Q75 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics

Let $\bar{a} = \hat{i} + \hat{j} + \hat{k}$, $\bar{b} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\bar{c} = 3\hat{i} - 2\hat{k}$. If a vector $\bar{p}$ satisfies the conditions $\bar{p} \cdot \bar{c} = 0$ and $\bar{p} \times \bar{a} = \bar{b} \times \bar{a}$, then the value of $|\bar{p}| = $...

  • A
    $\sqrt{13}$
  • B
    $\sqrt{14}$
  • Correct answer
    C
    $\sqrt{17}$
  • D
    $\sqrt{19}$

Solution

Given $\bar{p} \times \bar{a} = \bar{b} \times \bar{a}$

$\Rightarrow (\bar{p} - \bar{b}) \times \bar{a} = \bar{0}$

$\Rightarrow \bar{p} - \bar{b} = \lambda \bar{a}$

$\Rightarrow \bar{p} = \bar{b} + \lambda \bar{a}$

Also given $\bar{p} \cdot \bar{c} = 0$

$\Rightarrow (\bar{b} + \lambda \bar{a}) \cdot \bar{c} = 0$

$\Rightarrow \bar{b} \cdot \bar{c} + \lambda (\bar{a} \cdot \bar{c}) = 0$

Calculating the dot products:

$\bar{b} \cdot \bar{c} = (\hat{i} - 3\hat{j} + 2\hat{k}) \cdot (3\hat{i} - 2\hat{k}) = 3 - 4 = -1$

$\bar{a} \cdot \bar{c} = (\hat{i} + \hat{j} + \hat{k}) \cdot (3\hat{i} - 2\hat{k}) = 3 - 2 = 1$

Substituting these values:

$-1 + \lambda(1) = 0 \Rightarrow \lambda = 1$

Therefore, $\bar{p} = \bar{b} + \bar{a}$

$\bar{p} = (\hat{i} - 3\hat{j} + 2\hat{k}) + (\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} - 2\hat{j} + 3\hat{k}$

Magnitude of $\bar{p}$:

$|\bar{p}| = \sqrt{2^2 + (-2)^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$

Answer: $\sqrt{17}$

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

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