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a) You will need vector addition, which is just adding vectors componentwise.
b) Use the law of cosines
c) Use the formula $\displaystyle \frac{1}{2}|\vec{a} \times \vec{b}|$ where $\displaystyle \times$ is the cross-product and $\displaystyle |\qquad|$ is the magnitude.
d) Use the formula $\displaystyle \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$ where $\displaystyle \cdot$ is the dot-product.
If you don't understand those terms you will need to look them up. I believe your instructor intends for you to learn this on your own, so I won't be able to help you any further without an attempt at a solution.
You don't need Cramer's Rule. $\displaystyle \vec{i},\;\vec{j},\;\vec{k}$ are called cardinal directions and represent components of a vector.
$\displaystyle \vec{i} = \begin{bmatrix}1\\0\\0\end{bmatrix},\;\vec{j} = \begin{bmatrix}0\\1\\0\end{bmatrix},\;\vec{k} = \begin{bmatrix}0\\0\\1\end{bmatrix}$.
I would suggest using google or the search in the forums and look up the keywords I mentioned in my first post.
Would you call this a difficult problem?
I'm taking my final tomorrow and I'm having no luck in getting a grasp on these vector problems :/ :/ I'd greatly appreciate this problem worked out, or explained, I have others on an assignment sheet I want to model it after.
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.