Finding a perpendicular plane

Question

Question

Plane $B$ contains the points $(4, 2, 1)$ and $(4, 1, -6)$, and it is perpendicular to the plane $7x+9y+4z=18$ . What is the equation of the plane $B$?

What I tried:

I know that to find the equation of plane $B$ I need a point on the plane and a perpendicular vector. The point part is already covered $(4, 1, -6)$, and the cross product of $\langle4-4, 2-1, 1-(-6)\rangle$ and $\langle7, 9, 4\rangle$ should give me the perpendicular vector; then it's just a matter of plugging the numbers in the formula. However, this approach did not work and I don't know what I am doing wrong.

Answer 1

Your approach is both correct and simple.

Your plane has to be parallel to both the vectors

$$\begin{pmatrix}4\\2\\1\end{pmatrix}-\begin{pmatrix}4\\1\\-6\end{pmatrix}=\begin{pmatrix}0\\1\\7\end{pmatrix} \qquad\text{and}\qquad \begin{pmatrix}7\\9\\4\end{pmatrix}$$

so its normal vector can be computed as

$$\begin{pmatrix}0\\1\\7\end{pmatrix}\times \begin{pmatrix}7\\9\\4\end{pmatrix}= \begin{pmatrix}-59\\49\\-7\end{pmatrix}$$

Plug in one of your two points

$$-59\cdot4+49\cdot2-7\cdot1=-145$$

so if you also get rid of some minus signs, the equation of the plane is

$$59\,x-49\,y+7\,z=145$$

To verify the required properties, you can plug in both your points and see that they satisfy the equation, and you can also compute the dot product

$$\left<\begin{pmatrix}59\\-49\\7\end{pmatrix}, \begin{pmatrix}7\\9\\4\end{pmatrix}\right>= 59\cdot7-49\cdot9+7\cdot4=0$$

which proves that the normals are perpendicular, therefore the planes are perpendicular.

Answer 2

The quickest way is probably noticing that for a plane $\pi : ax+by+cz+d=0$, the vector $(a,b,c,d)^T$ is a basis for the direction of the orthogonal subspace of $\pi$.

If $B$ is perpendicular to the plane $7x+9y+4z=18$, then it contains the direction $\langle (7,9,4)\rangle$. If it passes through the two points $P_1$,$P_2$, then it contains the direction $\langle P_2-P_1\rangle =\langle (0,1,7)^T \rangle$. Since these two subspaces are linearly independent and a plane has dimension $2$, we are safe to say that

$$B= P_1+\left \langle (P_2-P_1), \pi^{\perp} \right \rangle= \left( \begin{array}{c} 4 \\ 2 \\ 1 \\ \end{array} \right)+\left \langle \left( \begin{array}{c} 7 \\ 9 \\ 4 \\ \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \\ 7 \\ \end{array} \right)\right \rangle$$.