For what value of the angle between the vectors \vec{a} and \vec{b} is the quantity |\vec{a}\times\vec{b}|+\sqrt{3}|\vec{a}\cdot\vec{b}| <strong>MAXIMUM</strong>?

  1. A. 0^{\circ}
  2. B. 30^{\circ}
  3. C. 45^{\circ}
  4. D. 60^{\circ}

Correct Answer: B. 30^{\circ}

Explanation

The expression can be written as |\vec{a}||\vec{b}|\sin\theta + \sqrt{3}|\vec{a}||\vec{b}|\cos\theta = |\vec{a}||\vec{b}|(\sin\theta + \sqrt{3}\cos\theta). We maximize f(\theta) = \sin\theta + \sqrt{3}\cos\theta = 2(\frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta) = 2\sin(\theta + 60^{\circ}). The maximum occurs when \theta + 60^{\circ} = 90^{\circ}, which yields \theta = 30^{\circ}.

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