Q216 / 270 · NDA 2025 (Phase 2) · Mathematics

Let $\vec{p} = \vec{a} - \vec{b}$, $\vec{q} = \vec{a} + \vec{b}$. If $|\vec{a}| = |\vec{b}| = 2$ and $\vec{a} \cdot \vec{b} = 2$, then what is the value of $|\vec{p} \times \vec{q}|$?

  • A
    $\sqrt{3}$
  • B
    $\sqrt{6}$
  • C
    $2\sqrt{3}$
  • Correct answer
    D
    $4\sqrt{3}$

Solution

$\vec{p} \times \vec{q} = (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b})$

$= \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}$

Since $\vec{a} \times \vec{a} = \vec{0}$, $\vec{b} \times \vec{b} = \vec{0}$, and $\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}$:

$\vec{p} \times \vec{q} = 2(\vec{a} \times \vec{b})$

$|\vec{p} \times \vec{q}| = 2|\vec{a} \times \vec{b}|$

Using Lagrange's identity:

$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$

Substituting the given values $|\vec{a}| = 2$, $|\vec{b}| = 2$, and $\vec{a} \cdot \vec{b} = 2$:

$|\vec{a} \times \vec{b}|^2 = (2)^2 (2)^2 - (2)^2$

$|\vec{a} \times \vec{b}|^2 = 16 - 4 = 12$

$|\vec{a} \times \vec{b}| = \sqrt{12} = 2\sqrt{3}$

Therefore, $|\vec{p} \times \vec{q}| = 2(2\sqrt{3}) = 4\sqrt{3}$

Answer: $4\sqrt{3}$

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

Relevant Questions