Let \(\vec{a}\) and \(\vec{b}\) be two vectors such that \(|2 \vec{a} + 3\vec{b}| = |3 \vec{a} + \vec{b}|\) and the angle between \(\vec{a}\) and \(\vec{b}\) is 60°. If \(\frac{1}{8}\vec{a}\) is a unit vector, then \(|\vec{b}|\) is equal to
(1) 4
(2) 6
(3) 5
(4) 8
Correct answer: (3) 5
Explanation:
\(|3 \vec{a} + \vec{b}|^2 = |2 \vec{a} + 3\vec{b}|^2\)
\((3 \vec{a} + \vec{b}).(3 \vec{a} + \vec{b})\) = \((2 \vec{a} + 3\vec{b}).(2 \vec{a} + 3\vec{b})\)
\(9\vec{a}. \vec{a} +6 \vec{a}.\vec{b} +\vec{b}. \vec{b}\) = \(4\vec{a}. \vec{a} +12 \vec{a}.\vec{b} +9\vec{b}. \vec{b}\)
\(5|\vec{a}|^2 -6 \vec{a}.\vec{b} = 8| \vec{b}|^2\)
5(8)2 - 6.8\(|\vec{b}|\)cos 60° = \(8|\vec{b}|^2\)
(∵ \(\frac{1}{8}|\vec{a}| = 1\)
⇒ \(|\vec{a}| = 8\))
40 - \(3|\vec{b}| = |\vec{b}|^2\)
⇒ \(|\vec{b}|^2 + 3|\vec{b}|-40\) = 0
\(|\vec{b}|\) = -8, \(|\vec{b}|\) = 5