Question

The equation of the plane passing through $(-2,2,2)$ and $(2,-2,-2)$ and perpendicular to the plane $9x-13y-3z=0$ is

• A. $5x-3y+2z=12$
• B. $5x+3y+2z=0$
• C. $5x+3y-2z+8=0$
• D. $5x-3y+2z+12=0$

Hint:

There is a massive shortcut here! Notice that point $A(-2,2,2)$ and point $B(2,-2,-2)$ are exact opposites (reflections across the origin). A line passing through both of them automatically passes straight through the origin $(0,0,0)$. Therefore, the plane containing them must also pass through the origin, meaning the constant term $d$ must be strictly zero! Only option (B) has $d=0$.

Answer 1

The Correct Option is B

Step 1: Gather what defines the plane.
Our plane passes through $A(-2,2,2)$ and $B(2,-2,-2)$ and is perpendicular to the plane $9x-13y-3z=0$, whose normal is $\langle 9,-13,-3\rangle$.
Step 2: Two vectors lie in our plane.
The line $AB$ lies in our plane, and because our plane is perpendicular to the given plane, the given normal $\langle 9,-13,-3\rangle$ also lies parallel to our plane.
Step 3: Find $AB$.
$\vec{AB}=\langle 4,-4,-4\rangle$, which we scale to $\langle 1,-1,-1\rangle$.
Step 4: Cross the two to get our normal.
$\langle 1,-1,-1\rangle\times\langle 9,-13,-3\rangle=\hat{i}((-1)(-3)-(-1)(-13))-\hat{j}((1)(-3)-(-1)(9))+\hat{k}((1)(-13)-(-1)(9))$, giving $\langle -10,-6,-4\rangle$, which scales to $\langle 5,3,2\rangle$.
Step 5: Build the plane.
Plane is $5x+3y+2z=d$. Using $B(2,-2,-2)$: $5(2)+3(-2)+2(-2)=10-6-4=0$, so $d=0$.
Step 6: State the answer.
The plane is $5x+3y+2z=0$, which is option (2). \[ \boxed{5x+3y+2z=0} \]