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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra2 Marks
Question
If either vector $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$, 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

If either $\vec{a}=0$ or $\vec{\mathrm{b}}=0$ then $\vec{a} . \vec{b}=0$
Now let $\vec{\mathrm{a}}=2 \hat{1}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} $ and $\vec{\mathrm{b}}=3 \hat{1}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \cdot(3 \hat{1}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$
$\Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=2 \hat{1} .3 \hat{1}-2 \hat{1} \cdot 6 \hat{\mathrm{j}}+2 \hat{1} \cdot 2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}} \cdot 3 \hat{1}$ - $3 \hat{0} \cdot 6 \hat{\jmath}+3 \hat{\jmath} .2 \hat{\mathrm{k}}+6 \hat{\mathrm{k}} .3 \hat{1}-6 \hat{\mathrm{k}} .6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} .2 \hat{\mathrm{k}}$
$\Rightarrow \vec{a} \cdot \vec{b}$ = 6 - 0 + 0 + 0 - 18 + 0 + 0 - 0 + 12 [$\hat{ {1}} \hat{ {j}}=\hat{ {j}} \cdot \hat{ {k}}=\hat{ {k}} \cdot \hat{ {\imath}}=0$]
$\Rightarrow \vec{a} \cdot \vec{b}$ = 18 - 18 = 0 [$\text { i. }\hat{\mathrm{i}}=\hat{\mathrm{j}} \cdot \hat{\mathrm{j}}=\hat{\mathrm{k}}. \hat{\mathrm{k}}=1$]
$\Rightarrow \vec{a} \cdot \vec{b}=0~ for ~\vec{\mathrm{a}} \neq 0, \vec{\mathrm{b}} \neq 0$
Hence, the converse of given statement need not be true.

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