Question

If the foot of the perpendicular from point $(4,3,8)$ on the line $L_1:\frac{[x–a]}{l}=\frac{[y–2]}{3}=\frac{[z–b]}{4}$, $l\ne 0$ is $(3,5,7)$, then the shortest distance between the line$L_1$ and line $L_2:\frac{[x–2]}{3}=\frac{[y–4]}{4}=\frac{[z–5]}{5}$ is equal to:

• A. $\frac{\sqrt{2}}{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{6}}$

Answer 1

The correct option is D

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

Explanation for the correct option:

Step 1: Solve for the equations of the lines

Given that the foot of the perpendicular from point $(4,3,8)$ on the line $L_1:\frac{[x–a]}{l}=\frac{[y–2]}{3}=\frac{[z–b]}{4}$ is $(3,5,7)$,

$(3,5,7)$ lies on the line $L_1$. Hence, its co-ordinates satisfy the equation of the line.

$\Rightarrow$$\frac{[3–a]}{l}=\frac{[5–2]}{3}=\frac{[7–b]}{4}$

$\Rightarrow$$\frac{[3–a]}{l}=1=\frac{[7–b]}{4}$

$\Rightarrow$$\frac{3-a}{l}=1$ and $\frac{7-b}{4}=1$

$\Rightarrow$ $a+l=3$ and $b=3$ $...\left(i\right)$

Let $A=(4,3,8)$ and $B=(3,5,7)$

$\Rightarrow$$\stackrel{\rightharpoonup }{AB}=\left(4-3\right)\hat{i}+\left(3-5\right)\hat{j}+\left(8-7\right)\hat{k}$

$\Rightarrow$$\stackrel{\rightharpoonup }{AB}=\hat{i}-2\hat{j}+\hat{k}$

For a line $\frac{[x–x_1]}{a_1}=\frac{[y–y_1]}{b_1}=\frac{[z–z_1]}{c_1}$, $\left(a_1,b_1,c_1\right)$ are the direction ratios.

The direction ratios of $L_1$ are $\left(l,3,4\right)$

$\stackrel{\rightharpoonup }{AB}$ is perpendicular to line $L_1$

$\Rightarrow$$\stackrel{\rightharpoonup }{AB}.\stackrel{\rightharpoonup }{L_1}=0$

$\Rightarrow$$1\times l-2\times 3+1\times 4=0$

$\Rightarrow$$l=2$

$\Rightarrow$$a=1$ …[From(i)]

Hence $L_1:\frac{[x–1]}{2}=\frac{[y–2]}{3}=\frac{[z–3]}{4}$ and $L_2:\frac{[x–2]}{3}=\frac{[y–4]}{4}=\frac{[z–5]}{5}$ are the equations of two skew lines.

Step 2: Solve for the required shortest distance

The shortest distance between two skew lines $\frac{[x–x_1]}{a_1}=\frac{[y–y_1]}{b_1}=\frac{[z–z_1]}{c_1}$ and $\frac{[x–x_2]}{a_2}=\frac{[y–y_2]}{b_2}=\frac{[z–z_2]}{c_2}$ is given as

$d=\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{{\left(b_1{c}_2-b_2{c}_1\right)}^2+{\left(c_1{a}_2-c_2{a}_1\right)}^2+{\left(a_1{b}_2-a_2{b}_1\right)}^2}}$

Substituting all the required values we get

$\Rightarrow$$d=\frac{\left|\begin{array}{ccc}2-1& 4-2& 5-3\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|}{\sqrt{{\left(15-16\right)}^2+{\left(12-10\right)}^2+{\left(8-9\right)}^2}}$

$\Rightarrow$$d=\frac{1\left(15-16\right)-2\left(10-12\right)+2\left(8-9\right)}{\sqrt{1+4+1}}$

$\Rightarrow$$d=\frac{1}{\sqrt{6}}$

Thus the shortest distance between the lines $L_1$ and $L_2$ is $\frac{1}{\sqrt{6}}units$ .

Hence the correct option is option(D)