Answer:
[tex]90^{\circ}[/tex]. In other words, these two vectors are perpendicular (orthogonal) to one another.
Step-by-step explanation:
Let [tex]\|a\|[/tex] and [tex]\|b\|[/tex] denote the magnitudes of vector [tex]a[/tex] and vector [tex]b[/tex]. The dot product between these two vectors is represented as [tex]a^{T}\, b[/tex]. Let [tex]\theta[/tex] denote the angle between the two vectors. The cosine of this angle would be equal to:
[tex]\begin{aligned}\cos(\theta) &= \frac{a^{T}\, b}{\|a\| \|b\|}\end{aligned}[/tex].
The dot product between vector [tex]a = \langle 1,\, 2\rangle[/tex] and [tex]b = \langle 1,\, -1/2\rangle[/tex] is:
[tex]\begin{aligned}a^{T}\, b &= {\begin{bmatrix}1 \\ 2\end{bmatrix}}^{T}\, \begin{bmatrix}1 \\ -1/2\end{bmatrix} \\ &= 1 \times 1 + 2 \times \left(-\frac{1}{2}\right) \\ &= 0\end{aligned}[/tex].
The magnitudes of the two vectors are:
[tex]\begin{aligned}\| a \| &= \sqrt{1^{2} + 2^{2}} \\ &= 5\end{aligned}[/tex].
[tex]\begin{aligned}\| b \| &= \sqrt{1^{2} + \left(-\frac{1}{2}\right)^{2}} \\ &= \frac{1}{2}\, \sqrt{5}\end{aligned}[/tex].
Therefore:
[tex]\begin{aligned}\cos(\theta) &= \frac{a^{T}\, b}{\|a\| \|b\|} \\ &= \frac{0}{5 \times (\sqrt{5} / 2)} \\ &= 0\end{aligned}[/tex].
Among all angles between [tex]0^{\circ}[/tex] and [tex]180^{\circ}[/tex], the only angle with a cosine of [tex]0[/tex] is [tex]90^{\circ}[/tex]. Therefore, the angle between vector [tex]a[/tex] and vector [tex]b[/tex] must be [tex]90^{\circ}\![/tex]. Hence, these two vectors are perpendicular to one another.
