Hey guys for number 59 is the force at the bottom the magnitude or what?
Can someone explain how to identify vectors that are parallel? I'm looking at the explanation of them on page 609 but not understanding anything from it. I get the orthogonal one though...
Chandan, You basically just have to plug in the vectors that they give you into the formula ((v*w)/(llvll*llwll)). Thats in the case that they give you the values for vectors v and w. If you plug it in the formula and your resulting answer is 1, then the two vectros are parallel. To find the magnitude of either one of them you just have to use the square root of a squared plus b squared, and you find a and b by using the coefficients of i and j in the equation for the vector.
How would you approach the first word problem in the 9.5 homework...I am getting completely lost and it's confusing me even more if I retry it (them). Please help if you can.
So in class we had made an attempt of number 19 from 9.5, and we got the answer. I'm somewhat sure about the process of the problem, but how do you know what Va, and Vw are? You need to know these 2 to start the problem off, but I'm not clear as to how they are found...
for number 59 in 9.4 the force at the bottom is the amount pulling down. Since the problem is a static equilibrium problem, the weight would represent the F(3) equation
For static equilibrium F(1) + F(2) + F(3) = 0
in this case F(3)= -1000j because the weight is pulling 1000 pounds down in the vertical direction, represented by the j component
Chandan, you can find Va and Vw using the formula v = ||v|| (cos theta i + sin theta j). So to find Va and Vw you just plug in numbers given in the equation. So to find Va, ||v|| = 550 and theta = 225 degrees. The ||v|| is just the speed and to find theta, if you set up a graph for the vectors, you would draw a ray extending from the origin in the SW direction (or into the III quadrant). If you measured this angle, starting at 0, you would get 225 for the angle measure. Then you just plug the numbers into the equation. So Va = 550 (cos 225 i + sin 225 j). Using your calculator and after some simplifying, you get that Va = -388.909i – 388.909j. So you use the same process to find Vw. ||v|| = 80 because that is the speed. If the jet stream comes from the west, it means it is moving east. If you graph this, the ray falls right on the x-axis moving in the positive direction. So theta = 0. Then, plugging in the numbers you get: Vw = 80 (cos 0 i + sin 0 j). Then: Vw = 80 (1i + 0j). Distributive property and Vw = 80i.
I don't know if this material will be covered in 9.6 or 9.7 later this week, but I was doing the review problems at the end of the chapter, and on page 632, number 79 and 80, what does the "x" mean?
It can't be multiplication; that's already defined by the dot.
**as I said, I don't know if this will be covered later in the chapter.
I guess you can just ignore that...the x's purpose is for the matrices, which will be covered in 9.7
Can someone go through again how to do numbers like 71 and 72 from section 9.6? It seems to be more complicated than I thought.
For 9.6 #16 and 19, how do you go about finding the other coordinates? They don't have a related example in the text.
You just have to plot the points that it is asking and really think about the other 2 remaining coordinates. It's that simple, but the process of finding the other coordinates isn't so easy. Numbers 15 and 16 are easy, but problems like 19 and 20 are a bit more challenging.
Just visualize the dimensions of the prism on the 3D plane and you should be able to find the values.
How would you do number 37 in 9.7, where you have to use a 3 term and a 2 term matrice?
For that one, you place a zero in for the variable that appears to be missing. so it would look like row 1; 2, -3, 1 and row 2; 1, 1, 0
Then you do the cross multiplying and everything to figure out the final answer, which should be -i + j + 5k
I noticed when doing the homework that for one of the problems v x w and w x v were equal but opposite. Is that true for all situations like that?
Wow I just realized we went over that in class and it's in the book...my bad.
In section 9,7 for numbers like 43 on onwards, how would you know what the adjacent sides are? For those problems it's always P1P2 and P1P3, but Mrs. Johnson was saying that that isn't the case for all problems.
So how would you know from looking/solving, which would be adjacent?
Chandan: Mrs. Johnson said she would look at the test and let us know tomorrow.
I also have a question, for one of the problems we did yesterday, when we had to find 6 additional points of a cube, when we were given two, I'm still slightly confused as to how we find those points, can someone explain this to me?
I remember her saying that as well, but still just in case in case the test gives us a curve ball I just want to be ready for it...
For those, you just have to visualize it and using what you know, just think about it and you should get it. The first few of that section are simple, but it gets challenging along the way. I get really stuck on those as well.
Does anyone know how to do number 4 on the review sheet that Mrs. Johnson gave us today during class?
Rumor has it the test is postponed until Monday, can anybody confirm or deny?
Regarding the "additional points of the prism"
Think of the prism as standing upright AND the initial point being the origin of its OWN x-, y-, and z-axes. Imaginary axes should simplify the process greatly.
You have to find coordinates in the following places:
~ the initial point's x-axis
~ the initial point's y-axis
~ the initial point's z-axis
~ the initial point's xy-plane
~ the initial point's yz-plane
~ the initial point's xz-plane
To find what's on the axes, take your initial point and replace ONE of each of the corresponding values from the end-point to the initial point (ex. z-axis, replace z coordinate).
To find what's on the axes, take the initial point and replace TWO of the corresponding values from the end-point to the initial point (ex. xy-plane, replace x and y coordinate).
For example, take points (1, 2, 3) and (5, 5, 5). The first point is your initial.
1st point - x-axis - (5, 2, 3)
2nd point - y-axis - (1, 5, 3)
3rd point - z-axis - (1, 2, 5)
4th point - xy-plane - (5, 5, 3)
5th point - yz-plane - (3, 5, 5)
6th point - xz-plane - (5, 2, 5)
Hope this helped. 8)
Regarding #4 on the Chapter Review Sheet
I assumed that since the question specifically said "Find AN equation on A plane", I just diagrammed a random perpendicular plane.
I didn't think there was a specific plane that was perpendicular.
In fact, there shouldn't be!
Just draw your own plane, perpendicular to the plane with the coordinates given, then derive THREE coordinates from there. With those, you should be able to find AN equation with the correct procedure.
Ok, so after you find another point on the same plane given in the problem, would you basically find the equation with the 3 points and then do the OPPOSITE of the equation? So take the equation from something like 3, and flip the signs for everything for what 4is asking for?
That's what I heard as well, since there were a few people who told me it was moved. I'm going to study for tomorrow though.
On the take home quiz, is the A coordinate the P0 in all the cases?
Regarding #4 on the Chapter Review Sheet
Whatever plane you constructed that is perpendicular to the original plane, you should NEVER have to negatize the equation that you get out of the 3x3 matrix's derivative.
If you're gonna keep the original two points in the equation, you're gonna have raise one of those points on the z-axis as a THIRD point.
Use those three points to find the derivative of the 3x3 matrix that you model out of AB x AC, which should end up to be your final answer.
And like I said, I'm ASSUMING that the answer varies due to the indefinite references to the equation and plane that we are supposed to find.
Ok, but how do you know which coordinate is P0?
p0 is always the point you use in both the ab and ac vectors so the a basically, its the inital
Yes, but how would you know which one it is if you weren't given the A, B, or C coordinates. For example...number 3 in the review sheet of problems. How would you know?
For number 3 you are given three points, een though they aren't labeled A, B, ad C. You would construct 2 planes out of those 3 points, as we have done in the other problems. Whichever point you use in both planes would be your P0.
In #3 on the review problems, that would most likely be (2,1,1)
I have the same question as Hunter, does anybody know if the math test is definitely postponed to Monday?
Okay, I have the same question as Hunter and Shivani.
Im really confused with the general process used to solve a work problem, specifically in 3D. Can someone help me through one, like the one on the review sheet we got today?
Heres an example from 9.5 # 31 for work problems:
I start by making a triangle labeling the hypotenuse 20 for the force it tells you in the problem and 30 degrees for the angle. Now break the vertical and horizontal parts into components like in physics vertical = 20sin30j and horizontal= 20cos30i.
Now combine them to get the force vector as 20(cos30i+sin30j) simplify to 10(sqrt(3)i+j)
finally you can plug that into the work equation W=F*vector(AB)
vector(AB) was given in the problem as 100 and its on the x axis so its 100i
the final equation to simplify to get work is 10(sqrt(3)i+j)*100i
the answer you should get is 1000sqrt(3)foot-pounds
hope this helps
For number 2 in the review sheet, how do you know what the units are for the work? Why is it ft lbs for that problem, but in the book work is in terms of joules?
joules really means newton-meters, so foot pounds is like the non-metric way of expressing joules, so don't worry about it.
and thanks christian, but im still stuck on #2 on the review page she gave us, but you might not have got it cuz you were gone
in number two im doing the projected work equation from the book and i am getting 1800 over 29, times 12j which does not make srnse cause it says the answer is 1800/ root 29
Find the unit vector of the equation given, and multiply it by 50, because that gives you the force vector in the given direction. Once you have that, find line segment -AB>
You would then multiply your force vector by the line segment to get your answer
Can anyone help with #1 on the review sheet??
For number 1 try making it into triangles and using law of sines anf cosines
If anyone has noticed....THE MATH TEST IS POSTPONED TO MONDAY! check the pre-calc tab if you don't believe me.
Alex: For number #1, you first need to find the equation of the 620 magnitude vector. You can set up a right triangle with the given information: an angle of 55 degrees (90-35), and a hypotenuse of 620. You can then use the trig identities to find the lengths each leg using 620cos55 and 620sin55. Each leg translates to i and j in the vector.
Once you have the 620 magnitude vector (I had about <356, 508>), you have to find the equation of the wind vector. Since the wind heads due south (from the north) at 25 mph, the vector is going to be a line going down 25, which translates to <0, -25>.
Once you have both vectors, you can find the final vector. Add the two together, and you have the equation of the third (I got around <356, 483>). The question asks for the speed and bearing of the plane. The speed is simply the magnitude, which you can find by taking √(356^2+483^2). (I got around 600 mph). To find the bearing, you can use the equation for the angle between two vectors ( [i1 x i2 + j1 x j2] / ||v1|| x ||v2|| = cos x). You would plug in numbers from the first and third vector, because you want to find the angle between them, so you can add it to the given 35 degrees.
This is how I plugged it in:
cos x = (356 x 356 + 508 x 483) / (620 x 600)
Solve for x (I got around 1.37 degrees), and add x to 35 degrees (the given) to find the actual bearing.
Can anyone help me with #8 on the review worksheet? It asks for the direction angles, but I don't really know what that's talking about...
Muhammed, to find the direction angles you just have to plug in A, B, and C into the formula. There are three direction angles, one for each A, B and C. In order to find them, you just have to take the value of A, B or C and divide it by the magnitude of the entire vector. So it would be (A/IIVII) or (B/IIVII) or (C/IIVII). Once you get that, you just do the inverse cosine of your answer and then you will get the direction angles for each.
Oh ok thanks Vish I just forgot about taking the inverse cosine of each
@Alex in response to Kyle
You could use Kyle's method, or you could just use simple law of cosines. once you know the figure, you just have to find the 3 known things, sides and angles combined, and then using law of cosines solve for the other side and angle. mrs. Johnson was saying that she would like to see the vector method, but she will give credit for the law of cosines. It's less complicated and much less work, but you still get the same answer.
Chandan, you have to use the vector method on the test. It's true you can use the law of cosines; however, Mrs. Johnson wants us to use vectors.
Really? She told our period that we only needed to show her we knew what it was on our quiz that we had...when someone asked her in our period (2nd), she was fine with the law of cosines method.
NOTE to everyone: I'm not saying don't understand the vector method; know it so just in case if she wants it on the test, you don't get points off.
never mind guys...the test said different than me.
how do i get to the hippocampus thing ms johnson wants us to go to?
the hippocampus link is under the Parent Corner tab on the left side of the screen. you have to scroll down and just click on the link.
Mrs. Johnson teaches math at Metea Valley High School in Aurora, IL.
Ap Ab Calculus
H Algebra 2 Trig
H Pre Calculus
Regular Pre Calculus