## FANDOM

49 Pages

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+bycz=D, where a, b, c, and d are constants)

MathHelp replies,

The method is

1. Find two vectors that are parallel to the plane.
2. Find the normal to the these two vectors.
3. Find the general equation of a plane perpendicular to the normal vector.
4. Substitute one of the points (A, B, or C) to get the specific plane required.
5. 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 BA is parallel to the plane, then so is any constant times BA. In this case, we can multiply BA by 5/2, and CB 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 10x+34y-11z=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 10x+34y-11z=100/5−44/5=56/5. Scaling the equation, we get our final answer,

50x+170y-55z=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.