MCQ
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=$
  • $\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
  • B
    $\big|\vec{\text{a}}+\vec{\text{b}}\big|^2$
  • C
    $\big|\vec{\text{a}}\big|^2+\big|\vec{\text{b}}\big|^2$
  • D
    $2\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

Answer

Correct option: A.
$\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
We have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\theta\big)^2+\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta\big)^2$
$=\big|\vec{\text{a}}\big|^2\big|{\text{b}}\big|^2\sin^2\theta+\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\cos^2\theta$
$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\big(\sin^2\theta+\cos^2\theta\big)$
$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$

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