Question

The distance between two lines $L_1:x=2-t,y=at,z=1+t,L_2:1+2t,y=3-4t,z=5-2t$ is $\sqrt{\frac{35}{6}}$,find $a$.

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

Given lines

$L_1:x=2-t,y=at,z=1+t$

changing in vector eq

$L_1:\overrightarrow{r}=(2-t)\hat{i}+at\hat{j}+(1+t)\hat{k}$

$L_1:\overrightarrow{r}=2\hat{i}+\hat{k}+t(-\hat{i}+a\hat{j}+\hat{k})$

position vector of above line

$\overrightarrow{a}=2\hat{i}+\hat{k}$

normal vector

$\overrightarrow{b}=-\hat{i}+a\hat{j}+\hat{k}$

$L_2:x=1+2t,y=3-4t,z=5-2t$

changing in vector eq

$L_2:\overrightarrow{r}=(1+2t)\hat{i}+(3-4t)\hat{j}+(5-2t)\hat{k}$

$L_2:\overrightarrow{r}=\hat{i}+3\hat{i}+5\hat{k}+t(2\hat{i}-4\hat{j}-2\hat{k})$

position vector of above line

$\overrightarrow{c}=\hat{i}+3\hat{j}+5\hat{k}$

normal vector

$\overrightarrow{d}=2\hat{i}-4\hat{j}-2\hat{k}$

$\overrightarrow{c}-\overrightarrow{a}=\hat{i}+3\hat{j}+5\hat{k}-(2\hat{i}+\hat{k})$

$\overrightarrow{c}-\overrightarrow{a}=-\hat{i}+3\hat{j}+4\hat{k}$

$\overrightarrow{b}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& a& 1\\ 2& -4& -2\end{array}\right|$

$\overrightarrow{b}\times \overrightarrow{d}=\hat{i}(-2a+4)-\hat{j}\left(0\right)+\hat{k}(4-2a)$

$\overrightarrow{b}\times \overrightarrow{d}=-2(a-2)\hat{i}-2(a-2)\hat{k}$

$distance=\sqrt{\frac{35}{6}}$

$\frac{\left|\right(\overrightarrow{c}-\overrightarrow{a})\cdot (\overrightarrow{b}\times \overrightarrow{d}\left)\right|}{|\overrightarrow{b}\times \overrightarrow{d}|}=\sqrt{\frac{35}{6}}$

$\frac{\left|\right(-\hat{i}+3\hat{j}+4\hat{k})\cdot (-2(a-2)\hat{i}-2(a-2)\hat{k}\left)\right|}{\sqrt{(-2(a-2){)}^2+(-2(a-2){)}^2}}=\sqrt{\frac{35}{6}}$

$\frac{\left|2\right(a-2)-8(a-2\left)\right|}{\sqrt{4(a-2{)}^2+4(a-2{)}^2}}=\sqrt{\frac{35}{6}}$

$\frac{2a-4-8a+16}{\sqrt{4(a-2{)}^2+4(a-2{)}^2}}=\sqrt{\frac{35}{6}}$

$\frac{-6a+12}{\sqrt{8(a-2{)}^2}}=\sqrt{\frac{35}{6}}$

$\frac{-6(a-2)}{\sqrt{8(a-2{)}^2}}=\sqrt{\frac{35}{6}}$

on squaring both sides

$\frac{36(a-2{)}^2}{8(a-2{)}^2}=\frac{35}{6}$

$36\times 6(a-2{)}^2=35\times 8(a-2{)}^2$

$9(a-2{)}^2=7(a-2{)}^2$

$2(a-2{)}^2=0$

$(a-2)=0$

$a=2$