MHT CET · Maths · Three Dimensional Geometry
The co-ordinates of the foot of the perpendicular drawn from the origin to the plane \(2 x+6 y-3 z=63\) are
- A \((4,2,-4)\)
- B \(\left(\frac{18}{7}, \frac{54}{7}, \frac{-27}{7}\right)\)
- C \(\left(\frac{2}{7}, \frac{6}{7}, \frac{-3}{7}\right)\)
- D \(\left(\frac{9}{7}, \frac{6}{7}, \frac{-3}{7}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{18}{7}, \frac{54}{7}, \frac{-27}{7}\right)\)
Step-by-step Solution
Detailed explanation
The co-ordinates of foot of perpendicular can be obtained by
\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-\left(a_1+b y_1+c_1-d\right)}{a^2+b^2+c^2} \)
\( \Rightarrow \frac{x-0}{2}=\frac{y-0}{6}=\frac{z-0}{-3}=\) \(\frac{-(2 \times 0+6 \times 0-3 \times 0-63)}{2^2+6^2+(-3)^2} \)
\( \Rightarrow \frac{x}{2}=\frac{y}{6}=\frac{z}{-3}=\frac{63}{49} \)
\( \Rightarrow x=\frac{18}{7}, y=\frac{54}{7}, z=\frac{-27}{7} \)
\( \Rightarrow\left(\frac{18}{7}, \frac{54}{7}, \frac{-27}{7}\right)\)
\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-\left(a_1+b y_1+c_1-d\right)}{a^2+b^2+c^2} \)
\( \Rightarrow \frac{x-0}{2}=\frac{y-0}{6}=\frac{z-0}{-3}=\) \(\frac{-(2 \times 0+6 \times 0-3 \times 0-63)}{2^2+6^2+(-3)^2} \)
\( \Rightarrow \frac{x}{2}=\frac{y}{6}=\frac{z}{-3}=\frac{63}{49} \)
\( \Rightarrow x=\frac{18}{7}, y=\frac{54}{7}, z=\frac{-27}{7} \)
\( \Rightarrow\left(\frac{18}{7}, \frac{54}{7}, \frac{-27}{7}\right)\)
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