Let a and b be two unit vectors such that | a+ b|=√3.

Question

Let $\vec a$ and $\vec b$ be two unit vectors such that $|\vec a+\vec b|=\sqrt3$. If $\vec c=\vec a+2\vec b+3(\vec a	imes\vec b)$, then $|2\vec c|$ is equal to:

Accepted Answer

$|\vec a+\vec b|^2=3 \implies 1+1+2\vec a\cdot\vec b=3 \implies \vec a\cdot\vec b=\dfrac12$ (angle $60^{\circ}$), and $|\vec a\times\vec b|=\sin60^{\circ}=\dfrac{\sqrt3}2$.

Since $\vec a\times\vec b$ is perpendicular to both $\vec a$ and $\vec b$:

$$|\vec c|^2 = |\vec a|^2+4|\vec b|^2+9|\vec a\times\vec b|^2+4(\vec a\cdot\vec b) = 1+4+9\cdot\frac34+4\cdot\frac12 = \frac{55}4$$

$$|\vec c|=\frac{\sqrt{55}}2 \implies |2\vec c|=\sqrt{55}$$

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