If $\mathbf{a}=\mathbf{i}$ and $\mathbf{b}=\mathbf{i}+\mathbf{j}$, the angle between $\mathbf{a}$ and $\mathbf{b}$ is:

Step-by-step solution

1. Express the vectors in component form: $\mathbf{a}=(1,0,0)$, $\mathbf{b}=(1,1,0)$.
2. Find the dot product: $\mathbf{a}\cdot \mathbf{b}=1\cdot 1+0\cdot 1+0\cdot 0=1$.
3. Find the magnitudes: $|\mathbf{a}|=\sqrt{1^2+0^2+0^2}=1$, $|\mathbf{b}|=\sqrt{1^2+1^2+0^2}=\sqrt{2}$.
4. Use $\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{1}{\sqrt{2}}$.
5. Therefore $\theta =\arccos \left(\frac{1}{\sqrt{2}}\right)=45^{\circ }$.
6. Select the option that matches: Option (B) $45^{\circ }$.

Concept tested: Angle between two vectors

Published 8 September 2026 · How this solution was prepared and checked