# Show that each of the given three vectors is a unit vector:

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.

Solution:

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

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

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

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

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

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

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

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

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

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

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