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

Step 1: Understanding the Question:
We need to find the point on the plane that is closest to the origin. This point is the foot of the perpendicular.
Step 2: Key Formula or Approach:
For a plane $ax + by + cz = d$, the foot of the perpendicular from origin $(0,0,0)$ is:
\[ \left( \frac{ad}{a^2+b^2+c^2}, \frac{bd}{a^2+b^2+c^2}, \frac{cd}{a^2+b^2+c^2} \right) \]
Step 3: Detailed Explanation:
From the plane equation $2x - 3y - 6z = 4$:
$a = 2, b = -3, c = -6$ and $d = 4$.

Calculate $a^2 + b^2 + c^2$:
\[ 2^2 + (-3)^2 + (-6)^2 = 4 + 9 + 36 = 49 \]

Now, calculate the coordinates:
$x = \frac{2 \times 4}{49} = \frac{8}{49}$
$y = \frac{-3 \times 4}{49} = -\frac{12}{49}$
$z = \frac{-6 \times 4}{49} = -\frac{24}{49}$
Step 4: Final Answer:
The coordinates are $\left(\frac{8}{49}, -\frac{12}{49}, -\frac{24}{49}\right)$.