The coordinates of the foot perpendicular from the point $(1,0,0)$ to the line $ \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8} ext { are } $

Question

The coordinates of the foot perpendicular from the point $(1,0,0)$ to the line $ \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8} 	ext { are } $

Aicharya · Accepted Answer

Let the equation of $A B$ is $ \frac{x-1}{2}=\frac{y-(-1)}{-3}=\frac{z-(-10)}{8}=k $ Let $L$ be the foot of the perpendicular drawn form $P(1,0,0)$.

$\therefore \quad L=(2 k+1,-3 \mathrm{k}-1,8 k-10)$. Now, direction ratio of $P L=(2 k,-3 k-1,8$ $-10)$ and direction ratio of $A B=(2,-3,8)$ Since, $P L$ is perpendicular to $A B$ $ \therefore \quad 2(2 k)-3(-3 k-1)+8(8 k-10)=0 $ Now, $k=\frac{2(1-1)+(-3)(0+1)+8(0+10)}{(2)^2+(-3)^2+(8)^2}$ $ =\frac{0-3+80}{4+9+64}=\frac{77}{77}=1 $ $\therefore \quad$ Required co-ordinate $ =L=(2+1,-3-1,8-10)=(3,-4,-2) \text {. } $

The coordinates of the foot perpendicular from the point $(1,0,0)$ to the line $…

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