Find the angle (in degrees) between vectors $\mathbf{a}=5\mathbf{i}-2\mathbf{j}+\mathbf{k}$ and $\mathbf{b}=-3\mathbf{i}+4\mathbf{j}+7\mathbf{k}$.

Step-by-step solution

1. Calculate the dot product: $5(-3)+(-2)(4)+1(7)=-15-8+7=-16$.
2. Find the magnitude of $\mathbf{a}$: $\sqrt{5^2+(-2{)}^2+1^2}=\sqrt{30}approx5.477$.
3. Find the magnitude of $\mathbf{b}$: $\sqrt{(-3{)}^2+4^2+7^2}=\sqrt{74}approx8.602$.
4. Compute $cos\theta$ using $\frac{\mathbf{a}cdotmathbfb}{|\mathbf{a}||\mathbf{b}|}=\frac{-16}{5.477\times 8.602}\approx -0.3399$.
5. Take $\theta =co{s}^{-1}(-0.3399)\approx 110.2^{\circ }$.
6. Rounded to the nearest integer, the angle is $110^{c}irc$.

Concept tested: Angle between two vectors using dot product

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked