Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be vectors with \(|\vec{a}| = 2\), \(|\vec{b}| = 3\) and \(|\vec{c}| = 6\), such that \(\vec{a}\) is perpendicular to \(\vec{b}+\vec{c}\), \(\vec{b}\) is perpendicular to \(\vec{c}+\vec{a}\) and \(\vec{c}\) is perpendicular to \(\vec{a}+\vec{b}\). Then \(|\vec{a}+\vec{b}+\vec{c}|\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 7
Each perpendicularity gives a zero scalar product:
\(\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c} = 0\)
\(\vec{b}\cdot\vec{c}+\vec{b}\cdot\vec{a} = 0\)
\(\vec{c}\cdot\vec{a}+\vec{c}\cdot\vec{b} = 0\)
Adding the three equations:
\(2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0\)
Now
\(|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})\)
\( = 4+9+36+0 = 49\)
\(|\vec{a}+\vec{b}+\vec{c}| = 7\)