Respuesta :
Explanation
let
[tex]\begin{gathered} u=-2\mleft(\cos 30\degree i+\sin 30\degree j\mright) \\ v=6\mleft(\cos 225\degree i+\sin 225\degree j\mright) \\ w=8\mleft(\cos 120\degree i+\sin 120\degree j\mright) \end{gathered}[/tex]Step 1
Part A)Find −7u • v.
the dot product is the sum of the products of the corresponding entries of the two sequences of numbers,and to multiply a vector by a scalar, multiply each component by the scalar.
[tex](ai+bj)\cdot(ci+dj)=ac+bd[/tex]so
[tex]\begin{gathered} -7u\cdot v \\ \text{hence} \\ -7u=-7(-2\mleft(\cos 30\degree i+\sin 30\degree j\mright))=14(\cos 30\degree i+\sin 30\degree j)=12.12i+7j \end{gathered}[/tex]therefore, replacing
[tex]\begin{gathered} -7u\cdot v \\ (12.12i+7j)\cdot6\mleft(\cos 225\degree i+\sin 225\degree j\mright) \\ (12.12i+7j)\cdot(-4.24i-4.24j) \\ (12.12i+7j)\cdot(-4.24i-4.24j)=(12.12\cdot-4.24)+(7\cdot-4.24)= \\ (12.12i+7j)\cdot(-4.24i-4.24j)=-51.43-29.690=-81.12 \end{gathered}[/tex]So,Part A
[tex]-7u\cdot v=-81.12[/tex]Step 2
Part B ,Use the dot product to determine if u and w are parallel, orthogonal, or neither
the dot product is also given by:
[tex]\begin{gathered} a\cdot b=\lvert a\rvert\lvert b\rvert\cos \theta \\ so \\ \cos \theta=\frac{a\cdot b}{\lvert a\rvert\lvert b\rvert} \end{gathered}[/tex]a) find the magnitude of the vectors
[tex]\begin{gathered} \text{ }u=-2\mleft(\cos 30\degree i+\sin 30\degree j\mright), \\ u=(-2\cos 30\degree i-2\sin 30\degree j) \\ \text{hence} \\ \lvert u\rvert=\sqrt[]{(-2\cos30)^2+(-2\sin30\degree)}^2 \\ \lvert u\rvert=\sqrt[]{3^{}+1\text{ }} \\ \lvert u\rvert=\sqrt[]{4}=2 \\ \lvert u\rvert=2 \end{gathered}[/tex]and
[tex]\begin{gathered} \text{ w}=8\mleft(\cos 120\degree i+\sin 120\degree j\mright) \\ w=8(\cos 120\degree i+\sin 120\degree j)=-4i+6.92j \\ \text{hence} \\ \lvert w\rvert=\sqrt[]{(-4)^2+(6.92)^2} \\ \lvert w\rvert=\sqrt[]{16^{}+48} \\ \lvert w\rvert=\sqrt[]{64} \\ \lvert w\rvert=8 \end{gathered}[/tex]c) dot product
[tex]\begin{gathered} u\cdot w=-2(\cos 30\degree i+\sin 30\degree j)\cdot8(\cos 120\degree i+\sin 120\degree j) \\ u\cdot w=-5.73+5.92 \\ u\cdot w=0.19 \end{gathered}[/tex]finally, replace in the formula and find the angle
[tex]\begin{gathered} \cos \theta=\frac{a\cdot b}{\lvert a\rvert\lvert b\rvert} \\ \cos \theta=\frac{0.19}{2\cdot8} \\ \cos \theta=0.011875 \\ \theta=\cos ^{-1}(0.011875) \\ \theta=90 \end{gathered}[/tex]so
the vectors are orthogonal ( make a 90° angle)
I hope this helps you
