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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra4 Marks
Question
Show that for any two vectors $\vec a $ and $\vec b$ , we always have $|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$ (triangle inequality).
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Answer

The inequality holds trivially in case either $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$.
So, let $|\vec{a}| \neq \vec{0} \neq|\vec{b}|$. Then
$|\vec{a}+\vec{b}|^{2}=(\vec{a}+\vec{b})^{2}=(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})$ 
= $\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}$ 
= $|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{b}|^{2}$ (scalar product is commuatative)
=$\leq|\vec{a}|^{2}+2|\vec{a} \cdot \vec{b}|+|\vec{b}|^{2}$ (since $x \leq|x|~ \forall x \in {R}$)
=$\leq|\vec{a}|^{2}+2|\vec{a}||\vec{b}|+|\vec{b}|^{2}$ (from Cauchy Schwartz Inequality) 
= $(|\vec{a}|+|\vec{b}|)^{2}$  
$\Rightarrow |\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$

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