Module 2: Now You Try V

1. Consider the vectors

$\vec{x} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \mbox{ and } \vec{y} = \begin{pmatrix} -3 \\ 4 \\ -1 \end{pmatrix}$

Find a vector perpendicular to both $\vec{x}$ and $\vec{y}$.

Solution:

Compute their cross product:
$\vec{x} \times \vec{y} = \begin{pmatrix} -7 \\ -5 \\ 1 \end{pmatrix}$
It is important to note that this is not the ONLY vector perpendicular to both...

2. Can you determine if

$\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \times \begin{pmatrix} 2 \\ 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$WITHOUT doing the cross product computation?

Solution:

Note that
$\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \\ -2 \end{pmatrix} = 4-2=2 \neq 0.$ So the vectors are not perpendicular. This means that $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ could not have been the result of the dot product.

3. Is there a value of $t$ so that the area of the parallelogram with edges given by

$\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \mbox{ and } \begin{pmatrix} t \\ 0 \\ 0 \end{pmatrix}$is exactly equal to one?

Solution:

We can determine the cross product of the two vectors then take its magnitude to find the area of the parallelogram:
$\left\| \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \times \begin{pmatrix} t \\ 0 \\ 0 \end{pmatrix} \right\| = \left\| \begin{pmatrix} 0 \\ 2t \\ t \end{pmatrix} \right\| = \sqrt{ 5t^2} = 1$ So $t = \frac{\pm 1}{\sqrt{5}}$ should both work.