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Scalar Or Dot Product — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsScalar Or Dot Product5 Marks
Question
If $\big|\vec{\text{a}}+\vec{\text{b}}\big|=60,\big|\vec{\text{a}}-\vec{\text{b}}\big|=40$ and $\big|\vec{\text{b}}\big|=46,$ find $|\vec{\text{a}}|$
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Answer

Here, $\big|\vec{\text{a}}+\vec{\text{b}}\big|=60$
Squaring both the sides,
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=(60)^2$
$(\vec{\text{a}}+\vec{\text{b}})=(60)^2$
$(\vec{\text{a}})^2+(\vec{\text{b}})^2+2\vec{\text{a}}\vec{\text{b}}=3600$
$|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}\vec{\text{b}}=3600\dots(1)$
Now, $\big|\vec{\text{a}}-\vec{\text{b}}\big|=40$
Squaring both the sides,
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=(40)^2$
$|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}\vec{\text{b}}=1600\dots(2)$
Adding (1) and (2),
$2|\vec{\text{a}}|^2+2\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}\vec{\text{b}}-2\vec{\text{a}}\vec{\text{b}}=3600-1600$
$2|\vec{\text{a}}|^2+2(46)^2=5200$
$2|\vec{\text{a}}|^2=5200-4232$
$2|\vec{\text{a}}|^2=968$
$|\vec{\text{a}}|^2=\frac{968}{2}$
$|\vec{\text{a}}|^2=484$
$|\vec{\text{a}}|=\sqrt{484}$
$|\vec{\text{a}}|=22$

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