The following is going to contain a very compressed version of what I would teach in about 1/4 to 1/2 a semester, so have fun.

**a)** We can write the vectors as

\(\displaystyle \vac{a} = \begin{bmatrix} 7\\-2\\3\end{bmatrix}\)

\(\displaystyle \vac{b} = \begin{bmatrix} 1\\4\\-6\end{bmatrix}\)

So adding them together we get

\(\displaystyle \begin{bmatrix} 7\\-2\\3\end{bmatrix} + \begin{bmatrix} 1\\4\\-6\end{bmatrix} = \begin{bmatrix} 7 + 1\\-2 + 4\\3 - 6\end{bmatrix} =\begin{bmatrix} 8\\2\\-3\end{bmatrix}\)

**b)** The law of cosines states \(\displaystyle \vec{a}\cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta\) where \(\displaystyle \theta\) is the angle between \(\displaystyle \vec{a} \text{ and } \vec{b}\). So to find the angle between two vectors we use the formula \(\displaystyle \theta = \cos^{-1}\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)\).

But, we need to discuss the dot product first. Say we have two vectors

\(\displaystyle \vec{u} = \begin{bmatrix}u_1\\u_2\\u_3\end{bmatrix},\; \vec{v} = \begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}\)

then the dot product is defined as

\(\displaystyle \vec{u}\cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3\).

Therefore, the dot product for our vectors would be

\(\displaystyle \vec{a}\cdot \vec{b} = 7\cdot 1 + (-2)\cdot 4 + 3\cdot (-6) = -19\).

We also need to know how to find the magnitude of our vectors. The magnitude of a vector \(\displaystyle \vec{u} = \begin{bmatrix}u_1\\u_2\\u_3\end{bmatrix}\) is given by

\(\displaystyle |\vec{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2}\).

Therefore, we have

\(\displaystyle |\vec{a}| = \sqrt{7^2 + (-2)^2 + 3^2} = \sqrt{62}\)

\(\displaystyle |\vec{b}| = \sqrt{1^2 + 4^2 + (-6)^2} = \sqrt{53}\)

and so

\(\displaystyle \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{-19}{\sqrt{62}\sqrt{53}} = -0.3314\).

Finally,

\(\displaystyle \theta = \cos^{-1} (-0.3314) = 109^{\circ}\)

**c)** Here we need to discuss cross-products. In my opinion the easiest way to do a cross product is by using the cardinal directions.

First, draw a triangle with one of each \(\displaystyle \vec{i},\; \vec{j},\; \vec{k}\) at each vertex, in that order. Now draw arrows from each vertex going clockwise. This will help you remember the products that I am about to tell you

\(\displaystyle \vec{i} \times \vec{j} = \vec{k}\)

What we did there was start at the \(\displaystyle \vec{i}\) vertex crossed with the \(\displaystyle \vec{j}\) vertex, which then sends us to the \(\displaystyle \vec{k}\) vertex. Notice we went in a clockwise direction. If we traveled in the counter clockwise (anti-clockwise) direction we would have got a negative answer.

Here are a couple more examples

\(\displaystyle \vec{k} \times \vec{i} = \vec{j}\)

\(\displaystyle \vec{j} \times \vec{i} = -\vec{k}\).

The next important thing is that

\(\displaystyle \vec{i}\times\vec{i} = \vec{0}\)

\(\displaystyle \vec{j}\times\vec{j} = \vec{0}\)

\(\displaystyle \vec{k}\times\vec{k} = \vec{0}\).

Now we can compute our cross-product

\(\displaystyle \vec{a}\times \vec{b} = (7\vec{i} - 2 \vec{j} + 3 \vec{k})\times (\vec{i} + 4\vec{j} - 6\vec{k}) \)

Now multiply these out similar to polynomials, except scalars multiply and vectors use cross-product. Also, in this case order of multiplication is important

\(\displaystyle \vec{a}\times \vec{b} = (7\cdot 1) (\vec{i}\times \vec{i}) + (7\cdot 4)(\vec{i}\times \vec{j}) + (7\cdot -6) (\vec{i}\times\vec{k}) \)

\(\displaystyle + (-2\cdot 1)(\vec{j}\times\vec{i}) + (-2\cdot 4)(\vec{j}\times \vec{j}) + (-2\cdot -6)(\vec{j}\times \vec{k}) \)

\(\displaystyle + (3\cdot 1)(\vec{k}\times\vec{i}) + (3\cdot 4)(\vec{k}\times \vec{j}) + (3\cdot -6)(\vec{k}\times \vec{k})\)

\(\displaystyle \vec{a}\times \vec{b} = 7\cdot \vec{0} + 28 \vec{k} + (-42)(-\vec{j}) + (-2)(-\vec{k}) + (-8)\cdot \vec{0} + 12\vec{i} + 3\vec{j} + 12(-\vec{i}) + (-18)\vec{0}\)

The rest of this you should be able to do on your own.