Question
If $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=4,\big|\vec{\text{a}}.\vec{\text{b}}\big|=2,$ then $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2=$
- 6
- 2
- 20
- 8
Solution:
We know
$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|62=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\dots(1)$
$\big|\vec{\text{a}}.\vec{\text{b}}\big|=2$ (Given)
$\Rightarrow\big|\vec{\text{a}}.\vec{\text{b}}\big|^2=\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
From (1), we get
$(2)^2+(4)^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2=20$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f(x) = \left\{ {\begin{array}{*{20}{l}}
{\frac{{k\cos x}}{{\pi - 2x}},}&{{\rm{ if }}\,x\, \ne \,\frac{\pi }{2}}\\
{3,}&{{\rm{ if }}\,x\, = \,\frac{\pi }{2}}
\end{array}} \right.$ at $x = \frac{\pi }{2}$