# Consider the following vectors: v = 3i + 4j, w = 4i + 3j, how do you find the angle between v and w?

Nov 5, 2015

I found: $\theta = {16.26}^{\circ}$

#### Explanation:

We could try using the SCALAR PRODUCT between them:

$\vec{v} \cdot \vec{w} = | \vec{v} | | \vec{w} | \cos \left(\theta\right) = {v}_{x} {w}_{x} + {v}_{y} {w}_{y}$

Where you have that the scalar product is equal to the product of the modulii times the cosine of the angle between the vectors AND equal to the sum of the products between the components of the two vectors, or:

$| \vec{v} | | \vec{w} | \cos \left(\theta\right) = {v}_{x} {w}_{x} + {v}_{y} {w}_{y}$

Now:

$| \vec{v} | = \sqrt{{3}^{2} + {4}^{2}} = \sqrt{25} = 5$
$| \vec{w} | = \sqrt{{4}^{2} + {3}^{2}} = \sqrt{25} = 5$
${v}_{x} {w}_{x} + {v}_{y} {w}_{y} = 3 \cdot 4 + 4 \cdot 3$

so:
$5 \cdot 5 \cdot \cos \left(\theta\right) = 3 \cdot 4 + 4 \cdot 3$
$25 \cos \left(\theta\right) = 24$
$\theta = a r \cos \left(\frac{24}{25}\right) = {16.26}^{\circ}$