Question

If \[ |\vec a|=3,\qquad |\vec b|=4 \] and the angle between \(\vec a\) and \(\vec b\) is \[ \frac{\pi}{6}, \] then \[ \left|(4\vec a+\vec b)\times(\vec a-3\vec b)\right| = \]

• A. \(66\)
• B. \(78\sqrt3\)
• C. \(78\)
• D. \(66\sqrt3\)

Hint:

For cross-product expressions involving linear combinations, expand first and immediately use \[ \vec a\times\vec a=0, \qquad \vec b\times\vec b=0, \qquad \vec b\times\vec a=-\vec a\times\vec b. \] This often reduces the entire expression to a single multiple of \(\vec a\times\vec b\).

Answer 1

The Correct Option is C

Concept: The cross product is distributive: \[ (\vec p+\vec q)\times\vec r = \vec p\times\vec r+\vec q\times\vec r. \] Also, \[ \vec a\times\vec a=\vec 0, \qquad \vec b\times\vec b=\vec 0, \] and \[ |\vec a\times\vec b| = |\vec a||\vec b|\sin\theta. \] These properties greatly simplify the computation.

Step 1: Expand the cross product. Consider \[ (4\vec a+\vec b)\times(\vec a-3\vec b). \] Using distributivity, \[ = 4\vec a\times\vec a -12\vec a\times\vec b +\vec b\times\vec a -3\vec b\times\vec b. \] Since \[ \vec a\times\vec a=0, \qquad \vec b\times\vec b=0, \] we obtain \[ = -12\vec a\times\vec b +\vec b\times\vec a. \] Using \[ \vec b\times\vec a = -\vec a\times\vec b, \] therefore \[ = -13(\vec a\times\vec b). \]

Step 2: Take magnitude. Hence, \[ \left| (4\vec a+\vec b)\times(\vec a-3\vec b) \right| = 13|\vec a\times\vec b|. \] Now, \[ |\vec a\times\vec b| = |\vec a||\vec b| \sin\frac{\pi}{6}. \] Substituting, \[ = 3\times4\times\frac12. \] \[ =6. \] Therefore, \[ 13\times6=78. \]

Step 3: Verify the result. The expression simplifies completely to a scalar multiple of \(\vec a\times\vec b\), making the magnitude calculation straightforward. Hence the value is \[ 78. \] Conclusion: \[ \boxed{78} \]