How do you find the angle between the vectors u=<1,0> and v=<0,-2>?

Jan 2, 2017

Explanation:

Two methods to dot-product to compute the dot product are:

$\text{[1] } \overline{u} \cdot \overline{v} = \left({u}_{x}\right) \left({v}_{x}\right) + \left({u}_{y}\right) \left({v}_{y}\right) + \ldots$

Where ${u}_{x}$ and ${v}_{x}$ are the x components, ${u}_{y}$ and ${v}_{y}$ are the y components, and so on.

$\text{[2] } \overline{u} \cdot \overline{v} = | \overline{u} | | \overline{v} | \cos \left(\theta\right)$

where the magnitudes are $| \overline{u} | = \sqrt{{u}_{x}^{2} + {u}_{y}^{2} + \ldots}$ and $| \overline{v} | = \sqrt{{v}_{x}^{2} + {v}_{y}^{2} + \ldots}$ and $\theta$ is the angle between the two vectors.

To do this problem, you compute the dot-product using method [1]:

$\overline{u} \cdot \overline{v} = \left(1\right) \left(0\right) + \left(0\right) \left(- 2\right)$

$\overline{u} \cdot \overline{v} = 0$

This is a special case where the dot-product is zero and we instantly know that $\theta = \frac{\pi}{2}$ but let's proceed as, if it were not this special case.

Next, compute the magnitudes:

$| \overline{u} | = \sqrt{{1}^{2} + {0}^{2}}$

$| \overline{u} | = 1$

$| \overline{v} | = \sqrt{{0}^{2} + {2}^{2}}$

$| \overline{v} | = 2$

Into the equation for method [2], substitute 0 for $\overline{u} \cdot \overline{v}$, 1 for $| \overline{u} |$, and 2 for $| \overline{v} |$:

$0 = \left(1\right) \left(2\right) \cos \left(\theta\right)$

Solve the equation for $\theta$

$\cos \left(\theta\right) = 0$

$\theta = \frac{\pi}{2}$

Jan 2, 2017

$\frac{\pi}{2} \mathmr{and}$ (, for the opposite sense of measurement, ) $\frac{3}{2} \pi$

Explanation:

The dot ( scalar ) product

$u . v = < 1 , 0 > . < 0 , - 2 \ge \left(1\right) \left(0\right) + \left(0\right) \left(- 2\right) = 0$..

So, the cosine of the angle, $\frac{u . v}{| u | | v |}$ is 0.

In $\theta \in \left[0 , 2 \pi\right]$, the angle is $\frac{\pi}{2}$, in the clockwise sense,

from $v \to u$, and $\frac{3}{2} \pi$, for measurement in the opposite sense.