Q102 / 150 · MHT CET 2026 (20 April Shift 1) · Physics

The area of parallelogram formed by vectors $\vec{P} = 2\hat{i} - \hat{j} + 5\hat{k}$ and $\vec{Q} = 3\hat{i} - 2\hat{j} + 4\hat{k}$ is

  • A
    $\sqrt{72}$
  • Correct answer
    B
    $\sqrt{86}$
  • C
    $\sqrt{104}$
  • D
    $\sqrt{240}$

Solution

The area of the parallelogram formed by vectors $\vec{P}$ and $\vec{Q}$ is given by $|\vec{P} \times \vec{Q}|$.

$\vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 5 \\ 3 & -2 & 4 \end{vmatrix}$

$= \hat{i}((-1)(4) - (5)(-2)) - \hat{j}((2)(4) - (5)(3)) + \hat{k}((2)(-2) - (-1)(3))$

$= \hat{i}(-4 + 10) - \hat{j}(8 - 15) + \hat{k}(-4 + 3)$

$= 6\hat{i} + 7\hat{j} - \hat{k}$

The magnitude of the cross product is:

$|\vec{P} \times \vec{Q}| = \sqrt{6^2 + 7^2 + (-1)^2}$

$= \sqrt{36 + 49 + 1} = \sqrt{86}$

Answer: $\sqrt{86}$

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

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