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,
$a_{1} =\hat{ i }+\hat{ j }$
$b_{1} =2 \hat{ i }-\hat{ j }+\hat{ k }$
$a_{2} -2 \hat{ i }+\hat{ j }-\hat{ k }$
$b_{2} =3 \hat{ i }-5 \hat{ j }+2 \hat{ k }$
$a _{2}- a _{1} =\hat{ i }-\hat{ k }$
$b _{1} \times b _{2}=\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\ 2 & -1 & 1 \\ 3 & -5 & 2
\end{vmatrix}=3 \hat{ i }-\hat{ j }-7 \hat{ k }$
$\left| b _{1} \times b _{2}\right|=\sqrt{9+1+49}=\sqrt{59}$
Shortest Distance $=\frac{\left( a _{2}- a _{1}\right) \cdot\left( b _{1} \times b _{2}\right)}{\left| b _{1} \times b _{2}\right|}$
$=\frac{( i -\hat{ k }) \cdot(3 \hat{ i }-\hat{ j }-7 \hat{ k })}{\sqrt{59}}=\frac{10}{\sqrt{59}}$