Q77 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
The shortest distance between the lines $\vec{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(\hat{i} + 4\hat{j} - 5\hat{k})$ is...
- A$\dfrac{1}{\sqrt{2}}$
- B$\dfrac{1}{2}$
- Correct answerC$\dfrac{1}{\sqrt{3}}$
- D$\dfrac{1}{3}$
Solution
The given lines are $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$, where:
$\vec{a}_1 = 4\hat{i} - \hat{j}$
$\vec{b}_1 = \hat{i} + 2\hat{j} - 3\hat{k}$
$\vec{a}_2 = \hat{i} - \hat{j} + 2\hat{k}$
$\vec{b}_2 = \hat{i} + 4\hat{j} - 5\hat{k}$
The shortest distance $d$ between two skew lines is given by:
$d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$
First, find $\vec{a}_2 - \vec{a}_1$:
$\vec{a}_2 - \vec{a}_1 = (\hat{i} - \hat{j} + 2\hat{k}) - (4\hat{i} - \hat{j}) = -3\hat{i} + 2\hat{k}$
Next, find the cross product $\vec{b}_1 \times \vec{b}_2$:
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix}$
$\vec{b}_1 \times \vec{b}_2 = \hat{i}(-10 + 12) - \hat{j}(-5 + 3) + \hat{k}(4 - 2) = 2\hat{i} + 2\hat{j} + 2\hat{k}$
Find the magnitude of $\vec{b}_1 \times \vec{b}_2$:
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}$
Now, find the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:
$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-3)(2) + (0)(2) + (2)(2) = -6 + 4 = -2$
Finally, calculate the shortest distance $d$:
$d = \dfrac{|-2|}{2\sqrt{3}} = \dfrac{2}{2\sqrt{3}} = \dfrac{1}{\sqrt{3}}$
Answer: $\dfrac{1}{\sqrt{3}}$
$\vec{a}_1 = 4\hat{i} - \hat{j}$
$\vec{b}_1 = \hat{i} + 2\hat{j} - 3\hat{k}$
$\vec{a}_2 = \hat{i} - \hat{j} + 2\hat{k}$
$\vec{b}_2 = \hat{i} + 4\hat{j} - 5\hat{k}$
The shortest distance $d$ between two skew lines is given by:
$d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$
First, find $\vec{a}_2 - \vec{a}_1$:
$\vec{a}_2 - \vec{a}_1 = (\hat{i} - \hat{j} + 2\hat{k}) - (4\hat{i} - \hat{j}) = -3\hat{i} + 2\hat{k}$
Next, find the cross product $\vec{b}_1 \times \vec{b}_2$:
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix}$
$\vec{b}_1 \times \vec{b}_2 = \hat{i}(-10 + 12) - \hat{j}(-5 + 3) + \hat{k}(4 - 2) = 2\hat{i} + 2\hat{j} + 2\hat{k}$
Find the magnitude of $\vec{b}_1 \times \vec{b}_2$:
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}$
Now, find the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:
$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-3)(2) + (0)(2) + (2)(2) = -6 + 4 = -2$
Finally, calculate the shortest distance $d$:
$d = \dfrac{|-2|}{2\sqrt{3}} = \dfrac{2}{2\sqrt{3}} = \dfrac{1}{\sqrt{3}}$
Answer: $\dfrac{1}{\sqrt{3}}$
Hence, the correct answer is option: 'C'.