Question
Show that the vectors
$\vec{\text{a}}=\frac{1}{7}(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}),\vec{\text{b}}=\frac{1}{7}(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}),\vec{\text{c}}=\frac{1}{7}(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})$ are mutually perpendicular unit vectors.

Answer

We have
$|\vec{\text{a}}|=\frac{1}{7}\sqrt{2^2+3^2+6^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1$
$\big|\vec{\text{b}}\big|=\frac{1}{7}\sqrt{3^2+(-6)^2+2^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1$
$|\vec{\text{c}}|=\frac{1}{7}\sqrt{6^2+2^2+(-3)^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1$
And
$\vec{\text{a}}.\vec{\text{b}}$
$=\frac{1}{7}(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}).\frac{1}{7}(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}})$
$=\frac{1}{49}(6-18+12)$
$=0$
$\vec{\text{b}}.\vec{\text{c}}$
$=\frac{1}{7}(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}).\frac{1}7{}(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})$
$=\frac{1}{49}(18-12-6)$
$=0$
$\vec{\text{c}}.\vec{\text{a}}$
$=\frac{1}{7}(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}).\frac{1}7{}(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})$
$=\frac{1}{49}(12+6-18)$
$=0$
So, $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=|\vec{\text{c}}|=1$ and $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
So, the given vectors are mutuaiiy perpendicular unit vectors.

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