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Vector or Cross Product — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector or Cross Product5 Marks
Question
Given $\vec{\text{a}}=\frac{1}{7}\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big),\vec{\text{b}}=\frac{1}{7}\big(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}\big),$$\vec{\text{c}}=\frac{1}{7}\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big),\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$
being a right handed orthogonal system of unit vector in spece, show that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ is also another system.
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Answer

Given:
$\vec{\text{a}}=\frac{1}{7}\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$
$\vec{\text{b}}=\frac{1}{7}\big(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}\big)$
$\vec{\text{c}}=\frac{1}{7}\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\vec{\text{a}}\times\vec{\text{b}}=\Big(\frac{1}{7}\Big)\Big(\frac{1}{7}\Big)\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&3&6\\3&-6&2 \end{vmatrix}$
$=\frac{1}{49}\big(42\hat{\text{i}}+14\hat{\text{j}}-21\hat{\text{k}}\big)$
$=\frac{1}{49}\big[7\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)\big]$
$=\frac{1}{7}\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$=\vec{\text{c}}$
$\vec{\text{b}}\times\vec{\text{c}}=\Big(\frac{1}{7}\Big)\Big(\frac{1}{7}\Big)\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-6&2\\6&2&-3 \end{vmatrix}$
$=\frac{1}{49}\big(14\hat{\text{i}}+21\hat{\text{j}}+42\hat{\text{k}}\big)$
$=\frac{1}{49}\big[7\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)\big]$
$=\frac{1}{7}\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$
$=\vec{\text{a}}$
$\vec{\text{c}}\times\vec{\text{a}}=\Big(\frac{1}{7}\Big)\Big(\frac{1}{7}\Big)\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\6&2&-3\\2&3&6 \end{vmatrix}$
$=\frac{1}{49}\big(21\hat{\text{i}}-42\hat{\text{j}}+14\hat{\text{k}}\big)$
$=\frac{1}{49}\big[7\big(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}\big)\big]$
$=\frac{1}{7}\big(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}\big)$
$=\vec{\text{b}}$
$|\vec{\text{a}}|=\frac{1}{7}\sqrt{4+9+36}$
$=\frac{7}{7}$
$=1$
$\big|\vec{\text{b}}\big|=\frac{1}{7}\sqrt{9+36+4}$
$=\frac{7}{7}$
$=1$
$|\vec{\text{c}}|=\frac{1}{7}\sqrt{36+4+9}$
$=\frac{7}{7}$
$=1$
Thus, $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ from a right handed orthogonal system of unit vectors.

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