Question

If the foot of the perpendicular drawn from the origin to a plane is \( P(2, -1, 4) \), then the equation of the plane is

• A. \( 2x + y + 4z - 19 = 0 \)
• B. \( x + y + z - 5 = 0 \)
• C. \( 2x - 2y - 3z + 6 = 0 \)
• D. \( 2x - y + 4z - 21 = 0 \)

Hint:

If origin is $(0,0,0)$ and foot is $(x_1, y_1, z_1)$, the plane is simply $x_1x + y_1y + z_1z = x_1^2 + y_1^2 + z_1^2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The vector from origin to the foot of the perpendicular is the normal to the plane. The plane passes through this point.
Step 2: Key Formula or Approach:
The equation of a plane through \( (x_1, y_1, z_1) \) with normal DRs \( (a, b, c) \) is:
\( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \).
Step 3: Detailed Explanation:
Normal vector \( \bar{n} = \vec{OP} = (2, -1, 4) \).
Point on plane \( P = (2, -1, 4) \).
Equation:
\[ 2(x - 2) + (-1)(y - (-1)) + 4(z - 4) = 0 \] \[ 2x - 4 - y - 1 + 4z - 16 = 0 \] \[ 2x - y + 4z - 21 = 0 \] Step 4: Final Answer:
The equation is \( 2x - y + 4z - 21 = 0 \).