If the foot of the perpendicular from point $(4,3,8)$ on the line

Question:

If the foot of the perpendicular from point $(4,3,8)$ on the line

$L_{1}: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4}, l \neq 0$ is $(3,5,7)$, then the shortest distance

between the line $\mathrm{L}_{1}$ and line

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

 

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

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

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

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


Correct Option: , 2

Solution:

$(3,5,7)$ satisfy the line $\mathrm{L}_{1}$

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

$\frac{3-\mathrm{a}}{\ell}=1 \quad \&$

$\frac{7-b}{4}=1$

$a+\ell=3 \quad \ldots(1)$

$\& \quad b=3 \ldots .(2)$

$\overrightarrow{\mathbf{v}}_{1}=<4,3,8>-<3,5,7>$

$\overrightarrow{\mathbf{v}}_{1}=\langle 1,-2,1\rangle$

$\overrightarrow{\mathrm{v}}_{2}=\langle\ell, 3,4\rangle$

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

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

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

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

$\mathrm{A}=<1,2,3>$

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

$\overrightarrow{\mathrm{AB}}=<1,2,2>$

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

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

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

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

 

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