Q96 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics

The point having position vector $4\hat{i} - 11\hat{j} + 2\hat{k}$ lies on the line

  • A
    $\bar{r} = (6\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(\hat{i} + 7\hat{j} + 3\hat{k})$
  • B
    $\bar{r} = (6\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(2\hat{i} + \hat{j} + 3\hat{k})$
  • C
    $\bar{r} = (6\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + \hat{k})$
  • Correct answer
    D
    $\bar{r} = (6\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(2\hat{i} + 7\hat{j} + 3\hat{k})$

Solution

Let the position vector of the given point be $\vec{p} = 4\hat{i} - 11\hat{j} + 2\hat{k}$.

All the given lines pass through the point with position vector $\vec{a} = 6\hat{i} - 4\hat{j} + 5\hat{k}$.

For the point to lie on the line $\vec{r} = \vec{a} + \lambda \vec{b}$, the vector $\vec{p} - \vec{a}$ must be a scalar multiple of the direction vector $\vec{b}$.

Calculating $\vec{p} - \vec{a}$:

$\vec{p} - \vec{a} = (4\hat{i} - 11\hat{j} + 2\hat{k}) - (6\hat{i} - 4\hat{j} + 5\hat{k})$

$\vec{p} - \vec{a} = -2\hat{i} - 7\hat{j} - 3\hat{k} = -1(2\hat{i} + 7\hat{j} + 3\hat{k})$

This shows that $\vec{p} - \vec{a}$ is a scalar multiple of $2\hat{i} + 7\hat{j} + 3\hat{k}$.

Thus, the point lies on the line $\vec{r} = (6\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(2\hat{i} + 7\hat{j} + 3\hat{k})$ with $\lambda = -1$.

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

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