Question

The coordinates of the foot of the perpendicular from the origin to the plane $2x - 3y - 6z = 4$ are

• A. $\left(\frac{8}{49}, -\frac{12}{49}, -\frac{24}{49}\right)$
• B. $\left(\frac{2}{7}, -\frac{3}{7}, -\frac{6}{7}\right)$
• C. $\left(\frac{8}{7}, -\frac{12}{7}, -\frac{24}{7}\right)$
• D. $\left(\frac{4}{49}, -\frac{6}{49}, -\frac{12}{49}\right)$

Hint:

For a plane \(ax+by+cz=d\), the foot of the perpendicular from the origin can be quickly found using \[ \left(\frac{ad}{a^2+b^2+c^2},\frac{bd}{a^2+b^2+c^2},\frac{cd}{a^2+b^2+c^2}\right) \] This shortcut saves time in coordinate geometry and vector problems.

Answer 1

The Correct Option is A

Concept:
The point on a plane that is closest to the origin is the foot of the perpendicular drawn from the origin to that plane. For a plane of the form: \[ ax + by + cz = d \] the coordinates of the foot of the perpendicular from the origin \((0,0,0)\) are given by: \[ \left(\frac{ad}{a^2+b^2+c^2},\frac{bd}{a^2+b^2+c^2},\frac{cd}{a^2+b^2+c^2}\right) \]
Step 1: {Identify the coefficients from the plane equation.} Given plane: \[ 2x - 3y - 6z = 4 \] Thus, \[ a=2,\quad b=-3,\quad c=-6,\quad d=4 \]
Step 2: {Compute \(a^2+b^2+c^2\).} \[ a^2+b^2+c^2 =2^2+(-3)^2+(-6)^2 \] \[ =4+9+36=49 \]
Step 3: {Substitute into the formula.} \[ x=\frac{2\times4}{49}=\frac{8}{49} \] \[ y=\frac{-3\times4}{49}=-\frac{12}{49} \] \[ z=\frac{-6\times4}{49}=-\frac{24}{49} \]
Step 4: {Write the coordinates of the point.} \[ \left(\frac{8}{49}, -\frac{12}{49}, -\frac{24}{49}\right) \]
Step 5: {Conclusion.} Hence, the coordinates of the foot of the perpendicular from the origin to the given plane are: \[ \left(\frac{8}{49}, -\frac{12}{49}, -\frac{24}{49}\right) \]