question_answer

            Find the angle between the vectors \[\hat{i}-2\hat{j}+3\hat{k}\] and \[3\hat{i}-2\hat{j}+\hat{k}.\]

Answer:

We know that the angle 6 between two vectors \[\vec{a}\]             And \[\vec{b}\] is given by                         \[\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\]                          ? (i) Here, let \[\vec{a}=\hat{i}-2\hat{j}+3\hat{k}\] and \[\vec{b}=3\hat{i}-2\hat{j}+\hat{k}\] Then, \[\vec{a}\cdot \vec{b}=(\hat{i}-2\hat{j}+\hat{k})\cdot (3\hat{i}-2\hat{j}+\hat{k})\]             \[=(1\cdot 3)+(-\,2)(-\,2)+(3\cdot 1)\]             \[=3+4+3=10\] \[\therefore \]      \[|\vec{a}|\,=\,\sqrt{1+4+9}=\sqrt{14}\] And      \[|\vec{b}|\,=\,\sqrt{9+4+1}=\sqrt{14}\] On putting these values in Eq. (i), we get             \[\cos \theta =\frac{10}{\sqrt{14}\sqrt{14}}=\frac{10}{14}=\frac{5}{7}\] \[\therefore \]      \[\theta ={{\cos }^{-1}}\left( \frac{5}{7} \right)\]