# SOLVEDVectors - help needed

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#### lvleph

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.

#### wiseguy

Is there a good page or something that defines this specific topic in vectors?

#### wiseguy

Is it as simple as adding the two equations together?

#### wiseguy

I attempted to solve as a system of equations with cramers rule with no success. :/

#### lvleph

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.

#### wiseguy

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.

#### Plato

MHF Helper
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.
I tell you this having taught this material for 30+ years.
If your exam is tomorrow, you are wasting time trying to learn this material overnight.
Spend your time on what you think the exam will cover that you know.

#### lvleph

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.

wiseguy