Module 2: Now You Try III

Consider the vectors

$\vec{x} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \mbox{ and } \vec{y} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$

1. Find the angle between vectors $\vec{x}$ and $\vec{y}$

Solution:

$\vec{x} \cdot \vec{y} = -7$. Also $\|\vec{x}\| = \sqrt{2}$ and $\|\vec{y} \| = 5$. So:
$\cos (\theta) = \frac{ | -7|}{(\sqrt{2}(5)} \Rightarrow \theta = \cos^{-1} \left(\frac{7\sqrt{2}}{10} \right).$
Once could use technology to find an approximate angle.

2. Are the vectors $\vec{x}$ and $\vec{y}$ perpendicular?

Solution:

3. Can you find a vector perpendicular to both $\vec{x}$ and $\vec{y}$?

Solution:

Suppose that the vector $\vec{z} = \begin{pmatrix}a//b\end{pmatrix}$is perpendicular to both $\vec{x}$ and $\vec{y}$. Then we would get zero by dotting each case. So we can create the following equations:
$\vec{z} \cdot \vec{x} = a-b = 0 \mbox{ and } \vec{z} \cdot \vec{y} = -3a+4b=0.$The only solution to these equations is $a=b=0$. Thus the only vector perpendicular to both is the zero vector. In particular: there is no non-zero vector perpendicular to both $\vec{x}$ and $\vec{y}$.