Let ( , , ) be the foot of perpendicular from the point (1,2,3) on the line (x+3)/(5)=(y-1…

Question

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 :

Accepted Answer

Let foot $P(5 k-3,2 k+1,3 k-4)$

DR's $$\rightarrow$$ 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 $$19(\alpha+\beta+\gamma)=\mathbf{1 0 1}$$

Solutions · 1