#### Question

Find the angle between the vectors with direction ratios proportional to 1, −2, 1 and 4, 3, 2.

#### Solution

\[\text{Let}\ \vec{a}\ \text{be a vector with direction ratios}\ 1, - 2, 1 . \]

\[ \Rightarrow \vec{a} =\hat{ i} - 2 \hat{j} + \hat {k} . \]

\[\text{Let} \ \vec{b}\ \text{be a vector with direction ratios} \ 4, 3, 2 . \]

\[ \Rightarrow \vec{b} = 4\hat{ i} + 3 \hat{j} + 2 \hat{k} . \]

\[\text{ Let }\ \theta \text{ be the angle between the given vectors } . \]

\[\text{ Now, } \]

\[\text{ cos }\theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} \]

\[ = \frac{\left( \hat{ i } - 2\hat { j } + \ \hat{k} \right) . \left( 4\hat { i } + 3 \ \hat{j}+ 2 \ \hat {k} \right)}{\left| \hat { i } - 2 \ \hat { j } +\ \hat {k} \right|\left| 4 \ \hat { i }+ 3\ \hat { j } + 2 \ \hat { k } \right|}\]

\[ = \frac{4 - 6 + 2}{\sqrt{1 + 4 + 1} \sqrt{16 + 9 + 4}} \]

\[ = \frac{0}{\sqrt{6} \sqrt{29}} \]

\[ = 0 \]

\[ \therefore \theta = \frac{\pi}{2}\]

\[\text{Thus, the angle between the given vectors measures }\frac{\pi}{2} .\]