The vector \(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\), and the angle between \(\vec{b}\) and \(\vec{c}\) is \(\displaystyle\frac{2\pi}{3}\). If \(|\vec{a}| = 2\), \(|\vec{b}| = 3\) and \(|\vec{c}| = 4\), then \(|\vec{a}+\vec{b}+\vec{c}|\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\sqrt{17}\)
Since \(\vec{a}\) is perpendicular to \(\vec{b}\) and \(\vec{c}\), \(\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c} = 0\). The remaining scalar product is
\( \displaystyle \vec{b}\cdot\vec{c} = 3\times 4\times\cos\frac{2\pi}{3} = 12\times\left(-\frac{1}{2}\right) = -6 \)
Using \(|\vec{v}|^2 = \vec{v}\cdot\vec{v}\):
\(|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})\)
\( = 4+9+16+2(0-6+0)\)
\( = 29-12 = 17\)
\(|\vec{a}+\vec{b}+\vec{c}| = \sqrt{17}\)
Why the other options are incorrect
Option (1) \(\sqrt{29}\) — Incorrect. It treats all three vectors as mutually perpendicular and leaves out \(2\,\vec{b}\cdot\vec{c} = -12\).
Option (2) \(\sqrt{41}\) — Incorrect. It takes \(\vec{b}\cdot\vec{c} = +6\), which is the value for the angle \(\frac{\pi}{3}\).
Option (3) 5 — Incorrect. \(5^2 = 25\), but \(|\vec{a}+\vec{b}+\vec{c}|^2 = 17\).