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VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA5 Marks
Question
Prove that: $\big(\vec{\text{a}}-\vec{\text{b}}\big).\big\{\big(\vec{\text{b}}-\vec{\text{c}}\big)\big\}=0$
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Answer

We have
$\big(\vec{\text{a}}-\vec{\text{b}}\big).\big\{\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\big(\vec{\text{c}}-\vec{\text{a}}\big)\big\}$
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}-\vec{\text{c}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$ (By distributive law)
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$\big(\therefore\vec{\text{c}}\times\vec{\text{c}}=0\big)$
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\vec{\text{a}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\vec{\text{a}}\big(\vec{\text{c}}\times\vec{\text{a}}\big)-\vec{\text{b}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{b}}\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\vec{\text{b}}\big(\vec{\text{c}}\times\vec{\text{a}}\big)$ (By distributive 1)
$=\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]+\big[\vec{\text{a}}\ \vec{\text{a}}\ \vec{\text{b}}\big]+\big[\vec{\text{a}}\ \vec{\text{c}}\ \vec{\text{a}}\big]-\big[\vec{\text{b}}\ \vec{\text{b}}\ \vec{\text{c}}\big]-\big[\vec{\text{b}}\ \vec{\text{a}}\ \vec{\text{b}}\big]-\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]$
$=\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]+0+0-0-0-\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]$
$\big(\therefore\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]=\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]\big)$
$=0$

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