go for it!

3. Well duhh!