Question

If \((2, -1, 3)\) is the foot of the perpendicular drawn from the origin \((0, 0, 0)\) to a plane then the equation of that plane is

• A. \(2x + y - 3z + 6 = 0 \)
• B. \(2x - y + 3z - 14 = 0 \)
• C. \(2x - y + 3z - 13 = 0 \)
• D. \(2x + y + 3z - 10 = 0 \)

Hint:

When the foot of the perpendicular from the origin to a plane is given as \((x_0, y_0, z_0)\), this point itself provides the direction ratios of the normal to the plane. Thus, the normal vector is \(\langle x_0, y_0, z_0 \rangle\). The equation of the plane is then \(x_0 x + y_0 y + z_0 z = x_0^2 + y_0^2 + z_0^2\). Using this quick tip, for \((2, -1, 3)\): \(2x + (-1)y + 3z = 2^2 + (-1)^2 + 3^2\) \(2x - y + 3z = 4 + 1 + 9\) \(2x - y + 3z = 14\) \(2x - y + 3z - 14 = 0\).

Answer 1

The Correct Option is B

Step 1: Determine the normal vector of the plane.
Let the origin be \(O(0, 0, 0)\) and the foot of the perpendicular from the origin to the plane be \(P(2, -1, 3)\).
The line segment \(OP\) is perpendicular to the plane. This means that the vector \(\vec{OP}\) serves as the normal vector to the plane.
The coordinates of the normal vector \(\vec{n}\) are given by the coordinates of point \(P\) relative to \(O\):
\[ \vec{n} = \vec{OP} = \langle 2-0, -1-0, 3-0 \rangle = \langle 2, -1, 3 \rangle \] The general equation of a plane is \(Ax + By + Cz + D = 0\), where \(\langle A, B, C \rangle\) is the normal vector.
Thus, the equation of the plane can be written as \(2x - y + 3z + D = 0\). 
Step 2: Use the given point to find the constant \(D\).
Since the foot of the perpendicular \(P(2, -1, 3)\) lies on the plane, its coordinates must satisfy the equation of the plane.
Substitute \(x=2\), \(y=-1\), and \(z=3\) into the plane equation: \[ 2(2) - (-1) + 3(3) + D = 0 \] \[ 4 + 1 + 9 + D = 0 \] \[ 14 + D = 0 \] \[ D = -14 \] Step 3: Write the final equation of the plane.
Substitute the value of \(D\) back into the plane equation: \[ 2x - y + 3z - 14 = 0 \] The final answer is $\boxed{2x - y + 3z - 14 = 0}$.