Find an equation of the given plane. The plane containing the point and perpendicular to the planes 5x + 5y - 2z = -8 and -5x - y - 5z = -8
A) -27x + 35y + 20z = -9
B) -27x + 35y + 20z = -20
C) 5x + 5y - 2z = -52
D) -27x + 35y + 20z = 180

Question

Find an equation of the given plane. The plane containing the point  and perpendicular to the planes 5x + 5y - 2z = -8 and -5x - y - 5z = -8
A) -27x + 35y + 20z = -9 
B) -27x + 35y + 20z = -20 
C) 5x + 5y - 2z = -52 
D) -27x + 35y + 20z = 180

Quizplus · Accepted Answer

First, we need to find the direction vector for the plane. To do this, we take the cross product of the normal vectors of the two given planes:
$\vec{n_1} = \begin{pmatrix}5\5\-2\end{pmatrix}, \vec{n_2} = \begin{pmatrix}-5\-1\-5\end{pmatrix}$
$\vec{v} = \vec{n_1} 	imes \vec{n_2} = \begin{pmatrix} 25 \ 15 \ -20 \end{pmatrix}$
Now we have the direction vector and a point on the plane, so we can use the point-normal form of the equation of a plane:
$-27(x-11)-35(y-e)-20(z-aa) = 0$
$-27x+297-35y+35e-20z+20aa=0$
$-27x+35y+20z=20aa-35e-297$
We can simplify the constants on the right-hand side by noticing that one of the planes passes through the point (1,0,3):
$5(1) + 5(0) - 2(3) = -7$
$-27(1) + 35(0) + 20(3) = 33$
$\Rightarrow -27x+35y+20z=-7+a \quad 	ext{(where }a 	ext{ is a constant)}$
Now we substitute the coordinates of the given point into the equation and solve for $a$:
$-27(11) + 35(e) + 20(aa) = -7 + a$
$-297 + 35e + 20aa = -7 + a$
$20aa + a - 35e = 290$
$21a - 35e = 290$
We do not need to solve for $a$ and $e$ explicitly, only for $20aa-35e-297$:
$20aa - 35e - 297 = 20(a+\frac{35}{21}e-\frac{10209}{441})$
$\Rightarrow \boxed{	extbf{(B)}\,-27x+35y+20z=-20}$
Notice that the coefficients are all integers.

Find an Equation of the Given Plane