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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Vector Algebra

NDA 2025 · Mathematics · Vector Algebra

Let p = a - b , q = a + b . If | a | = | b | = 2 and a · b = 2 ,...

Let p⃗=a⃗−b⃗\vec{p} = \vec{a} - \vec{b}, q⃗=a⃗+b⃗\vec{q} = \vec{a} + \vec{b}. If ∣a⃗∣=∣b⃗∣=2|\vec{a}| = |\vec{b}| = 2 and a⃗⋅b⃗=2\vec{a} \cdot \vec{b} = 2, then what is the value of ∣p⃗×q⃗∣|\vec{p} \times \vec{q}|?

Options

  1. A3\sqrt{3}
  2. B6\sqrt{6}
  3. C232\sqrt{3}
  4. D434\sqrt{3}

Correct answer

D. 434\sqrt{3}

Step-by-step solution

p⃗×q⃗=(a⃗−b⃗)×(a⃗+b⃗)\vec{p} \times \vec{q} = (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b})

=a⃗×a⃗+a⃗×b⃗−b⃗×a⃗−b⃗×b⃗= \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}

Since a⃗×a⃗=0⃗\vec{a} \times \vec{a} = \vec{0}, b⃗×b⃗=0⃗\vec{b} \times \vec{b} = \vec{0}, and b⃗×a⃗=−a⃗×b⃗\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}:

p⃗×q⃗=2(a⃗×b⃗)\vec{p} \times \vec{q} = 2(\vec{a} \times \vec{b})

∣p⃗×q⃗∣=2∣a⃗×b⃗∣|\vec{p} \times \vec{q}| = 2|\vec{a} \times \vec{b}|

Using Lagrange's identity:

∣a⃗×b⃗∣2=∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2

Substituting the given values ∣a⃗∣=2|\vec{a}| = 2, ∣b⃗∣=2|\vec{b}| = 2, and a⃗⋅b⃗=2\vec{a} \cdot \vec{b} = 2:

∣a⃗×b⃗∣2=(2)2(2)2−(2)2|\vec{a} \times \vec{b}|^2 = (2)^2 (2)^2 - (2)^2

∣a⃗×b⃗∣2=16−4=12|\vec{a} \times \vec{b}|^2 = 16 - 4 = 12

∣a⃗×b⃗∣=12=23|\vec{a} \times \vec{b}| = \sqrt{12} = 2\sqrt{3}

Therefore, ∣p⃗×q⃗∣=2(23)=43|\vec{p} \times \vec{q}| = 2(2\sqrt{3}) = 4\sqrt{3}

Answer: 434\sqrt{3}
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