Question

If \(|\overset{̄}{a}| = |\overset{̄}{b}| = 1,|\overset{̄}{c}| = 2\) and \(\overset{̄}{a}\times (\overset{̄}{a}\times \overset{̄}{c})+\overset{̄}{b} = \overset{̄}{0}\), then the acute angle between \(\overset{̄}{a}\) and \(\overset{̄}{c}\) is \(\ldots\)

• A. \(\frac{π}{2}\)
• B. \(\frac{π}{3}\)
• C. \(\frac{π}{4}\)
• D. \(\frac{π}{6}\)

Hint:

Expand the triple product to (a.c)a - c, then take magnitudes of b = c - (a.c)a.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We use the vector triple product expansion \(\bar a\times(\bar a\times\bar c) = (\bar a\cdot\bar c)\bar a - (\bar a\cdot\bar a)\bar c\).

Step 2: Key Formula or Approach:
With \(|\bar a| = 1\), we get \(\bar a\cdot\bar a = 1\), so the triple product equals \((\bar a\cdot\bar c)\bar a - \bar c\).

Step 3: Detailed Explanation:
The given equation \(\bar a\times(\bar a\times\bar c) + \bar b = \bar 0\) becomes
\[ \bar b = \bar c - (\bar a\cdot\bar c)\,\bar a \]
Take the magnitude squared. Let \(t = \bar a\cdot\bar c\):
\[ |\bar b|^2 = |\bar c|^2 - 2t(\bar a\cdot\bar c) + t^2|\bar a|^2 = 4 - 2t^2 + t^2 = 4 - t^2 \]
Since \(|\bar b| = 1\), we get \(4 - t^2 = 1\), so \(t^2 = 3\).
But \(t = |\bar a||\bar c|\cos\theta = 2\cos\theta\). So
\[ 4\cos^2\theta = 3 \Rightarrow \cos\theta = \frac{\sqrt3}{2} \]
The acute angle is \(\theta = \dfrac\pi6\).
Option (A) \(\frac\pi2\) would give \(t = 0\) and \(|\bar b| = 2\). Options (B) and (C) give \(t^2 = 1\) and \(t^2 = 2\), neither of which equals 3.

Final Answer:
The acute angle between \(\bar a\) and \(\bar c\) is \(\dfrac\pi6\), option (D). \[ \boxed{\frac{\pi}{6} \text{ (D)}} \]