Question

The shortest distance between the lines
$\frac{x-7}{3} = \frac{y+4}{-16} = \frac{z-6}{7}$
and $\frac{x-10}{3} = \frac{y-30}{8} = \frac{4-z}{5}$ is

• A. $\frac{234}{7} $ units
• B. $\frac{288}{21} $ units
• C. $\frac{221}{3} $ units
• D. $\frac{234}{21} $ units

Answer 1

Answer: (B)

Given, lines are $\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6 }{7}$
and $\frac{x-10}{3}=\frac{y-30}{8}=\frac{4-z}{5}$
The vector form of given equations are
$ r =7 \hat{ i }-4 \hat{ j }+6 \hat{ k }+\lambda(3 \hat{ i }-16 \hat{ j }+7 \hat{ k }) $
and $r =10 \hat{ i }+30 \hat{ j }+4 \hat{ k }+\mu(3 \hat{ i }+8 \hat{ j }-5 \hat{ k }) $
On comparing these equations with
$r = a _{1}+\lambda b _{1}$ and $r = a _{2}+\mu b _{2}$, we get
$a _{1}=7 \hat{ i }-4 \hat{ j }+6 \hat{ k }$
$a _{2}=10 \hat{ i }+30 \hat{ j }+4 \hat{ k }$
$b _{1}=3 \hat{ i }-16 \hat{ j }+7 \hat{ k }$
and $b _{2}=3 \hat{ i }+8 \hat{ j }-5 \hat{ k }$
$b _{1} \times b _{2}=\begin{vmatrix}\hat{ i } & \hat{ j } & \mathbf { k } \\ 3 & -16 & 7 \\ 3 & 8 & -5\end{vmatrix}$
$=(80-56) \hat{ i }+(21+15) \hat{ j }+(24+48) \hat{ k } $
$=24 \hat{ i }+36 \hat{ j }+72 \hat{ k } $
Then $\mid b _{1} \times b _{2} \mid=\sqrt{(24)^{2}+(36)^{2}+(72)^{2}} $
$=\sqrt{576+1296+5184}$
$=\sqrt{7056}=84 $
$ \therefore $ Shortest distance$=\left|\frac{\left( a _{2}- a _{1}\right) \cdot\left( b _{1} \times b _{2}\right)}{\left| b _{1} \times b _{2}\right|}\right| $
$=\left|\frac{(3 \hat{ i }+34 \hat{ j }-2 \hat{ k }) \cdot(24 \hat{ i }+36 \hat{ j }+72 \hat{ k })}{84}\right| $
$=\left|\frac{72+1224-144}{84}\right|=\left|\frac{1152}{84}\right|=\frac{288}{21}$ units