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: Understanding the Question:
We need to determine the Cartesian equation of a plane that contains two specific points and is strictly perpendicular to another given plane.

Step 2: Key Formula or Approach:
To construct a plane equation, we need a point on the plane and its normal vector ($\vec{n}$).
Let the two given points on our required plane be $A(-2, 2, 2)$ and $B(2, -2, -2)$. The vector $\vec{AB}$ lies completely flat on the required plane.
Because the required plane is perpendicular to the given plane $9x - 13y - 3z = 0$, the normal vector of the given plane, $\vec{n}_1 = \langle 9, -13, -3 \rangle$, must lie perfectly parallel to (or directly on) our required plane.
Thus, the normal vector $\vec{n}$ of our required plane must be perpendicular to both $\vec{AB}$ and $\vec{n}_1$. We find it using the cross product: $\vec{n} = \vec{AB} \times \vec{n}_1$.

Step 3: Detailed Explanation:
Find the direction vector $\vec{AB}$:
$$\vec{AB} = \langle 2 - (-2), -2 - 2, -2 - 2 \rangle = \langle 4, -4, -4 \rangle$$ To make calculations easier, we can scale this down by a factor of 4 to get a simpler parallel direction vector: $\vec{v} = \langle 1, -1, -1 \rangle$.
Now, calculate the cross product to find the normal vector $\vec{n}$:
$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 9 & -13 & -3 \end{vmatrix}$$ $$\vec{n} = \hat{i}(3 - 13) - \hat{j}(-3 - (-9)) + \hat{k}(-13 - (-9))$$ $$\vec{n} = -10\hat{i} - 6\hat{j} - 4\hat{k}$$ We can factor out $-2$ to get a simplified normal vector: $\vec{n}_{simple} = \langle 5, 3, 2 \rangle$.
The general equation of the plane is $5x + 3y + 2z = d$.
Substitute point $B(2, -2, -2)$ to find $d$:
$$5(2) + 3(-2) + 2(-2) = d$$ $$10 - 6 - 4 = 0 \implies d = 0$$ The final equation is $5x + 3y + 2z = 0$.

Step 4: Final Answer:
The equation of the plane is $5x+3y+2z=0$, matching option (B).