Given that x = \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\) calculate, correct to the nearest degree, the angle between the vectors
Use \(\cos\theta=\dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|\,|\mathbf{y}|}\) with \(\mathbf{x}=\begin{pmatrix}-4\\3\end{pmatrix},\ \mathbf{y}=\begin{pmatrix}-9\\15\end{pmatrix}\).
Dot product: \(\mathbf{x}\cdot\mathbf{y}=(-4)(-9)+(3)(15)=36+45=81\).
Magnitudes: \(|\mathbf{x}|=\sqrt{(-4)^{2}+3^{2}}=\sqrt{25}=5\); \(|\mathbf{y}|=\sqrt{(-9)^{2}+15^{2}}=\sqrt{306}\approx17.49\).
\[\cos\theta=\frac{81}{5\times17.49}=\frac{81}{87.46}=0.9261\]
\[\theta=\cos^{-1}(0.9261)\approx22^{\circ}\ \text{(nearest degree)}\]