Q77 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics

If $p$ is the shortest distance between the lines $\dfrac{x+1}{7} = \dfrac{y+1}{-6} = z+1$ and $\vec{r} = (3\hat{i} + 5\hat{j} + 7\hat{k}) + \mu(\hat{i} - 2\hat{j} + \hat{k})$ then $[p]$ is... ,(where $[\,.\,]$ denotes the greatest integer function.)

  • A
    $5$
  • B
    $20$
  • Correct answer
    C
    $10$
  • D
    $8$

Solution

The shortest distance $p$ between two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by $p = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.

For the first line, $\vec{a}_1 = -\hat{i} - \hat{j} - \hat{k}$ and $\vec{b}_1 = 7\hat{i} - 6\hat{j} + \hat{k}$.

For the second line, $\vec{a}_2 = 3\hat{i} + 5\hat{j} + 7\hat{k}$ and $\vec{b}_2 = \hat{i} - 2\hat{j} + \hat{k}$.

$\vec{a}_2 - \vec{a}_1 = 4\hat{i} + 6\hat{j} + 8\hat{k}$

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = -4\hat{i} - 6\hat{j} - 8\hat{k}$

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (4)(-4) + (6)(-6) + (8)(-8) = -16 - 36 - 64 = -116$

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-4)^2 + (-6)^2 + (-8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116}$

$p = \dfrac{|-116|}{\sqrt{116}} = \sqrt{116}$

Since $100 < 116 < 121$, we have $10 < \sqrt{116} < 11$.

Therefore, $[p] = 10$.

Answer: $10$

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

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