Determine if the plane given by -x + 2z = 10 and the line given by r = (5,2-t,10+4t) are orthogonal, parallel

Determine if the plane given by -x + 2z = 10 and the line given by r = (5,2-t,10+4t) are orthogonal, parallel or neither.

Notice that you can write the equation of the plane as:
-x + 0y + 2z = 10.

At this point, you can see that a normal vector to this plane is:
<-1, 0, 2>.

On the other hand, the direction vector of the line <5, 2 - t, 10 + 4t> is:
<0, -1, 4>.

Since <1, 0, 2> and <0, -1, 4> are not constant multiples of each other---in other words, there doesn't exist a value of t such that <1, 0, 2> = t<0, -1, 4>---the plane and the line are not parallel to each other.

By computing the cross-product <-1, 0, 2> x <0, -1, 4> and showing that the result is not the zero vector, they are not perpendicular to each other either.

Therefore, the plane and the line are neither parallel or perpendicular to each other.

I hope this helps!