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Vector Algebra — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVector Algebra1 Mark
Question
Assertion $(A) :$ If $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0},|\vec{a}|=3$, $|\vec{b}|=4,|\vec{c}|=5$, then $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is equal to $-25 .$
Reason $(R) :$ If $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, then the angle $\theta$ between $\vec{b}$ and $\vec{c}$ is given by
$\cos \theta=\frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|^2} .$
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Answer

$(b) :$ We have, $|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5$ and
$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \Rightarrow(\vec{a}+\vec{b}+\vec{c})^2=0$
$\Rightarrow|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\Rightarrow(3)^2+(4)^2+(5)^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-\frac{1}{2}[9+16+25]=-\frac{1}{2}(50)=-25$
$\text { Now, } \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \Rightarrow \vec{b}+\vec{c}=-\vec{a}$
$\Rightarrow(\vec{b}+\vec{c})^2=(-\vec{a})^2 \Rightarrow|\vec{b}|^2+|\vec{c}|^2+2 \vec{b} \cdot \vec{c}=|\vec{a}|^2$
$\Rightarrow|\vec{b}|^2+|\vec{c}|^2+2|\vec{b}||\vec{c}| \cos \theta=|\vec{a}|^2$
$\Rightarrow \cos \theta=\frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}$
Hence, both assertion and reason are true but reason is not the correct explanation of assertion.

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