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

The correct answer is: $θ=cos^{-1}(\frac{5}{7})$
The given vectors are $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}+\hat{k}$
$|\vec{a}|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$
$|\vec{b}|=\sqrt{3^2+(-2)^2+1^2}=\sqrt{9+4+1}=\sqrt{14}$
Now,$\vec{a}.\vec{b}=(\hat{i}-2\hat{j}+3\hat{k})(3\hat{i}-2\hat{j}+\hat{k})$
$=1.3+(-2)(-2)+3.1$
$=3+4+3$
$=10$
Also,we know that $\vec{a}.\vec{b}=|\vec{a}||\vec{b}|cosθ$
$∴10=\sqrt{14}\sqrt{14}cosθ$
$⇒cosθ=\frac{10}{14}$
$⇒θ=cos^{-1}(\frac{5}{7})$