Question

Find the shortest distance between the following lines :

$\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (2\hat{i}+3\hat{j}+4\hat{k})$ and $\overrightarrow{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu (4\hat{i}+6\hat{j}+8\hat{k})$

OR

Find the equation of the plane passing through the line of intersection of the planes 2x + y - z = 3 and 5x - 3y + 4z + 9 = 0 and is parallel to the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{5-z}{-5}.$

Answer 1

Let $L_1:\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (2\hat{i}+3\hat{j}+4\hat{k})and{L}_2:\overrightarrow{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu (4\hat{i}+6\hat{j}+8\hat{k})$

Here ${\overrightarrow{a}}_1=\hat{i}+2\hat{j}+3\hat{k},{\overrightarrow{b}}_1=2\hat{i}+3\hat{j}+4\hat{k};{\overrightarrow{a}}_2=2\hat{i}+4\hat{j}+5\hat{k},{\overrightarrow{b}}_2=4\hat{i}+6\hat{j}+8\hat{k}$

Note that, ${\overrightarrow{b}}_2=2{\overrightarrow{b}}_1$ that implies ${\overrightarrow{b}}_1||{\overrightarrow{b}}_2$ so, the lines are parallel.

Therefore, S.D. between two parallel lines, $d=\frac{\left|\begin{array}{c}({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)\times \overrightarrow{b}\end{array}\right|}{\left|\begin{array}{c}\overrightarrow{b}\end{array}\right|}$

$\Rightarrow d=\frac{\left|\begin{array}{c}(\hat{i}+2\hat{j}+2\hat{k})\times (2\hat{i}+3\hat{j}+4\hat{k})\end{array}\right|}{\left|\begin{array}{c}2\hat{i}+3\hat{j}+4\hat{k}\end{array}\right|}=\frac{\left|\begin{array}{c}2\hat{i}-\hat{k}\end{array}\right|}{\sqrt{4+9+16}}=\frac{\sqrt{5}}{\sqrt{29}}=\sqrt{\frac{5}{29}}$ units.

OR

Plane passing through the intersection of the planes 2x + y - z = 3 and 5x - 3y + 4z + 9 = 0 is : $\pi :2x+y-z-3+\lambda (5x-3y+4z+9)=0\Rightarrow \pi :x(2+5\lambda )+y(1-3\lambda )+z(-1+4\lambda )-3+9\lambda =0$

Since plane $\pi$ is parallel to the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{5-z}{5}$ with d.r's 2, 4, 5 so, the normal to the plane will be perpendicular to the line.

Therefore, $2(2+5\lambda )+4(1-3\lambda )+5(-1+4\lambda )=0\Rightarrow 10\lambda -12\lambda +20\lambda +4+4-5=0$

$\Rightarrow 18\lambda =-3\therefore \lambda =-\frac{1}{6}$

Therefore the required plane is $\pi :x(2-\frac{5}{6})+y(1+\frac{1}{2})+z(-1-\frac{2}{3})-3-\frac{3}{2}=0$

That is 7x + 9y - 10z - 27 = 0.