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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA3 Marks
Question
If either vector $\vec{a}=\vec{0}\ \text{or}\ \vec{b}=\vec{0},\ \text{then}\ \vec{a}\cdot\vec{b}=0.$ But the converse need not be true. Justify your answer with an example.
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Answer

Case I: $\ \text{Vector}\ \vec{a}=\vec{0}.$ Therefore by definition of zero vector, $\ \big|\vec{a}\big|=0\ \ ....\text{(i)}$$\therefore \ \vec{a}.\vec{b}=\big|\vec{a}\big|.\Big|\vec{b}\Big|\text{cos}\theta=0.\Big|\vec{b}\Big|\text{cos}\theta$ [Form eq. (i)]
$\Rightarrow\ \ \vec{a}.\vec{b}=0$
Case II: $\ \text{Vector}\ \vec{b}=\vec{0}.$ Therefore by definition of zero vector, $\ \big|\vec{b}\big|=0\ \ \ .....\text{(ii)}$
$\therefore\ \ \vec{a}.\vec{b}=\big|\vec{a}\big|.\Big|\vec{b}\Big|\text{cos}\theta=\vec{a}.0.\text{cos}\theta$ [Form eq. (ii)]
$\Rightarrow\ \ \vec{a}.\vec{b}=0$
But the converse is not true. Justification: $\\ \text{Let}\ \vec{a}=\hat{i}+\hat{j}+\hat{k}$$\text{Therefore},\ \big|\vec{a}\big|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}\neq0$
$\text{Therefore},\ \ \vec{a}\neq\vec{0}$
$\text{Again let}\ \ \ \vec{b}=\hat{i}+\hat{j}-2\hat{k}$
$\therefore\ \ \ \Big|\vec{b}\Big|=\sqrt{(1)^2+(1)^2+(-2)^2}=\sqrt{6}\neq0$
$\text{Therefore},\ \vec{b}\neq\vec{0}$
$\text{But}\ \ \ \vec{a}.\vec{b}=1(1)+1(1)+1(-2)=1+1+-2=0$
$\text{Hence, here}\ \vec{a}.\vec{b}=0,\ \text{but}\ \vec{a}\neq\vec{0}\ \text{and}\ \vec{b}\neq\vec{0}.$

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