જો $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ અને $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k}$ હોય,તો નીચેનાની દિશામાં એકમ સદિશ શોધો:
$(i)$ $6 \vec{b}$
(ii) $2 \vec{a}-\vec{b}$

આપેલ છે કે $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ અને $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k}.$
$(i)$ $6 \vec{b} = 6(2 \hat{i}+\hat{j}+2 \hat{k}) = 12 \hat{i}+6 \hat{j}+12 \hat{k}.$
માન $|6 \vec{b}| = \sqrt{12^2+6^2+12^2} = \sqrt{144+36+144} = \sqrt{324} = 18.$
એકમ સદિશ = $\frac{6 \vec{b}}{|6 \vec{b}|} = \frac{12 \hat{i}+6 \hat{j}+12 \hat{k}}{18} = \frac{2}{3} \hat{i} + \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k}.$
(ii) $2 \vec{a}-\vec{b} = 2(\hat{i}+\hat{j}+2 \hat{k}) - (2 \hat{i}+\hat{j}+2 \hat{k}) = (2 \hat{i}+2 \hat{j}+4 \hat{k}) - (2 \hat{i}+\hat{j}+2 \hat{k}) = 0 \hat{i} + 1 \hat{j} + 2 \hat{k} = \hat{j} + 2 \hat{k}.$
માન $|2 \vec{a}-\vec{b}| = \sqrt{0^2+1^2+2^2} = \sqrt{5}.$
એકમ સદિશ = $\frac{2 \vec{a}-\vec{b}}{|2 \vec{a}-\vec{b}|} = \frac{\hat{j}+2 \hat{k}}{\sqrt{5}} = \frac{1}{\sqrt{5}} \hat{j} + \frac{2}{\sqrt{5}} \hat{k}.$