Question

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

Answer 1

Let $a=\hat{i}-2\hat{j}+3\hat{k}$ and $b=3\hat{i}-2\hat{j}+\hat{k}$
Magnitude of a, $|a|=\sqrt{1^2+(-2{)}^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$
Magnitude of b, $|b|=\sqrt{3^2+(-2{)}^2+1^2}=\sqrt{9+4+1}=\sqrt{14}$
Now,$a.b=(\hat{i}-2\hat{j}+3\hat{k}).(3\hat{i}-2\hat{j}+\hat{k})$
$=1.3+(-2).(-2)+3.1=3+4+3=10$
(Dot product of two vectors is equal to the sum of the products of their corresponding components)
Let $\theta$ be the required angle between a and b, then
$cos\theta =\frac{a.b}{|a||b|}=\frac{10}{\sqrt{14}\sqrt{14}}=\frac{10}{14}=\frac{5}{7}\Rightarrow \theta =co{s}^{-1}\left(\frac{5}{7}\right)$