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VECTOR ALGEBRA — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVECTOR ALGEBRA5 Marks
Question
If  $\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}},\vec{\text{b}}=-3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}},$ compute $\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\vec{\text{c}}$ and $\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{c}}\big)$ and verify that these are not equal.
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Answer

Given:
$\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}$
$\vec{\text{b}}=-3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&5&-7\\-3&4&1 \end{vmatrix}$
$=(5+28)\hat{\text{i}}-(2-21)\hat{\text{j}}+(8+15)\hat{\text{k}}$
$=33\hat{\text{i}}+19\hat{\text{j}}+23\hat{\text{k}}$
$\Rightarrow\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\vec{\text{c}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\33&19&23\\1&-2&-3 \end{vmatrix}$
$=(-57+46)\hat{\text{i}}-(-99-23)\hat{\text{j}}+(-66-19)\hat{\text{k}}$
$\Rightarrow\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\vec{\text{c}}=-11\hat{\text{i}}+122\hat{\text{j}}-85\hat{\text{k}}\dots(1)$
$\therefore\vec{\text{b}}\times\vec{\text{c}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-3&4&1\\1&-2&-3 \end{vmatrix}$
$=(-12+2)\hat{\text{i}}-(9-1)\hat{\text{j}}+(6-4)\hat{\text{k}}$
$=-10\hat{\text{i}}-8\hat{\text{j}}+2\hat{\text{k}}$
$\Rightarrow\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{c}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&5&-7\\-10&-8&2 \end{vmatrix}$
$=(10-56)\hat{\text{i}}-(4-70)\hat{\text{j}}+(-16+50)\hat{\text{k}}$
$\Rightarrow\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{c}}\big)=-46\hat{\text{i}}+66\hat{\text{j}}+34\hat{\text{k}}\dots(2)$
From (1) and (2), we get
$\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\vec{\text{c}}\neq\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{c}}\big)$

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