Question

Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors such that \(|\vec{a} + \vec{b}| = \sqrt{3}\). If \(\vec{c} = \vec{a} + 2\vec{b} + (\vec{a} \times \vec{b})\), then \(2|\vec{c}|\) is equal to:

• A. \(\sqrt{51}\)
• B. \(\sqrt{37}\)
• C. \(\sqrt{43}\)
• D. \(\sqrt{55}\)

Hint:

Get \(\vec{a}\cdot\vec{b} = \frac{1}{2}\); \(\vec{a}\times\vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), so add squared magnitudes.

Answer 1

The Correct Option is D

NOTE: the official image shows \(\vec{c} = \vec{a} + 2\vec{b} + (\vec{a}\times\vec{b})\) with no coefficient; with that, \(2|\vec{c}| = \sqrt{31}\), which is not an option. The original JEE Main 2015 question has \(3(\vec{a}\times\vec{b})\), which gives option (4); that version is solved here.
\(|\vec{a}+\vec{b}|^2 = 1 + 1 + 2\vec{a}\cdot\vec{b} = 3\), so \(\vec{a}\cdot\vec{b} = \frac{1}{2}\).
\(|\vec{a}\times\vec{b}|^2 = 1 - \frac{1}{4} = \frac{3}{4}\), and \(\vec{a}\times\vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\).
\(|\vec{a} + 2\vec{b}|^2 = 1 + 4 + 4\cdot\frac{1}{2} = 7\).
With \(\vec{c} = \vec{a} + 2\vec{b} + 3(\vec{a}\times\vec{b})\): \(|\vec{c}|^2 = 7 + 9\cdot\frac{3}{4} = \frac{55}{4}\), so \(2|\vec{c}| = \sqrt{55}\).
Answer: (4) \(\sqrt{55}\)