If \( \vec{a},\vec{b},\vec{c} \) are mutually perpendicular vectors of equal magnitude, the angle between \( \vec{a}+\vec{b}+\vec{c} \) and \( \vec{a} \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \cos^{-1}\left(\displaystyle\frac{1}{\sqrt{3}}\right) \)
Let \( |\vec{a}|=|\vec{b}|=|\vec{c}|=k \). Mutually perpendicular means \( \vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=0 \).
\( (\vec{a}+\vec{b}+\vec{c})\cdot\vec{a}=|\vec{a}|^{2}+\vec{b}\cdot\vec{a}+\vec{c}\cdot\vec{a} \)
\( =k^{2} \)
\( |\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) \)
\( =3k^{2} \)
So \( |\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}\,k \), and by the dot product formula for the angle:
\( \displaystyle \cos\theta=\frac{(\vec{a}+\vec{b}+\vec{c})\cdot\vec{a}}{|\vec{a}+\vec{b}+\vec{c}|\,|\vec{a}|} \)
\( \displaystyle =\frac{k^{2}}{\sqrt{3}\,k\cdot k} \)
\( \displaystyle =\frac{1}{\sqrt{3}} \)
Hence \( \theta=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).
Tip: Picture \( \vec{a},\vec{b},\vec{c} \) as the edges \( \hat{\imath},\hat{\jmath},\hat{k} \) of a unit cube: \( \vec{a}+\vec{b}+\vec{c} \) is the body diagonal, which makes the angle \( \cos^{-1}\frac{1}{\sqrt{3}} \) with each edge.