Question

If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors satisfying \(|\vec{a}|=1, |\vec{b}|=2, |\vec{c}|=1\) and \(\vec{a}\times(\vec{a}\times\vec{b}) + \vec{c} = 0\), then what is the angle between the vectors \(\vec{a}\) and \(\vec{b}\)?

• A. \(30^\circ\)
• B. \(45^\circ\)
• C. \(60^\circ\)
• D. \(75^\circ\)

Hint:

Use \(\vec a\times(\vec a\times\vec b)=(\vec a\cdot\vec b)\vec a - \vec b\) and take magnitudes.

Answer 1

The Correct Option is A

Step 1: Use the vector triple product identity \(\vec{a}\times(\vec{a}\times\vec{b}) = \vec{a}(\vec{a}\cdot\vec{b}) - \vec{b}(\vec{a}\cdot\vec{a})\). Since \(|\vec a|=1\), \(\vec a\cdot\vec a=1\), so \[ \vec{a}\times(\vec{a}\times\vec{b}) = (\vec{a}\cdot\vec{b})\vec{a} - \vec{b} \]

Step 2: Substitute into the given equation: \[ (\vec{a}\cdot\vec{b})\vec{a} - \vec{b} + \vec{c} = 0 \implies \vec{c} = \vec{b} - (\vec{a}\cdot\vec{b})\vec{a} \]

Step 3: Take the magnitude squared of both sides, writing \(k=\vec a\cdot\vec b\): \[ |\vec c|^2 = |\vec b|^2 - 2k(\vec a\cdot\vec b) + k^2|\vec a|^2 = |\vec b|^2 - 2k^2+k^2 = |\vec b|^2-k^2 \] Since \(|\vec c|=1\) and \(|\vec b|=2\): \[ 1 = 4-k^2 \implies k^2=3 \implies k = \vec a\cdot\vec b = \pm\sqrt3 \]

Step 4: Since \(\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta = 2\cos\theta\), we get \[ 2\cos\theta = \pm\sqrt3 \implies \cos\theta = \pm\frac{\sqrt3}{2} \] This gives \(\theta = 30^\circ\) or \(\theta=150^\circ\). Among the given options, the matching value is \(30^\circ\). \[ \boxed{\theta = 30^\circ} \]