Question

If \[ \vec a=i+2j-2k,\quad \vec b=6i-3j+2k, \] and \(\vec c\perp \vec a\), \(\vec c\times \vec b=i-2j-6k\), then angle between \(\vec b\) and \(\vec c\) is

• A. \(\frac{\pi}{3}\)
• B. \(\cos^{-1}\frac{29}{21\sqrt2}\)
• C. \(\frac{\pi}{4}\)
• D. \(\cos^{-1}\frac{23}{21\sqrt2}\) \bigskip

Hint:

When cross product is given, always convert into \( |\vec a||\vec b|\sin\theta \).

Answer 1

The Correct Option is D

Concept: Use: \[ |\vec c\times \vec b|=|\vec c||\vec b|\sin\theta \]

Step 1: Magnitudes.
\[ |\vec b|=7,\quad |\vec c\times \vec b|=\sqrt{41} \]

Step 2: Find \(|\vec c|\).
Using \(\vec c\perp \vec a\), solve system gives: \[ |\vec c|=\frac{3\sqrt2}{7} \]

Step 3: Compute angle.
\[ \sin\theta=\frac{\sqrt{41}}{7|\vec c|} \] \[ \cos\theta=\frac{23}{21\sqrt2} \] \[ \boxed{(D)} \]