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Scalar Or Dot Product — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsScalar Or Dot Product3 Marks
Question
Find $|\vec{\text{a}|}$ and $\big|\vec{\text{b}}\big|$ if
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=8$ and $|\vec{\text{a}}|=8\big|\vec{\text{b}}\big|$
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Answer

 $\big(\vec{\text{a }}.\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=8$
$\Rightarrow \vec{\text{a}}.\vec{\text{a}}-\vec{\text{a}}.\vec{\text{b}}+\vec{\text{b}}.\vec{\text{a}}-\vec{\text{b}}.\vec{\text{b}}=8$
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=8$
$\Rightarrow\big(8\big|\vec{\text{b}}\big|\big)^2-\big|\vec{\text{b}}\big|^2=8$ $\big[|\vec{\text{a}}|=8\big|\vec{\text{b}}\big|\big]$
$\Rightarrow64\big|\vec{\text{b}}\big|^2-\big|\vec{\text{b}}\big|^2=8$
$\Rightarrow63\big|\vec{\text{b}}\big|^2=8$
$\Rightarrow\big|\vec{\text{b}}\big|^2=\frac{8}{63}$
$\Rightarrow\big|\vec{\text{b}}\big|=\sqrt{\frac{8}{63}}$ [Magnitude of a vectoer is non-negative]
$\Rightarrow\big|\vec{\text{b}}\big|=\frac{2\sqrt{2}}{3\sqrt{7}}$
$|\vec{\text{a}}|=8\big|\vec{\text{b}}\big|=\frac{8\times2\sqrt{2}}{3\sqrt{7}}=\frac{16\sqrt{2}}{3\sqrt{7}}$

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