Question

If the foot of the perpendicular drawn from the origin to a plane is P(-1, -1, 2), then the equation of the plane is ______.

• A. $x + y - 2z + 6 = 0$
• B. $2x + y + z + 1 = 0$
• C. $x + y + 2z - 2 = 0$
• D. $x - y - z + 2 = 0$

Hint:

Shortcut: If the foot of the perpendicular from the origin is $(x_1, y_1, z_1)$, the plane equation is incredibly simple: $x_1 x + y_1 y + z_1 z = x_1^2 + y_1^2 + z_1^2$.
Here: $-1x - 1y + 2z = 1 + 1 + 4 \implies -x - y + 2z = 6 \implies x + y - 2z + 6 = 0$. Extremely fast!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the precise coordinates of the foot of a perpendicular dropped from the origin (0,0,0) to a plane. We must use this single point to derive the entire equation of that plane.

Step 2: Detailed Explanation:
Let the origin be $O(0, 0, 0)$ and the foot of the perpendicular on the plane be $P(-1, -1, 2)$.
A crucial geometric property is that the line segment $OP$ is perfectly perpendicular (normal) to the plane itself.
Therefore, the direction vector of the line segment $\vec{OP}$ serves perfectly as the Normal Vector ($\vec{n}$) to the plane.
Calculate the normal vector $\vec{n}$:
$\vec{n} = \vec{OP} = (-1 - 0)\hat{i} + (-1 - 0)\hat{j} + (2 - 0)\hat{k}$
$\vec{n} = -1\hat{i} - 1\hat{j} + 2\hat{k}$
The direction ratios of the normal are $(a, b, c) = (-1, -1, 2)$.
Because point $P(-1, -1, 2)$ physically lies on the plane, we can use the point-normal form of the plane equation:
$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$
Substitute the normal vector and the point $P$:
$-1(x - (-1)) - 1(y - (-1)) + 2(z - 2) = 0$
$-1(x + 1) - 1(y + 1) + 2(z - 2) = 0$
Expand the brackets:
$-x - 1 - y - 1 + 2z - 4 = 0$
Combine the constant terms:
$-x - y + 2z - 6 = 0$
To match standard formatting (and the options), multiply the entire equation by $-1$:
$x + y - 2z + 6 = 0$

Step 3: Final Answer:
The equation of the plane is $x + y - 2z + 6 = 0$, matching option (a).