Question

Let $ \bar{a}, \bar{b}, \bar{c} $ be three vectors such that $ \bar{a} $ is perpendicular to $ \bar{b} $ and $ \bar{b} $ is perpendicular to $ \bar{c} $. If $ |\bar{a}| = 2, |\bar{b}| = 3, |\bar{c}| = 5 $ and $ |\bar{a} + \bar{b} + \bar{c}| = 4\sqrt{3} $, then the angle between $ \bar{a} $ and $ \bar{c} $ is

• A. $ \cos^{-1} \left(\frac{2}{5}\right) $
• B. $ \frac{\pi}{3} $
• C. $ \cos^{-1} \left(\frac{2}{3}\right) $
• D. $ \frac{\pi}{6} $

Hint:

Remember that the dot product of perpendicular vectors is zero. Use the expansion of the squared magnitude of a vector sum and the definition of the dot product in terms of the angle between vectors.

Answer 1

The Correct Option is B

Step 1: Use perpendicularity to establish dot products.
$ \bar{a} \cdot \bar{b} = 0 $ $ \bar{b} \cdot \bar{c} = 0 $
Step 2: Square the magnitude of the sum of vectors.
$ |\bar{a} + \bar{b} + \bar{c}|^2 = 48 $
Step 3: Expand the squared magnitude.
$ |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} + \bar{b} \cdot \bar{c}) = 48 $
Step 4: Substitute given values.
$ 4 + 9 + 25 + 2(0 + \bar{a} \cdot \bar{c} + 0) = 48 $ $ 38 + 2(\bar{a} \cdot \bar{c}) = 48 $
Step 5: Solve for $ \bar{a \cdot \bar{c} $.}
$ 2(\bar{a} \cdot \bar{c}) = 10 $ $ \bar{a} \cdot \bar{c} = 5 $
Step 6: Use the dot product formula to find the angle $ \theta $.
$ |\bar{a}| |\bar{c}| \cos \theta = 5 $ $ (2)(5) \cos \theta = 5 $ $ 10 \cos \theta = 5 $ $ \cos \theta = \frac{1}{2} $
Step 7: Determine the angle $ \theta $.
$ \theta = \frac{\pi}{3} $
Step 8: Conclusion.
The angle between $ \bar{a} $ and $ \bar{c} $ is $ \frac{\pi}{3} $.