Q117 / 180 · KCET 2016 · Mathematics
Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane \( 5 y+8 \)
\( =0 \) :
\( =0 \) :
- A\( \left(0,-\frac{18}{5}, 2\right) \)
- B\( \left(0, \frac{8}{5}, 0\right) \)
- C\( \left(\frac{8}{25}, 0,0\right) \)
- Correct answerD\( \left(0,-\frac{8}{5}, 0\right) \)
Solution
Given equation of plane: \( 5 y+8=0 \)
So, \( \vec{N}=(0,5,0) \)
Equation of line passes through origin and is perpendicular to this plane is given by
\[ \frac{x}{0}=\frac{y}{5}=\frac{z}{0}=\lambda \]
Foot of the perpendicular is found by intersection of line with plane. So,
\[ x=0, y=5 \lambda, z=0 \]
Now, \( 25 \lambda+8=0 \)
\[ \Rightarrow \lambda=\frac{-8}{25} \]
Therefore, point is \( \left(0, \frac{-8}{5}, 0\right) \)
So, \( \vec{N}=(0,5,0) \)
Equation of line passes through origin and is perpendicular to this plane is given by
\[ \frac{x}{0}=\frac{y}{5}=\frac{z}{0}=\lambda \]
Foot of the perpendicular is found by intersection of line with plane. So,
\[ x=0, y=5 \lambda, z=0 \]
Now, \( 25 \lambda+8=0 \)
\[ \Rightarrow \lambda=\frac{-8}{25} \]
Therefore, point is \( \left(0, \frac{-8}{5}, 0\right) \)
Hence, the correct answer is option: 'D'.