A surveyor records two displacement vectors a and b with lvert arvert = 3 metres and lvert brvert =…

Step 1. Expand. $(\vec a + 2\vec b)\cdot(\vec a - \vec b) = \vec a\cdot\vec a - \vec a\cdot\vec b + 2\vec b\cdot\vec a - 2\vec b\cdot\vec b = \lvert\vec a\rvert^2 + \vec a\cdot\vec b - 2\lvert\vec b\rvert^2$.

Step 2. Put in the numbers. $9 + \vec a\cdot\vec b - 8 = 0$, so $\vec a\cdot\vec b = -1$ square metre.

Step 3. The angle. $\cos\theta = \dfrac{-1}{3 \times 2} = -\dfrac{1}{6}$, so $\theta = \cos^{-1}\left(-\dfrac{1}{6}\right) \approx 99.59^\circ$.