Given $\mathbf{a}=2\mathbf{i}+3\mathbf{j}-\mathbf{k}$ and $\mathbf{b}=-\mathbf{i}+4\mathbf{j}+2\mathbf{k}$, find the angle between them in degrees (nearest integer).

Step-by-step solution

1. Write $\mathbf{a}=(2,,3,,-1)$ and $\mathbf{b}=(-1,,4,,2)$.
2. Compute dot product: $\mathbf{a}cdotmathbfb=2(-1)+3(4)+(-1)(2)=8$.
3. Compute magnitudes: $|\mathbf{a}|=\sqrt{2^2+3^2+(-1{)}^2}=\sqrt{14}$ and $|\mathbf{b}|=\sqrt{(-1{)}^2+4^2+2^2}=\sqrt{21}$.
4. Use $cos\theta =\frac{\mathbf{a}cdotmathbfb}{|\mathbf{a}||\mathbf{b}|}=\frac{8}{\sqrt{14},\sqrt{21}}=\frac{8}{\sqrt{294}}\approx 0.4669$.
5. Take $\theta =\arccos (0.4669)\approx 62.2^{\circ }$, round to $62^{\circ }$.
6. Hence the angle between $\mathbf{a}$ and $\mathbf{b}$ is $62^{c}irc$.

Concept tested: Angle between vectors

Published 31 August 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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