Q72 / 160 · TS EAMCET 2019 (03 May Shift 2) · Mathematics

If $\vec{r} = \hat{i} + \hat{j} + t (2 \hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2 \hat{i} - \hat{j} - \hat{k} + t (3 \hat{i} - 5 \hat{j} + 2 \hat{k})$ are the vector equations of two lines $L_{1}$ and $L_{2}$ then the shortest distance between them is

A
$\frac{9}{\sqrt{59}}$
Correct answer
B
$\frac{10}{\sqrt{59}}$
C
$\frac{11}{\sqrt{59}}$
D
$0$

Solution

Here, ${\vec{a}}_{1} = \hat{i} + \hat{j} & {\vec{b}}_{1} = 2 \hat{i} - \hat{j} + \hat{k}$ and ${\vec{a}}_{2} = 2 \hat{i} + \hat{j} - \hat{k} & {\vec{b}}_{2} = 3 \hat{i} - 5 \hat{j} + 2 \hat{k}$ .

$\Rightarrow {\vec{a}}_{2} - {\vec{a}}_{2} = \hat{i} - \hat{k} & {\vec{b}}_{1} \times {\vec{b}}_{2} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 1 & 1 \\ 3 & - 5 & 2 \end{matrix}| = 3 \hat{i} - \hat{j} - 7 \hat{k}$

The shortest distance between lines $L_{1}$ and $L_{2} = |\frac{({\vec{a}}_{2} - {\vec{a}}_{2}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2})}{{\vec{b}}_{1} \times {\vec{b}}_{2}}|$

$= |\frac{(\hat{i} - \hat{k}) \cdot (3 \hat{i} - \hat{j} - 7 \hat{k})}{3 \hat{i} - \hat{j} - 7 \hat{k}}|$

$= \frac{10}{\sqrt{59}}$

Hence, the correct answer is option: 'B'.

If r→=i^+j^+t(2i^-j^+k^) and r→=2i^-j^-k^+t(3i^-5j^+2k^) are the vector equations of two lines L1 and L2 then the shortest distance between them is

Solution

Relevant Questions

If $\vec{r} = \hat{i} + \hat{j} + t (2 \hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2 \hat{i} - \hat{j} - \hat{k} + t (3 \hat{i} - 5 \hat{j} + 2 \hat{k})$ are the vector equations of two lines $L_{1}$ and $L_{2}$ then the shortest distance between them is