Three Dimensional Geometry Question 16
Question 16 - 2024 (30 Jan Shift 1)
Let $(\alpha, \beta, \gamma)$ be the foot of perpendicular from the point $(1,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$. then
$19(\alpha+\beta+\gamma)$ is equal to :
(1) 102
(2) 101
(3) 99
(4) 100
Show Answer
Answer (2)
Solution
$(1,2,3)$
Let foot $P(5 k-3,2 k+1,3 k-4)$
$\mathrm{DR}^{\prime} \mathrm{s} \rightarrow \mathrm{AP}: 5 \mathrm{k}-4,2 \mathrm{k}-1,3 \mathrm{k}-7$
DR’s $\rightarrow$ Line: $5,2,3$
Condition of perpendicular lines $(25 k-20)+(4 k-2)+(9 k-21)=0$
Then $\mathrm{k}=\frac{43}{38}$
Then $19(\alpha+\beta+\gamma)=\mathbf{1 0 1}$
