Question
If $|\vec{\text{a}}|=10,\big|\vec{\text{b}}\big|=2$ and $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=16,$ find $\vec{\text{a}}.\vec{\text{b}}.$

Answer

We know
$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+(16)^2=(10)^2\times2^2$ $\big(\because\big|\vec{\text{a}}\times\vec{\text{b}}\big|=16,|\vec{\text{a}}|=10\text{ and }\big|\vec{\text{b}}\big|=2\big)$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+256=400$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=144$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=\pm12$

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