# If the foot of the perpendicular from point

Question:

If the foot of the perpendicular from point $(4,3,8)$ on the line $\mathrm{L}_{1}: \frac{\mathrm{x}-\mathrm{a}}{l}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-\mathrm{b}}{4}$,

$l \neq 0$ is $(3,5,7)$, then the shortest distance between the line $L_{1}$ and line

$\mathrm{L}_{2}: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-4}{4}=\frac{\mathrm{z}-5}{5}$ is equal to :

1. $\frac{1}{2}$

2. $\frac{1}{\sqrt{6}}$

3. $\sqrt{\frac{2}{3}}$

4. $\frac{1}{\sqrt{3}}$

Correct Option: , 2

Solution:

$(3,5,7)$ satisfy the line $L_{1}$

$\frac{3-\mathrm{a}}{\ell}=\frac{5-2}{3}=\frac{7-\mathrm{b}}{4}$

$\frac{3-a}{\ell}=1 \quad \& \quad \frac{7-b}{4}=1$

$\mathrm{a}+\ell=3 \quad \ldots(1) \quad \& \quad \mathrm{~b}=3$......(2)

$\overrightarrow{\mathrm{v}}_{1}=\langle 4,3,8\rangle-\langle 3,5,7\rangle$

$\overrightarrow{\mathrm{v}}_{1}=<1,-2,1>$

$\overrightarrow{\mathrm{v}}_{2}=<\ell, 3,4>$

$\overrightarrow{\mathrm{v}}_{1} \cdot \overrightarrow{\mathrm{v}}_{2}=0 \quad \Rightarrow \quad \ell-6+4=0 \quad \Rightarrow \quad \ell=2$

$a+\ell=3 \Rightarrow a=1$

$\mathrm{L}_{1}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}$

$\mathrm{L}_{2}: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-4}{4}=\frac{\mathrm{z}-5}{5}$

$\mathrm{A}=\langle 1,2,3\rangle$

$\mathrm{B}=\langle 2,4,5\rangle$

$\overrightarrow{\mathrm{AB}}=\langle 1,2,2\rangle$

$\overrightarrow{\mathrm{p}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{q}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$

Shortest distance $=\left|\frac{\overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right|=\frac{1}{\sqrt{6}}$