KCET · Maths · Three Dimensional Geometry
Foot of the perpendicular drawn from the point \( (1,3,4) \) to the plane \( 2 x-y+z+3=0 \) is
- A \( (0,-4,-7) \)
- B \((-3,5,2) \)
- C \((-1,4,3) \)
- D \( (1,2,-3) \)
Answer & Solution
Correct Answer
(C) \((-1,4,3) \)
Step-by-step Solution
Detailed explanation
(C)
The dr's of PA are \( x_{1}-1, y_{1}-3, z_{1}-4 \)
The dr's \( n \) are \( 2,-1,1 \)
The dr's of PA and n are parallel
\[
\begin{array}{l}
\therefore \frac{x_{1}-1}{2}=\frac{y_{1}-3}{-1}=\frac{z_{1}-4}{1}=\lambda \\
x_{1}=2 \lambda+1, y_{1}=-\lambda+3, z_{1}=\lambda+4 \\
A=(2 \lambda+1,-\lambda+3, \lambda+4) \text { lies on } 2 x-y+z+3=0 \\
\Rightarrow-4 \lambda+2-\lambda-3-\lambda+4+3=0 \\
\Rightarrow 6 \lambda=-6 \\
\Rightarrow \lambda=-1 \\
\therefore P=(-1,4,3)
\end{array}
\]
The dr's of PA are \( x_{1}-1, y_{1}-3, z_{1}-4 \)
The dr's \( n \) are \( 2,-1,1 \)
The dr's of PA and n are parallel
\[
\begin{array}{l}
\therefore \frac{x_{1}-1}{2}=\frac{y_{1}-3}{-1}=\frac{z_{1}-4}{1}=\lambda \\
x_{1}=2 \lambda+1, y_{1}=-\lambda+3, z_{1}=\lambda+4 \\
A=(2 \lambda+1,-\lambda+3, \lambda+4) \text { lies on } 2 x-y+z+3=0 \\
\Rightarrow-4 \lambda+2-\lambda-3-\lambda+4+3=0 \\
\Rightarrow 6 \lambda=-6 \\
\Rightarrow \lambda=-1 \\
\therefore P=(-1,4,3)
\end{array}
\]
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