Question

Let \(\overset{̄}{a},\overset{̄}{b},\overset{̄}{c}\) be three vectors having magnitudes 1, 1 and 2 respectively. If \(\overset{̄}{a}\times (\overset{̄}{a}\times \overset{̄}{c})+\overset{̄}{b} = \overset{̄}{0}\), then the acute angle between \(\overset{̄}{a}\) and \(\overset{̄}{c}\) is

• A. \(\frac{π}{4}\)
• B. \(\frac{π}{6}\)
• C. \(\frac{π}{3}\)
• D. \(\frac{π}{8}\)

Hint:

Use the vector triple product expansion to relate b with a and c, then take magnitudes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The vector triple product rule is \(\bar{a}\times(\bar{a}\times\bar{c}) = (\bar{a}\cdot\bar{c})\bar{a} - (\bar{a}\cdot\bar{a})\bar{c}\).

Step 2: Expand.
With \(|\bar{a}| = 1\), we have \(\bar{a}\cdot\bar{a} = 1\), so
\[ (\bar{a}\cdot\bar{c})\bar{a} - \bar{c} + \bar{b} = \bar{0} \Rightarrow \bar{b} = \bar{c} - (\bar{a}\cdot\bar{c})\bar{a} \]

Step 3: Take the magnitude squared.
\[ |\bar{b}|^2 = |\bar{c}|^2 - 2(\bar{a}\cdot\bar{c})^2 + (\bar{a}\cdot\bar{c})^2|\bar{a}|^2 = 4 - (\bar{a}\cdot\bar{c})^2 \]
Since \(|\bar{b}| = 1\): \((\bar{a}\cdot\bar{c})^2 = 3\).

Step 4: Find the angle.
\(\bar{a}\cdot\bar{c} = |\bar{a}||\bar{c}|\cos\theta = 2\cos\theta\). So \(4\cos^2\theta = 3\), giving \(\cos\theta = \pm\dfrac{\sqrt{3}}{2}\). The acute angle has \(\cos\theta = \dfrac{\sqrt{3}}{2}\), so \(\theta = \dfrac{\pi}{6}\).

Final Answer:
The acute angle is \(\dfrac{\pi}{6}\), option (B). \[ \boxed{\frac{\pi}{6}} \]