Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have} \\ \left|\overrightarrow{a}\right|=\frac{1}{7}\sqrt{2^2+3^2+6^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1 \\ \left|\overrightarrow{b}\right|=\frac{1}{7}\sqrt{3^2+{\left(-6\right)}^2+2^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1 \\ \left|\overrightarrow{c}\right|=\frac{1}{7}\sqrt{6^2+2^2+{\left(-3\right)}^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1 \\ \mathrm{And} \\ \overrightarrow{a}.\overrightarrow{b} \\ =\frac{1}{7}\left(2\hat{i}+3\hat{j}+6\hat{k}\right).\frac{1}{7}\left(3\hat{i}-6\hat{j}+2\hat{k}\right) \\ =\frac{1}{49}\left(6-18+12\right) \\ =0 \\ \overrightarrow{b}.\overrightarrow{c} \\ =\frac{1}{7}\left(3\hat{i}-6\hat{j}+2\hat{k}\right).\frac{1}{7}\left(6\hat{i}+2\hat{j}-3\hat{k}\right) \\ =\frac{1}{49}\left(18-12-6\right) \\ =0 \\ \overrightarrow{c}.\overrightarrow{a} \\ =\frac{1}{7}\left(6\hat{i}+2\hat{j}-3\hat{k}\right).\frac{1}{7}\left(2\hat{i}+3\hat{j}+6\hat{k}\right) \\ =\frac{1}{49}\left(12+6-18\right) \\ =0 \\ \text{So,}\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1\text{and}\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0 \\ \text{So, the given vectors are mutually perpendicular unit vectors.} \end{gathered}$$