Answer:
The angle between vector [tex]\vec{u} = 5\, \vec{i} - 8\, \vec{j}[/tex] and [tex]\vec{v} = 5\, \vec{i} + \, \vec{j}[/tex] is approximately [tex]1.21[/tex] radians, which is equivalent to approximately [tex]69.3^\circ[/tex].
Step-by-step explanation:
The angle between two vectors can be found from the ratio between:
To be precise, if [tex]\theta[/tex] denotes the angle between [tex]\vec{u}[/tex] and [tex]\vec{v}[/tex] (assume that [tex]0^\circ \le \theta < 180^\circ[/tex] or equivalently [tex]0 \le \theta < \pi[/tex],) then:
[tex]\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}[/tex].
The first component of [tex]\vec{u}[/tex] is [tex]5[/tex] and the first component of [tex]\vec{v}[/tex] is also
The second component of [tex]\vec{u}[/tex] is [tex](-8)[/tex] while the second component of [tex]\vec{v}[/tex] is [tex]1[/tex]. The product of these two second components is [tex](-8) \times 1= (-8)[/tex].
The dot product of [tex]\vec{u}[/tex] and [tex]\vec{v}[/tex] will thus be:
[tex]\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}[/tex].
Apply the Pythagorean Theorem to both [tex]\vec{u}[/tex] and [tex]\vec{v}[/tex]:
Let [tex]\theta[/tex] represent the angle between [tex]\vec{u}[/tex] and [tex]\vec{v}[/tex]. Apply the formula[tex]\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}[/tex] to find the cosine of this angle:
[tex]\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}[/tex].
Since [tex]\theta[/tex] is the angle between two vectors, its value should be between [tex]0\; \rm radians[/tex] and [tex]\pi \; \rm radians[/tex] ([tex]0^\circ[/tex] and [tex]180^\circ[/tex].) That is: [tex]0 \le \theta < \pi[/tex] and [tex]0^\circ \le \theta < 180^\circ[/tex]. Apply the arccosine function (the inverse of the cosine function) to find the value of [tex]\theta[/tex]:
[tex]\displaystyle \cos^{-1}\left(\frac{17}{\sqrt{89}\cdot \sqrt{26}}\right) \approx 1.21 \;\rm radians \approx 69.3^\circ[/tex] .
