SOLVED Vectors - help needed

Mar 2009
378
68
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.
 
Jul 2010
101
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Is there a good page or something that defines this specific topic in vectors?
 
Jul 2010
101
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Is it as simple as adding the two equations together?
 
Jul 2010
101
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I attempted to solve as a system of equations with cramers rule with no success. :/
 
Mar 2009
378
68
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.
 
Jul 2010
101
0
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
Aug 2006
22,507
8,664
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.
 
Mar 2009
378
68
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.
 
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