KCET · Maths · Three Dimensional Geometry
Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane \( 5 y+8 \)
\( =0 \) :
- A \( \left(0,-\frac{18}{5}, 2\right) \)
- B \( \left(0, \frac{8}{5}, 0\right) \)
- C \( \left(\frac{8}{25}, 0,0\right) \)
- D \( \left(0,-\frac{8}{5}, 0\right) \)
Answer & Solution
Correct Answer
(D) \( \left(0,-\frac{8}{5}, 0\right) \)
Step-by-step Solution
Detailed explanation
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 \( \left(0, \frac{-8}{5}, 0\right) \)
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