Jee Main Jan 30 Shift 1 Mathematics - Question2
Question 2
If the foot of perpendicular from $(1,2,3)$ to the line $\frac{x+1}{2}=\frac{y-2}{5}=\frac{z-4}{1}$ is $(\alpha, \beta, \gamma)$ then find $\alpha+\beta+\gamma$
(1) 6
(2) 5.8
(3) 4.8
(4) 5
Show Answer
Answer: (2)
Solution:
$(\alpha-1) \times 2+(\beta-2) \times 5+(\gamma-3) \times 1=0$
$2 \alpha+5 \beta+\gamma-15=0$
Also, $P$ lie on line
$\Rightarrow \alpha+1=2 \lambda$
$\beta-2=5 \lambda$
$$ \begin{aligned} & \gamma-4=\lambda \\ \Rightarrow & 2(2 \lambda-1)+5(5 \lambda+2)+\lambda+4-15=0 \\ \Rightarrow & 4 \lambda+25 \lambda+\lambda-2+10+4-15=0 \\ & 30 \lambda-3=0 \\ \Rightarrow & \lambda=\frac{1}{10} \\ \Rightarrow & \alpha+\beta+\gamma=(2 \lambda-1)+(5 \lambda+2)+(\lambda+4) \\ & 8 \lambda+5=\frac{8}{10}+5=5.8 \end{aligned} $$
BETA
