Given the following vectors:
[tex]\begin{gathered} u=-4\mleft(\cos 60\degree i+\sin 60\degree j\mright) \\ v=8\mleft(\cos 135\degree i+\sin 135\degree j\mright) \\ w=12\mleft(\cos 150\degree i+\sin 150\degree j\mright) \end{gathered}[/tex]We will find the following:
a) -7u • v
The dot product of two vectors will be as follows:
[tex]A\cdot B=A_xB_x+A_yB_y[/tex]So, for the given product, the answer will be as follows:
[tex]\begin{gathered} -7u\cdot v=-7(-4)(8)(\cos 60\cdot\cos 135+\sin 60\cdot\sin 135) \\ =224(\frac{1}{2}\cdot(-\frac{1}{\sqrt[]{2}})+\frac{\sqrt[]{3}}{2}\cdot\frac{1}{\sqrt[]{2}}) \\ \\ =224(-\frac{1}{2\sqrt[]{2}}+\frac{\sqrt[]{3}}{2\sqrt[]{2}})\cdot\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ \\ =224(-\frac{\sqrt[]{2}}{4}+\frac{\sqrt[]{6}}{4})=56(\sqrt[]{6}-\sqrt[]{2}) \end{gathered}[/tex]Part B: Use the dot product to determine if u and w are parallel, orthogonal, or neither
The vectors will be parallel if the dot product = the product of the magnitudes which means the angle between the vectors = 0 or 180
And the vectors will be orthogonal of the dot product = 0
This means the angle between them = 90
The dot product of the vectors (u) and (w) will be as follows:
[tex]\begin{gathered} u\cdot w=(-4)(12)(\cos 60\cos 150+\sin 60\sin 150) \\ =(-48)(\frac{1}{2}\cdot\frac{-\sqrt[]{3}}{2}+\frac{\sqrt[]{3}}{2}\cdot\frac{1}{2}) \\ \\ =(-48)(\frac{-\sqrt[]{3}}{4}+\frac{\sqrt[]{3}}{4})=(-48)(0)=0 \end{gathered}[/tex]So, as the result of the dot product = 0
The vectors (u) and (w) are Orthogonal.
