Question
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ prove that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2=\begin{vmatrix}\vec{\text{a}}.\vec{\text{a}}&\vec{\text{a}}.\vec{\text{b}}\\\vec{\text{b}}.\vec{\text{a}}&\vec{\text{b}} .\vec{\text{b}}\end{vmatrix}.$

Answer

$\text{RHS}=\begin{vmatrix}\vec{\text{a}}.\vec{\text{a}}&\vec{\text{a}}.\vec{\text{b}}\\\vec{\text{b}}.\vec{\text{a}}&\vec{\text{b}} .\vec{\text{b}}\end{vmatrix}.$
$=\begin{vmatrix}|\vec{\text{a}}|^2&|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\\|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta&\big|\vec{\text{b}}\big|^2 \end{vmatrix}$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\cos^2\theta$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2(1-\cos^2\theta)$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\sin^2\theta$
$=\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\big)^2$
$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$
$=\text{LHS}$
Hence proved.

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