Vector Cross Product (with Right Triangle)

[d-rock]

Homework Statement

For the vectors in the figure below, with a = 8, b = 7, c = sqrt(113), give the magnitude and direction of the following cross products.

(See attachment for figure/right triangle).

(a) a x b


(b) a x c


(c) b x c

Homework Equations

[tex]\vec{A} \cdot \vec{B} = ABcos(\Theta)[/tex]

[tex]\vec{C} = ABsin(\Theta)[/tex]

The Attempt at a Solution

(a) For part one, I was able to do the Cross Product since they gave the magnitudes A and B and in the figure, vectors A and B are straight lines which means the magnitude is also the X or Y (depending on the direction).

So, I did the Cross Product and got 56k, with it coming out of the computer screen.



Now, for the last two, I have no real clue how to do this. We just learned Cross Products today and haven't gone in-depth, and I've read the book on it but it does not help. Take note I am in a college-level/duel-enrollment level course which I'm in high school, and have had no background with physics before.

Can anyone just help me on how to figure out the X and Y of vector C? That's most likely why I think I keep getting this wrong.

 

[gabbagabbahey]

Homework Statement

For the vectors in the figure below, with a = 8, b = 7, c = sqrt(113), give the magnitude and direction of the following cross products.

(See attachment for figure/right triangle).

(a) a x b


(b) a x c


(c) b x c

Homework Equations

[tex]\vec{A} \cdot \vec{B} = ABcos(\Theta)[/tex]

[tex]\vec{C} = ABsin(\Theta)[/tex]

The Attempt at a Solution

(a) For part one, I was able to do the Cross Product since they gave the magnitudes A and B and in the figure, vectors A and B are straight lines which means the magnitude is also the X or Y (depending on the direction).

So, I did the Cross Product and got 56k, with it coming out of the computer screen.



Now, for the last two, I have no real clue how to do this. We just learned Cross Products today and haven't gone in-depth, and I've read the book on it but it does not help. Take note I am in a college-level/duel-enrollment level course which I'm in high school, and have had no background with physics before.

Can anyone just help me on how to figure out the X and Y of vector C? That's most likely why I think I keep getting this wrong.

I can't see the figure. Does [tex]\vec{c}=\vec{a} +\vec{b}[/tex]?

 

[d-rock]

I think the answer to that is no.

The figure is like this:

VectorA is straight to the right.

VectorB is straight up.

VectorC starts at the end of VectorB and goes back to the origin (so the head of VectorC touches the tail of VectorA).


So VectorC is like the opposite of what it would be in addition, if my knowledge of vectors so far is right.

 

[gabbagabbahey]

I think the answer to that is no.

The figure is like this:

VectorA is straight to the right.

VectorB is straight up.

VectorC starts at the end of VectorB and goes back to the origin (so the head of VectorC touches the tail of VectorA).So VectorC is like the opposite of what it would be in addition, if my knowledge of vectors so far is right.

If [tex]\vec{c}[/tex] goes form the tip of [tex]\vec{b}[/tex] to the tail of [tex]\vec{a}[/tex], then

[tex]\vec{a} +\vec{b} +\vec{c} = 0 \Rightarrow \vec{c}=-\vec{a} - \vec{b} [/tex]

and you should be able to easily obtain the components of [tex]\vec{c}[/tex] by simple addition.

 

[HallsofIvy]

a+ b= -c


a= <8, 0, 0>
b= <0, 7, 0>
c= <-8, -7, 0>

It should be easy to do the cross products using the component formula.

If you must use the "length- angle" formula, start by writing it correctly! [itex]\vec{A} \times\vec{B}= |\vec{A}||\vec{B}|cos(\theta)[/itex] is clearly wrong as the right hand side is a number, not a vector. You mean [itex]|\vec{A} \times\vec{B}|= |\vec{A}||\vec{B}|cos(\theta)[/itex] and the result is a vector of that length perpendicular to both A and B. Since A and B are perpendicular, [itex]cos(\theta)= 0[/itex] so A x B is easy.

C is a vector with length [itex]\sqrt{64+ 49}= \sqrt{113}[/itex] as you say. The vector from the base of A to the tip of B makes angle, with A, of arctan(7/8) so for C itself you can either take the angle to be 180 minus that, or just ignore it and multiply the final answer by -1.

The angle C makes with B is, of course, 90 degrees minus arctan(7/8).

 

[d-rock]

I wrote it that way because the physics book wrote it that way, they just specified it (i.e. on the right side A and B stood for the magnitudes of the vectors, etc.).


Also I had that process similarly done already:

part B: A x C

8 0 0 8 0 0

-7 -8 0 -7 -8 0

Doing that process gave an answer of -64. However, on WebAssign, on which my homework is, it told me I was wrong.

I also chose the direction of the angle as "Into the page" which it said was correct, however.


So I do not understand why the -64 is wrong?

 

[gabbagabbahey]

I wrote it that way because the physics book wrote it that way, they just specified it (i.e. on the right side A and B stood for the magnitudes of the vectors, etc.).Also I had that process similarly done already:

part B: A x C

8 0 0 8 0 0

-7 -8 0 -7 -8 0

Doing that process gave an answer of -64. However, on WebAssign, on which my homework is, it told me I was wrong.

I also chose the direction of the angle as "Into the page" which it said was correct, however.So I do not understand why the -64 is wrong?

[tex] \vec{c} = (-8,-7,0) [/tex] not [tex](-7,8-0)[/tex] And so the cross product is:

[tex]\vec{a} \times \vec{c} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 8 & 0 & 0 \\ -8 & -7 & 0 \end{array} \right| = -56 \hat{k}[/tex]

Which has a magnitude of 56 and is directed into the page (since [tex]- \hat{k} [/tex] is into the page)

 

[d-rock]

Alright. Well, to find the i and j of the third vector earlier, I had used arctan to find an angle and then did magnitude * cos(angle) and so on.


But turns out I used the wrong angle. I had done the angle from the origin, instead I should've used the one between VectorB and VectorC.



Well anyway, I understand your way too. Thanks for explaining, I got it right!