Question

Show that each of the given three vectors is a unit vector: $\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}),\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}),\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})$ .
Also, show that they are mutually perpendicular to each other.

Answer 1

Let $\overrightarrow{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})=\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}+\frac{6}{7}\hat{k}$,

$\overrightarrow{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})=\frac{3}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{2}{7}\hat{k}$,

and $\overrightarrow{c}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})=\frac{6}{7}\hat{i}+\frac{2}{7}\hat{j}-\frac{3}{7}\hat{k}$.

$|\overrightarrow{a}|=\sqrt{{\left(\frac{2}{7}\right)}^2+{\left(\frac{3}{7}\right)}^2+{\left(\frac{6}{7}\right)}^2}=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1$

$|\overrightarrow{b}|=\sqrt{{\left(\frac{3}{7}\right)}^2+{(-\frac{6}{7})}^2+{\left(\frac{2}{7}\right)}^2}=\sqrt{\frac{9}{49}+\frac{36}{49}+\frac{9}{49}}=1$

$|\overrightarrow{c}|=\sqrt{{\left(\frac{6}{7}\right)}^2+{\left(\frac{2}{7}\right)}^2+{(-\frac{3}{7})}^2}=\sqrt{\frac{36}{49}+\frac{4}{49}+\frac{9}{49}}=1$

Thus, each of the given three vectors is a unit vector.

$\overrightarrow{a}\cdot \overrightarrow{b}=\frac{2}{7}\times \frac{3}{7}+\frac{3}{7}\times \left(\frac{-6}{7}\right)+\frac{6}{7}\times \frac{2}{7}=\frac{6}{49}-\frac{18}{49}+\frac{12}{49}=0$

$\overrightarrow{b}\cdot \overrightarrow{c}=\frac{3}{7}\times \frac{6}{7}+\left(\frac{-6}{7}\right)\times \frac{2}{7}+\frac{2}{7}\times \left(\frac{-3}{7}\right)=\frac{18}{49}-\frac{12}{49}-\frac{6}{49}=0$

$\overrightarrow{c}\cdot \overrightarrow{a}=\frac{6}{7}\times \frac{2}{7}+\frac{2}{7}\times \frac{3}{7}+\left(\frac{-3}{7}\right)\times \frac{6}{7}=\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=0$

Hence, the given three vectors are mutually perpendicular to each other.