Question

Find the angle between vectors a=(1,2) and b=(1,-1/2).

Answer 1

Answer:

$90^{\circ}$. In other words, these two vectors are perpendicular (orthogonal) to one another.

Step-by-step explanation:

Let $\|a\|$ and $\|b\|$ denote the magnitudes of vector $a$ and vector $b$. The dot product between these two vectors is represented as $a^{T}\, b$.  Let $\theta$ denote the angle between the two vectors. The cosine of this angle would be equal to:

$\begin{aligned}\cos(\theta) &= \frac{a^{T}\, b}{\|a\| \|b\|}\end{aligned}$.

The dot product between vector $a = \langle 1,\, 2\rangle$ and $b = \langle 1,\, -1/2\rangle$ is:

$\begin{aligned}a^{T}\, b &= {\begin{bmatrix}1 \\ 2\end{bmatrix}}^{T}\, \begin{bmatrix}1 \\ -1/2\end{bmatrix} \\ &= 1 \times 1 + 2 \times \left(-\frac{1}{2}\right) \\ &= 0\end{aligned}$.

The magnitudes of the two vectors are:

$\begin{aligned}\| a \| &= \sqrt{1^{2} + 2^{2}} \\ &= 5\end{aligned}$.

$\begin{aligned}\| b \| &= \sqrt{1^{2} + \left(-\frac{1}{2}\right)^{2}} \\ &= \frac{1}{2}\, \sqrt{5}\end{aligned}$.

Therefore:

$\begin{aligned}\cos(\theta) &= \frac{a^{T}\, b}{\|a\| \|b\|} \\ &= \frac{0}{5 \times (\sqrt{5} / 2)} \\ &= 0\end{aligned}$.

Among all angles between $0^{\circ}$ and $180^{\circ}$, the only angle with a cosine of $0$ is $90^{\circ}$. Therefore, the angle between vector $a$ and vector $b$ must be $90^{\circ}\!$. Hence, these two vectors are perpendicular to one another.