Lecture 1/28/21

 

Lecture 1/26/21

 

10.4#1

 Vector a <-1,-3,5> Vector b <0, 0, -1> Did the cross product and got <3-0, 1-0, 0-0> = <3, 1, 0> My y-value is apparently wrong. Did I multiply incorrectly? I thought it would have been j(az-cx)

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Looks like you figured it out.  You were forgetting about the - sign in the template 

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10.4#5

The question wants me to find two vectors orthogonal to <5,3,3> and <1,-1,-3>. I first tried the cross-product only to get the vectors <6,-18,8> and it's opposite <-6,18,-8> but after some research I had to include 1/sqrt(6^2+(-18)^2+8^2) = 1/sqrt(424) and multiply it by the vectors so they would come out as 1/sqrt(424)<6,-18,8> and to it's negative counterpart. Why did I have to include the magnitude of the new vector to get the exact answer?

 

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See that part I circled in red on the problem?  You need to divide by the magnitude of the new vector to get a unit vector.

The right hand rule ( section 10.4)

Dear Dr. Taylor

In the picture I provided, I am confused as to how to solve as this does not give any number to calculate. I also searched online and it states to use the right hand rule in which I am having trouble understanding how to know which finger points which direction? 

Regards,

 

10.3#7

It is saying that the scalar projection is incorrect, while it is also saying that the vector projection is correct. I do not understand why it is saying this or why it is even happening?

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Well....in this case the reason it's happening is because you computed the scalar projection incorrectly.

You've correctly computed that the vector projection of b on a is [(a.b)/(a.a)] a=[-11/38]a. But then you tried to use [(a.b)/(a.a)] for the scalar projection, which only works if a is a unit vector. You could get this by replacing a by the unit vector a^ = a/|a| = a/√38; since |a^|=1 this is the same thing as saying that the scalar projection is a^.b which I think is the way we talked about it in class, or what is the same thing as a.b/√38. which is the way the textbook talks about it in the subsection of 10.3 that talks about projections. 

(ps y'all: yah gotta read the textbook. Really you do.)

Lecture 1/21/21

 

Video Part 2 (too large to upload)

Lecture 1/19/21

 

10.2#10

 Hi Professor Can you help me? what am I doing wrong on this question? I started by using pythagorean theorem and arctan for the second

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In the first case the Pythagorean theorem gives that √(18^2+3^2)=√333=18.248 so that if you round to the nearest 0.1 the number you should get 18.2 because the second decimal is less than 5; maybe you rounded first with respect the third decimal place before you rounded with respect to the second one. 

In the second case you gave me so little information about what you did that I really can't help you except to say that clearly you took the arctan of the wrong number. I suggest that you draw a labeled picture of the situation and figure out which angle you'd like to find.

Attendance Last Week

Just so you can see how his goes, here is your attendance last week. Click on the pic to enlarge 

Q10.2#2

Hey professor. I solved this problem, but I wanted to understand it a little bit more. Because I tried earlier to just multiply the vector of <4,-4,1> by 2, giving the vector <8,-8,2>. Since it's just a scaled vector, wouldn't it still be pointing in the same direction? Why would 4/(\sqrt(33)) be the answer instead of 8? Does scaling it by 2 make the vector point in a different direction? 

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Ok, as discussed in the lecture last Thursday and in section 10.2 of the textbook (the top of p. 548 in my copy), a unit vector is a vector of length equal to 1. The way you get a Unit vector u^ pointing in the same direction as a vector u is by multiply u by one over it's magnitude.  Since ||<4,-4,1>||=√[4^2+(-4)^2+1^2]=√(16+16+1)=√33, the unit vector is

u^=(1/√33)u=(1/√33)<4,-4,1>=<4/√33,-4/√33,1/√33>

To a couple other points of confusion: yes, multiplying a vector by *ANY* positive number gives you another vector pointing in the same direction.  Since the magnitude of u is √33, though, 2u has a magnitude of 2√33, which is not 1, so 2u is not a unit vector.