Equation of Planes

Question

Find an equation of a plane through the point (5,4,1) which is orthogonal (perpendicular) to the line:

x=t-1 , y=3-4t, z= 1-4t

Can you explain how you were able to solve this? Thanks!

Answer 1

There are several ways of doing it, I will show you just one.

The general, equation of a plane is of the form

ax + by + cz = d

You need to find the constants a, b, c, d.

I assume you had the following fact in class:

If you have a function f(x, y, z),

then the gradient ∇f is a vector perpendicular to the surface defined by f(x, y, z) = 0.

This fact will give us the coefficients a, b, c:

Define

f(x, y, z) = ax + by + cz

Then

∇f = (a, b, c)

If we set a, b, c so that the vector (a, b, c) is parallel to the given line,

then the plane f(x, y, z) = 0 will be perpendicular to the given line.

We can get a vector parallel to the given line by taking any two points on the line.

For example, the points corresponding to t=0 and t=1.

That gives us the two points

(-1, 3, 1), (0, -1, -3)

The difference between them is then a vector parallel to the line

(-1, 4, 4)

Thus

-x + 4y + 4z = 0

is a plane perpendicular to the line.

All the planes -x + 4y + 4z = d (for different values of d)

are all parallel to each other; we can chose d so that we get a plane passing though

the given point (5, 4, 1):

-1×5 + 4×4 + 4×1 = d

Thus

d = 15

So the solution is

-x + 4y + 4z = 15