Question

Determine which set of vectors has an angle of 90° between them.

Answer 1

Explanation:

The angle between two vectors is represented by the following equation:

`\cos \theta=\frac{\vec{u}\cdot\vec{v}}{\lvert\vec{u}\rvert\lvert\vec{v}\rvert}`

Notice that this equation involves a trigonometric function, the dot product of two vectors, and the magnitude of two vectors.

For v=<10,2>, w=<-5,-1>:

As a first step let's determine the dot product of the two vectors:

`\begin{gathered} \vec{v}\cdot\vec{w}=10\cdot-5+2\cdot-1 \\ \vec{v}\cdot\vec{w}=-52 \end{gathered}`

Then, calculate the magnitudes of the vectors:

`\begin{gathered} \lvert\vec{v}\rvert=\sqrt[]{(10)^2+(2)^2}=2\sqrt[]{26} \\ \lvert\vec{w}\rvert=\sqrt[]{(-5)^2+(-1)^2}=\sqrt[]{26} \end{gathered}`

Now, substitute the values into the equation:

`\begin{gathered} \cos \theta=\frac{-52}{2\sqrt[]{26}\cdot\sqrt[]{26}} \\ \cos \theta=-1 \\ \theta=\cos ^(-1)(-1)=\text{ 180\degree} \end{gathered}`

For v=<3,1>, w=<2,-6>:

`\begin{gathered} \vec{v}\cdot\vec{w}=3\cdot2+1\cdot-6 \\ \vec{v}\cdot\vec{w}=0 \\ \lvert\vec{v}\rvert=\sqrt[]{(3)^2+(1)^2}=\sqrt[]{10} \\ \lvert\vec{w}\rvert=\sqrt[]{(2)^2+(-6)^2}=2\sqrt[]{10} \\ \cos \theta=\frac{0}{2\sqrt[]{10}\cdot\sqrt[]{10}} \\ \theta=\cos ^(-1)(0)=\text{ 90\degree} \end{gathered}`

For v=<-10,5>, w=<1,-2>: