Question

If (2, 3, -1) is the foot of the perpendicular from (4, 2, 1) to a plane, then the equation of the plane is

• A. 2x - y + 2z = 0
• B. 2x - y + 2z + 1 = 0
• C. 2x + y + 2z - 5 = 0
• D. 2x + y + 2z - 1 = 0

Hint:

To find the equation of the plane when you know a point and the foot of the perpendicular, use the point-normal form of the plane equation. The direction of the perpendicular vector is normal to the plane, and you can use it as the normal vector for the plane equation.

Answer 1

The Correct Option is B

The correct answer is: (B): $2x - y + 2z + 1 = 0$

We are given that $(2, 3, -1)$ is the foot of the perpendicular from the point $(4, 2, 1)$ to a plane, and we are tasked with finding the equation of the plane.

Step 1: Find the direction vector of the perpendicular

The direction of the perpendicular from the point $(4, 2, 1)$ to the plane is given by the vector connecting the point $(4, 2, 1)$ to the foot of the perpendicular $(2, 3, -1)$. The direction vector $\vec{v}$ is:

$\vec{v} = (2 - 4, 3 - 2, -1 - 1) = (-2, 1, -2)$

Step 2: Use the point-normal form of the plane equation

The equation of a plane can be written as:

$\vec{n} \cdot (\vec{r} - \vec{P}) = 0$

Here, $\vec{n}$ is the normal vector to the plane, $\vec{r} = (x, y, z)$ is a point on the plane, and $\vec{P} = (2, 3, -1)$ is the foot of the perpendicular.

Since $\vec{v} = (-2, 1, -2)$ is the direction of the perpendicular, it is also normal to the plane. Thus, the normal vector to the plane is $\vec{n} = (-2, 1, -2)$.

Step 3: Set up the equation of the plane

Now, use the point-normal form of the plane equation with $\vec{n} = (-2, 1, -2)$ and $\vec{P} = (2, 3, -1)$:

-2(x - 2) + 1(y - 3) - 2(z + 1) = 0

Step 4: Simplify the equation

Now, expand and simplify the equation:

-2x + 4 + y - 3 - 2z - 2 = 0

-2x + y - 2z - 1 = 0

Multiplying through by -1 gives:

2x - y + 2z + 1 = 0

Conclusion:
The equation of the plane is $2x - y + 2z + 1 = 0$, so the correct answer is (B): $2x - y + 2z + 1 = 0$.