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 \vec{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})$$=\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}+\frac{6}{7}\hat{k}
\vec{b}=\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})$$=\frac{3}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{2}{7}\hat{k}
\vec{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}.
$|\vec{a}|=\sqrt{(\frac{2}{7})^2+(\frac{3}{7})^2+(\frac{6}{7})^2}=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1$
$|\vec{b}|=\sqrt{(\frac{3}{7})^2+(\frac{6}{7})^2+(\frac{2}{7})^2}=\sqrt{\frac{9}{49}+\frac{36}{49}+\frac{4}{49}}=1$
$|\vec{c}|=\sqrt{(\frac{6}{7})^2+(\frac{2}{7})^2+(\frac{3}{7})^2}=\sqrt{\frac{36}{49}+\frac{4}{49}+\frac{9}{49}}=1$
Thus,each of the given three vectors is a unit vector.
$\vec{a}.\vec{b}=\frac{2}{7}×\frac{3}{7}+\frac{3}{7}×(\frac{-6}{7})+\frac{6}{7}×\frac{2}{7}=\frac{6}{49}-\frac{18}{49}+\frac{12}{49}=0$
$\vec{b}.\vec{c}=\frac{3}{7}×\frac{6}{7}+\frac{-6}{7}×(\frac{2}{7})+\frac{2}{7}×\frac{-3}{7}=\frac{18}{49}-\frac{12}{49}+\frac{6}{49}=0$
$\vec{c}.\vec{a}=\frac{6}{7}×\frac{2}{7}+\frac{2}{7}×(\frac{3}{7})+\frac{-3}{7}×\frac{6}{7}=\frac{12}{49}-\frac{6}{49}+\frac{18}{49}=0$
Hence the given three vectors are mutually perpendicular to each other.