Question

If $r=\hat{i}+\hat{j}+t(2\hat{i}-\hat{j}+\hat{k})$ and $r=2\hat{i}+\hat{j}-\hat{k}+s(3\hat{i}-5\hat{j}+2\hat{k})$ are the vector equations of two lines $L_1$ and $L_2$, then the shortest distance between them is

• A. $\frac{9}{\sqrt{59}}$
• B. $\frac{10}{\sqrt{59}}$
• C. $\frac{11}{\sqrt{59}}$
• D. 0

Answer 1

Answer: (B)

We have,
$r=\hat{i}+\hat{j}+t(\hat{2}-\hat{j}+\hat{k})$
$r=2\hat{i}+\hat{j}-\hat{k}+s(3\hat{i}-5\hat{j}+2\hat{k})$
Here,
$a_1=\hat{i}+\hat{j}$
$b_1=2\hat{i}-\hat{j}+\hat{k}$
$a_2-2\hat{i}+\hat{j}-\hat{k}$
$b_2=3\hat{i}-5\hat{j}+2\hat{k}$
$a_2-a_1=\hat{i}-\hat{k}$
$b_1\times b_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 1\\ 3& -5& 2<br/><br/>\end{array}\right|=3\hat{i}-\hat{j}-7\hat{k}$
$\left|b_1\times b_2\right|=\sqrt{9+1+49}=\sqrt{59}$
Shortest Distance $=\frac{\left(a_2-a_1\right)\cdot \left(b_1\times b_2\right)}{|b_1\times b_2|}$
$=\frac{(i-\hat{k})\cdot (3\hat{i}-\hat{j}-7\hat{k})}{\sqrt{59}}=\frac{10}{\sqrt{59}}$