The perpendicular distance of the line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$ from the point $P(2,-10,1)$ is:
- (1) 6
- (2) $5\sqrt{2}$
- (3) $3\sqrt{5}$
- (4) $4\sqrt{3}$
Solution:
Let $A$ be the foot of the perpendicular from $P$ to the line.
General point on the line: $A(2\lambda+1, -\lambda-2, 2\lambda-3)$.
Direction ratios of the line: $\vec{n} = \langle 2, -1, 2 \rangle$.
Vector $\vec{PA} = \langle (2\lambda+1)-2, (-\lambda-2)-(-10), (2\lambda-3)-1 \rangle$
$$ \vec{PA} = \langle 2\lambda-1, -\lambda+8, 2\lambda-4 \rangle $$
Since $\vec{PA} \perp \vec{n}$, their dot product is 0:
$$ 2(2\lambda-1) – 1(-\lambda+8) + 2(2\lambda-4) = 0 $$
$$ 4\lambda – 2 + \lambda – 8 + 4\lambda – 8 = 0 $$
$$ 9\lambda – 18 = 0 \Rightarrow \lambda = 2 $$
Coordinates of $A$: $(2(2)+1, -2-2, 2(2)-3) = (5, -4, 1)$.
Distance $PA = \sqrt{(5-2)^2 + (-4+10)^2 + (1-1)^2}$
$$ = \sqrt{3^2 + 6^2 + 0} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} $$
Ans. (3)