Question

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

Answer 1

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

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) &= (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 * 1 + 2 * \left(-(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(-(1)/(2)\right)^(2)} \\ &= (1)/(2)\, √(5)\end{aligned}`.

Therefore:

`\begin{aligned}\cos(\theta) &= (a^(T)\, b)/(\|a\| \|b\|) \\ &= (0)/(5 * (√(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.