MHT CET · Maths · Three Dimensional Geometry
The co-ordinates of the foot of the perpendicular drawn form the origin to the plane \(3 x+2 y+6 z=56\) is
- A \(\left(\frac{48}{7}, \frac{24}{7}, \frac{16}{7}\right)\)
- B \(\left(\frac{24}{7}, \frac{48}{7}, \frac{16}{7}\right)\)
- C \(\left(\frac{16}{7}, \frac{24}{7}, \frac{48}{7}\right)\)
- D \(\left(\frac{24}{7}, \frac{16}{7}, \frac{48}{7}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{24}{7}, \frac{16}{7}, \frac{48}{7}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{x-0}{3}=\frac{y-0}{2}=\frac{z-0}{6}=\) \(\frac{-(3 \times 0+2 \times 0+1 \times 0-56)}{3^2+2^2+6^2}\)
\(\Rightarrow \frac{x}{3}=\frac{y}{2}=\frac{z}{6}=\frac{56}{49} \Rightarrow x=\frac{24}{7}, y=\frac{16}{7},\) \(\mathrm{z}=\frac{48}{7}\)
\(\Rightarrow\left(\frac{24}{7}, \frac{16}{7}, \frac{48}{7}\right)\)
\(\Rightarrow \frac{x}{3}=\frac{y}{2}=\frac{z}{6}=\frac{56}{49} \Rightarrow x=\frac{24}{7}, y=\frac{16}{7},\) \(\mathrm{z}=\frac{48}{7}\)
\(\Rightarrow\left(\frac{24}{7}, \frac{16}{7}, \frac{48}{7}\right)\)
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