- A student asks,

Let *P* be the plane in $ \mathbb{R}^3 $ which contains the points

$ \begin{split} A&=\left(\frac{-4}{5}, \quad \frac{8}{5}, \quad \frac{16}{5}\right) \\ B&=\left(2, \quad 0, \quad \frac{4}{5}\right) \\ C&=\left(\frac{2}{5}, \quad \frac{3}{5}, \quad \frac{6}{5}\right) \end{split} $

Find the equation of *P*. (The equation should be entered in the form *ax*+*by*−*cz*=*D*, where *a*, *b*, *c*, and *d* are constants)

- MathHelp replies,

The method is

- Find two vectors that are parallel to the plane.
- Find the normal to the these two vectors.
- Find the general equation of a plane perpendicular to the normal vector.
- Substitute one of the points (
*A*,*B*, or*C*) to get the specific plane required. - Check the answer by plugging points A, B, and C into this equation.

Let's take them one at a time.

1. To find two vectors parallel to the plane, think of A and B not as points, but as vectors from the origin to those points. Then, the difference B-A is a vector from A to B, which is parallel to the plane. Do the same with B and C. Then you get

$ \begin{split} B-A&=\langle 2-\frac{-4}{5}, \quad 0-\frac{8}{5}, \quad \frac{4}{5}-\frac{16}{5}\rangle \\ C-B&=\langle \frac{2}{5}-2, \quad \frac{3}{5}-0, \quad \frac{6}{5}-\frac{4}{5}\rangle \end{split} $

Simplifying,

$ \begin{split} B-A&=\langle \frac{14}{5}, \quad \frac{-8}{5}, \quad \frac{-12}{5}\rangle \\ C-B&=\langle \frac{-8}{5}, \quad \frac{3}{5}, \quad \frac{2}{5}\rangle \end{split} $

Now, scale the vectors to make them easier to work with. Remember, if *B*−*A* is parallel to the plane, then so is any constant times *B*−*A*. In this case, we can multiply *B*−*A* by 5/2, and *C*−*B* by 5, giving us two new vectors,

$ \begin{split} D&=\langle 7, \quad -4, \quad -6\rangle \\ E&=\langle -8, \quad 3, \quad 2\rangle \end{split} $

2. Find the normal to the these two vectors. The vector cross product automatically gives you a normal vector, as long as the vectors are not parallel. (What if the vectors *were* parallel? What does that mean regarding points *A*, *B*, and *C*?)

Recall the vector cross product of $ a=\langle a_1, \quad a_2, \quad a_3 \rangle $ and $ b=\langle b_1, b_2, b_3 \rangle $ is $ a\times b = \langle a_2 b_3 - a_3 b_2, \quad a_3 b_1 - a_1 b_3, \quad a_1 b_2 - a_2 b_1 \rangle $, so

$ \begin{split} D \times E & = \langle (-4)(2)-(-6)(3), \quad (-6)(-8)-(7)(2), \quad (7)(3)-(-4)(-8)\rangle \\ & = \langle 10, \quad 34, \quad -11 \rangle \end{split} $

3. Find the general equation of a plane perpendicular to the normal vector.

The equation of a plane perpendicular to vector $ \langle a, \quad b, \quad c \rangle $ is *ax*+*by*+*cz*=*d*, so the equation of a plane perpendicular to $ \langle 10, \quad 34, \quad -11 \rangle $ is 10*x*+34*y*-11*z*=*d*, for some constant, *d*.

4. Substitute one of the points (*A*, *B*, or *C*) to get the specific plane required.

In this case, the easiest point is *B*, so we let *x*=2, *y*=0, and *z*=4/5, giving us 10*x*+34*y*-11*z*=100/5−44/5=56/5. Scaling the equation, we get our final answer,

- 50
*x*+170*y*-55*z*=56

5. Let's check our answer by plugging points A, B, and C into this equation, to verify that each point indeed satisfies the equation.

```
```50*-0.8 + 170*1.6 - 55*3.2 = 56

50*2 + 170*0 - 55*0.8 = 56

50*0.4 + 170*0.6 - 55*1.2 = 56

So, indeed, it checks out.