If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are three mutually perpendicular unit vectors and $\vec{d}$ is a unit vector which makes equal angles with $\vec{a}$, $\vec{b}$ and $\vec{c}$, then $|\vec{a}+\vec{b}+\vec{c}+\vec{d}|^2$ is equal to:
Step-by-Step Solution
Key Concept: $\vec{d}$ can make equal angles in two opposite directions, giving $\cos\theta=\pm\frac{1}{\sqrt{3}}$ from the unit vector constraint.
Since $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually perpendicular unit vectors, we set up coordinates: $\vec{a}=(1,0,0)$, $\vec{b}=(0,1,0)$, $\vec{c}=(0,0,1)$. Since $\vec{d}$ is a unit vector making equal angles $\theta$ with each, we have $\vec{d}=(\cos\theta, \cos\theta, \cos\theta)$. From $|\vec{d}|=1$: $3\cos^2\theta=1$, so $\cos\theta=\pm\frac{1}{\sqrt{3}}$. Now $\vec{a}+\vec{b}+\vec{c}+\vec{d}=(1+\cos\theta, 1+\cos\theta, 1+\cos\theta)$. Therefore $|\vec{a}+\vec{b}+\vec{c}+\vec{d}|^2=3(1+\cos\theta)^2=3(1+2\cos\theta+\cos^2\theta)=3(1+2\cos\theta+\frac{1}{3})=4+6\cos\theta$. With $\cos\theta=\frac{1}{\sqrt{3}}$: we get $4+6/\sqrt{3}=4+2\sqrt{3}$; with $\cos\theta=-\frac{1}{\sqrt{3}}$: we get $4-2\sqrt{3}$.
Correct Answer: 3,4