# Vectors

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• Jul 30th 2011, 09:35 PM
Veronica1999
Vectors
1. Let A= (-6,-4), B=(3,2) and c=(6,4)

(a) These points lie on a line through the origin. Find its slope.

2/3

(b) Let u be the vector whose components are the x-coordinates of A,B and C, and let v be the vector whose components are the y- coordinates of A,B,and C. Show that u and v point in the same direction.

u = (-6,3,6) v= (-4,2,4)

3u = 2v

(c) Explain why the ratio of l v l to l u l is the slope you found in part (a).

36/81 = 2/3

I can see that the ratio of the magnitudes of the vectors are the same as the slope in (a) but I am not sure what the problem wants me to be seeing.....(Thinking)

(d) What would the vectors u and v have looked like if A,B,and C had not been collinear with the origin? Hmmm... (Thinking) They would be skew vectors?

2. Let A= (-3,-2) B= (-1,-1) C= (4,3), u=[-3,-1,4] and v=[-2,-1,3]

(a) Show that u and v point in different directions. Let w be the vector that results when v is projected onto u. Show that w is approximately [-2.19,-0.73,2.92]

cosx= u.v/uu therefore 19/26 = 0.73

0.73[-3, -1,4] = [ -2.19, -0.73, 2.92]

(b)Make a scatter plot. Verify that A,B, and C are not collinear. Notice that the x-coordinates of these points are the components of u and the y coordinates are the components of v, suggesting that u and v are like lists in your calculator.

I made a scatter plot and I see that the x,y coordinates are the components of each vector but what does like lists in your calculator mean?

(c) Verify that the points A' = (-3,-2.19) B' = (-1,-0.73) and C' =(4,2.92) lie on a line that goes through the origin. Notice that the y-coordinates of A',B' and C' appear as the components of w , and that they are proportional to the components of u.

I understand this part.

(d) You calculated w by first finding that it is m times as long as u, where m is uv/uu. Notice that m is also the slope of the the line through A',B' and C'. Now use your calculator to find an equation for the so called regression line(or LinReg)for the data points A,B,and C. The slope should look familiar.

This problem threw me off track because now I have to study
(Regression and Correlation relying mostly on google).
I am slowly starting to see the relation between the regression line and vectors but this will take some time. Can I get some insights, if possible?

(b) Verify that the vector r=v-w is perpendicular to u, then explain why this should have been expected. Is is customary to call r a residual vector, because it is really just a list of residuals. Review the meaning of this data analysis term if you need to.

r is perpendicular to u because the vector v-w is a projection of v onto u.
But why is it called a residual vector?

(c) The regression line is sometimes called the least squares line of best fit, because w was chosen to make r as short as possible. Explain this terminology. You will need to refer to the Pythagorean formula for calculating the length of a vector.

I think this is related to the error sum of squares and I think I understand SSE but i am not sure how to relate this to vectors.

I find this problem absolutely fascinating. Only if I could understand more....
I will be studying this problem for the next few weeks so any help will be really appreciated.

Thanks.

• Jul 30th 2011, 10:52 PM
earboth
Re: Vectors
Quote:

Originally Posted by Veronica1999
1. Let A= (-6,-4), B=(3,2) and c=(6,4)

(a) These points lie on a line through the origin. Find its slope.

2/3

(b) Let u be the vector whose components are the x-coordinates of A,B and C, and let v be the vector whose components are the y- coordinates of A,B,and C. Show that u and v point in the same direction.

u = (-6,3,6) v= (-4,2,4)

3u = 2v

(c) Explain why the ratio of l v l to l u l is the slope you found in part (a).

sqrt(36)/sqrt(81) = 2/3 <-- you forgot to calculate the square-root

I can see that the ratio of the magnitudes of the vectors are the same as the slope in (a) but I am not sure what the problem wants me to be seeing.....(Thinking)

...

to a): (Yes)

to b) Let m denotes the common slope of the vectors. Then

$\vec v = m \cdot \vec u$ . That means: $\vec v = \langle v_1, v_2, v_3 \rangle = \langle m \cdot u_1, m \cdot u_2, m \cdot u_3 \rangle$

to c) You are asked to calculate: $\dfrac{|\vec v|}{|\vec u|}=\dfrac{\sqrt{m^2 \cdot u_1^2+ m^2 \cdot u_2^2+ m^2 \cdot u_3^2}}{\sqrt{ u_1^2+ u_2^2+ u_3^2}}=\dfrac{\sqrt{m^2} \cdot |\vec u|}{|\vec u|} = m$

to d) Draw a sketch, do the same calculations as in a) to c) and look what happens with the results.