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 $\vec{a}\times \vec{b}=|\vec{a}||\vec{b}|\sin \theta \hat{n}$
$\therefore \left|\vec{a}\times \vec{b}\right|=\left|\left|\vec{a}\right|\left|\vec{b}\right|\sin \theta \hat{n}\right|$
$=\left|\left|\vec{a}\right|\left|\vec{b}\right|\sqrt{1-{\cos }^2\theta }\hat{n}\right|$
$=\left|\left|\vec{a}\right|\left|\vec{b}\right|\sqrt{1-{\left(\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}\right)}^2}\right|.\left|\hat{n}\right|$
$=\left|\left|\vec{a}\right|\left|\vec{b}\right|\frac{\sqrt{{\left(|\vec{a}||\vec{b}|\right)}^2-{\left(\vec{a}.\vec{b}\right)}^2}}{|\vec{a}||\vec{b}|}\right|\left(\therefore \left|\hat{n}\right|=1\right)$
$\left|\vec{a}\times \vec{b}\right|=\left|\sqrt{{\left|\vec{a}\right|}^2{\left|\vec{b}\right|}^2-{\left(\vec{a}.\vec{b}\right)}^2}\right|$
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 ${\left|\vec{a}\right|}^2{\left|\vec{b}\right|}^2={\left|\vec{a}\times \vec{b}\right|}^2+\left(\vec{a}.\vec{b}\right).\left(\vec{a}.\vec{b}\right)$
$={\left(\left|\vec{a}\times \vec{b}\right|\right)}^2+{\left(\left|\vec{a}.\vec{b}\right|\right)}^2$
$={\left(5\right)}^2+{\left(3\right)}^2=25+9=34$