# Thread: Discrete Math Exam Review - Planes

1. ## Discrete Math Exam Review - Planes

Determine a plan detailing on how you would find the value(s) of k, if any, that would make the planes $\displaystyle kx + 2ky + 5z - 6 = 0$ and $\displaystyle -x + (k - 1)y - 7z - 9 = 0$ perpendicular. Implement your plan and solve for k.

I tried use the dot product to solve this problem, but it just got messy and got a headache ...

Is there an easier way to solve for it and if so, please show in full steps?

Answer:

$\displaystyle k = 5$
$\displaystyle k = - \frac{7}{2}$

2. Recall that a plane given by the equation $\displaystyle ax + by + cz + d = 0$ has the vector (a,b,c) perpendicular to it.

Now looking at your two planes, we see that the normals for each plane respectively are (k,2k,5) and (-1, k-1, -7). If the planes are perpendicular to each other, then so should their normals. So, take the dot product of these two vectors and set it to 0:

$\displaystyle (k,2k,5) \cdot (-1, k-1, -7) = 0$
$\displaystyle -k + 2k(k-1) - 35 = 0$

etc. etc. and solve the quadratic.