Q81 / 180 · KCET 2019 · Mathematics
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\((-3,5,2) \)
- Correct answerC\((-1,4,3) \)
- D\( (1,2,-3) \)
Solution
(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
\[ \begin{array}{l} \therefore \frac{x_{1}-1}{2}=\frac{y_{1}-3}{-1}=\frac{z_{1}-4}{1}=\lambda \\ x_{1}=2 \lambda+1, y_{1}=-\lambda+3, z_{1}=\lambda+4 \\ A=(2 \lambda+1,-\lambda+3, \lambda+4) \text { lies on } 2 x-y+z+3=0 \\ \Rightarrow-4 \lambda+2-\lambda-3-\lambda+4+3=0 \\ \Rightarrow 6 \lambda=-6 \\ \Rightarrow \lambda=-1 \\ \therefore P=(-1,4,3) \end{array} \]
Hence, the correct answer is option: 'C'.