Question

Find the angle between the given vectors to the nearest degree.
a.
110.7
c.
108.4
b.
71.6
d.
112.5

Answer 1

Given:

Given that the two vectors are u = (7,-4) and v = (1,5)

We need to determine the angle between these two vectors.

Dot product of u and v:

The dot product of u and v is given by

$u \cdot v=(7 \times 1)+(-4 \times 5)$

       $=7-20$

$u \cdot v=-13$

Magnitude of u:

The magnitude of u is given by

$\| u \|=\sqrt{(7)^2+(-4)^2}$

$\| u \|=\sqrt{49+16}$

$\| u \|=\sqrt{65}$

Magnitude of v:

The magnitude of v is given by

$\| v \|=\sqrt{(1)^2+(5)^2}$

$\| v \|=\sqrt{1+25}$

$\| v \|=\sqrt{26}$

Angle between the two vectors:

The angle between the two vectors can be determined using the formula,

$cos \ \theta=\frac{u \cdot v}{\| u \| \| v \|}$

Substituting the values, we get;

$cos \ \theta=\frac{-13}{\sqrt{65} \sqrt{26} }$

$cos \ \theta=\frac{-13}{41.11 }$

$cos \ \theta=-0.316$

     $\theta=cos ^{-1}(-0.316)$

     $\theta=108.4^{\circ}$

Thus, the angle between the two vectors is 108.4°