# How do you find the angle between the vectors u=2i-j and v=i-2j?

Nov 2, 2016

Please see the explanation for a description of the process.

#### Explanation:

Compute the dot-product by multiplying the $\hat{i}$ coefficients and then adding the product of the $\hat{j}$ coefficients:

$\overline{u} \cdot \overline{v} = \left(2\right) \left(1\right) + \left(- 2\right) \left(- 1\right) = 4$

A second way to compute the dot-product uses the magnitude of the two vectors and the cosine of the angle between the two vectors:

$\overline{u} \cdot \overline{v} = | \overline{u} | | \overline{v} | \cos \left(\theta\right)$ [1]

We know that the dot-product is 4, therefore, we need to compute the magnitude of the two vectors and then solve for the angle between, $\left(\theta\right)$;

$| \overline{u} | = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{5}$

$| \overline{v} | = \sqrt{{1}^{2} + {\left(- 2\right)}^{2}} = \sqrt{5}$

Substitute into equation [1] and then solve for $\theta$:

$4 = \sqrt{5} \sqrt{5} \cos \left(\theta\right)$

$4 = 5 \cos \left(\theta\right)$

$\cos \left(\theta\right) = \frac{4}{5}$

$\theta = {\cos}^{-} 1 \left(\frac{4}{5}\right)$

theta ~~ 37°

Nov 2, 2016

The angle is =36.9º

#### Explanation:

The angle between two vectors is given by the dot product definition.
vecu.vecv=∥vecu∥*∥vecv∥costheta
where $\theta$ is the angle between the 2 vectors

so, costheta= (vecu.vecv) / (∥vecu∥*∥vecv∥)
The dot product is =〈2,-1〉〈1,-2〉=2+2=4
The modulus of vecu=∥vecu∥=sqrt(4+1)=sqrt5
The modulus of vecv=∥vecv∥=sqrt(1+4)=sqrt5

So, $\cos \theta = \frac{4}{\sqrt{5} \cdot \sqrt{5}} = \frac{4}{5}$
theta=arccostheta=36.9º