Q100 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
The vector equation of two lines are
$\begin{aligned}
& \vec{r}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-2 t) \hat{k} \\
& \vec{r}=(s+1) \hat{\imath}+(2 s-1) \hat{\jmath}-(2 s+1) \hat{k}
\end{aligned}$
Then the shortest distance between them is
- A$\dfrac{8}{29}$
- B$\dfrac{4}{\sqrt{29}}$
- Correct answerC$\dfrac{8}{\sqrt{29}}$
- D$\dfrac{4}{29}$
Solution
The first line is $\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + t(-\hat{i} + \hat{j} - 2\hat{k})$. Let $\vec{a}_1 = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b}_1 = -\hat{i} + \hat{j} - 2\hat{k}$.
The second line is $\vec{r} = (\hat{i} - \hat{j} - \hat{k}) + s(\hat{i} + 2\hat{j} - 2\hat{k})$. Let $\vec{a}_2 = \hat{i} - \hat{j} - \hat{k}$ and $\vec{b}_2 = \hat{i} + 2\hat{j} - 2\hat{k}$.
The shortest distance $d$ between two lines $\vec{r} = \vec{a}_1 + t\vec{b}_1$ and $\vec{r} = \vec{a}_2 + s\vec{b}_2$ is given by $d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.
Calculate $\vec{a}_2 - \vec{a}_1 = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = 0\hat{i} + \hat{j} - 4\hat{k}$.
Calculate $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1) = 2\hat{i} - 4\hat{j} - 3\hat{k}$.
The magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$.
The dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (0\hat{i} + \hat{j} - 4\hat{k}) \cdot (2\hat{i} - 4\hat{j} - 3\hat{k}) = 0 - 4 + 12 = 8$.
Therefore, $d = \dfrac{|8|}{\sqrt{29}} = \dfrac{8}{\sqrt{29}}$.
Answer: $\dfrac{8}{\sqrt{29}}$
The second line is $\vec{r} = (\hat{i} - \hat{j} - \hat{k}) + s(\hat{i} + 2\hat{j} - 2\hat{k})$. Let $\vec{a}_2 = \hat{i} - \hat{j} - \hat{k}$ and $\vec{b}_2 = \hat{i} + 2\hat{j} - 2\hat{k}$.
The shortest distance $d$ between two lines $\vec{r} = \vec{a}_1 + t\vec{b}_1$ and $\vec{r} = \vec{a}_2 + s\vec{b}_2$ is given by $d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.
Calculate $\vec{a}_2 - \vec{a}_1 = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = 0\hat{i} + \hat{j} - 4\hat{k}$.
Calculate $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1) = 2\hat{i} - 4\hat{j} - 3\hat{k}$.
The magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$.
The dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (0\hat{i} + \hat{j} - 4\hat{k}) \cdot (2\hat{i} - 4\hat{j} - 3\hat{k}) = 0 - 4 + 12 = 8$.
Therefore, $d = \dfrac{|8|}{\sqrt{29}} = \dfrac{8}{\sqrt{29}}$.
Answer: $\dfrac{8}{\sqrt{29}}$
Hence, the correct answer is option: 'C'.