Question

If $|\vec{a} \times \vec{b}| = 5$ and $|\vec{a} .\vec{b}| = 3$ , then $|\vec{a}|^2 |\vec{b}|^2 $ is equal to

• A. 16
• B. 31
• C. 25
• D. 34

Answer 1

Answer: (D)

Given , $|\vec{a} \times \vec{b}| = 5 $ and $| \vec{a}. \vec{b}|= 3$
We know that $\vec{a} \times \vec{b} = | \vec{a}||\vec{b}| \sin \, \theta \hat{n}$
$\therefore \:\:\: \left|\vec{a} \times \vec{b}\right|=\left|\left|\vec{a}\right|\left|\vec{b}\right|\sin\theta \hat{n} \right| $
$= \left|\left|\vec{a}\right|\left|\vec{b}\right|\sqrt{1 -\cos^{2} \theta} \hat{n} \right| $
$= \left|\left|\vec{a}\right| \left|\vec{b}\right| \sqrt{1-\left( \frac{\vec{a} .\vec{b}}{\left|\vec{a}\right|\left|\vec{b}\right|}\right)^{2}} \right|. \left|\hat{n}\right| $
$= \left|\left|\vec{a}\right|\left|\vec{b}\right| \frac{\sqrt{\left(\left|\vec{a}\right|\left|\vec{b}\right|\right)^{2} -\left(\vec{a} .\vec{b}\right)^{2}}}{\left|\vec{a}\right|\left|\vec{b}\right|} \right| \left(\therefore \left|\hat{n} \right| =1\right)$
$ \left|\vec{a} \times \vec{b}\right| = \left|\sqrt{\left|\vec{a}\right|^{2} \left|\vec{b}\right|^{2} - \left(\vec{a}.\vec{b}\right)^{2}}\right| $
Squaring both sides, we get
$\left|\vec{a} \times \vec{b}\right|^{2} = \left|\vec{a}\right|^{2} \left|\vec{b}\right|^{2} -\left(\vec{a} .\vec{b}\right)^{2} $
or $ \left|\vec{a}\right|^{2} \left|\vec{b}\right|^{2} = \left|\vec{a} \times \vec{b}\right|^{2} +\left(\vec{a}.\vec{b}\right).\left(\vec{a}.\vec{b}\right) $
$= \left(\left|\vec{a} \times \vec{b}\right|\right)^{ 2} +\left(\left|\vec{a}. \vec{b}\right|\right)^{2} $
$=\left(5\right)^{2} + \left(3\right)^{2} =25 + 9 = 34$