Question

The foot of the perpendicular drawn from the origin to the plane \( x + y + 3z - 4 = 0 \) is

• A. \( \left( \frac{2}{11}, \frac{2}{11}, \frac{9}{11} \right) \)
• B. \( \left( \frac{4}{11}, \frac{4}{11}, \frac{12}{11} \right) \)
• C. \( \left( \frac{1}{7}, \frac{1}{7}, \frac{6}{7} \right) \)
• D. \( \left( \frac{1}{5}, \frac{1}{5}, \frac{3}{5} \right) \)

Hint:

To find the foot of the perpendicular from a point to a plane, use the formula that involves the coordinates of the point and the plane equation coefficients.

Answer 1

The Correct Option is B

Step 1: Formula for the foot of the perpendicular.
The foot of the perpendicular from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \) is given by the formula: \[ \left( x_1 - \frac{A}{A^2 + B^2 + C^2} \cdot D, y_1 - \frac{B}{A^2 + B^2 + C^2} \cdot D, z_1 - \frac{C}{A^2 + B^2 + C^2} \cdot D \right) \] Here, \( (x_1, y_1, z_1) = (0, 0, 0) \) and the equation of the plane is \( x + y + 3z - 4 = 0 \). Thus, \( A = 1, B = 1, C = 3, D = -4 \).
Step 2: Apply the formula.
Substituting the values into the formula, we get: \[ x = \frac{1}{1^2 + 1^2 + 3^2} \cdot 4 = \frac{4}{11}, \quad y = \frac{4}{11}, \quad z = \frac{12}{11} \]
Step 3: Conclusion.
Thus, the foot of the perpendicular is \( \left( \frac{4}{11}, \frac{4}{11}, \frac{12}{11} \right) \), corresponding to option (B).