Confusing Vector Questions: How Do I Solve Them?

[Taylor_1989]

Homework Statement

Hi guys, I have am having a problem with the questions below.

what do they mean by show $B_{perp}=B sin\theta$ I mean I know it a silly question, and my first thought is just use socahtoa, or am I missing something. The question from there on is straight forward enough,. I just have a feeling I am missing something or am I over thinking it completely?

Homework Equations

The Attempt at a Solution

 

[Doc Al]

my first thought is just use socahtoa, or am I missing something.

I think that's all there is to it.

 

[FactChecker]

Do you have another way to calculate B_⊥? If so, you should show that it gives the same answer.

 

[Taylor_1989]

Do you have another way to calculate B_⊥? If so, you should show that it gives the same answer.

I was thinking of showing it with an orthogonal projection, would this be correct?

 

[Ray Vickson]

I was thinking of showing it with an orthogonal projection, would this be correct?

Try it out, to see whether it works.

 

[FactChecker]

I was thinking of showing it with an orthogonal projection, would this be correct?

The exercise does say that the first step is to decompose B into the projection on A and the rejection (orthogonal projection) on A.

 

[Taylor_1989]

first thanks for the responses I have used the orthogonal projection, ]which I feel is the best way to show this relationship.

 

[Taylor_1989]

Hi guys, my perivous comment is bogus ^^^ I have been looking ove this question again and I now fully confusing myself. The way I thought I could show it which is I think i worng now is that I could say that:

$|| Proj_ba+ \perp Proj_ba=b$

By rearanging this in to: $|| b-Proj_ba= \perp Proj_ba$ I would be able to show what they wanted, when I did it before for some reason I thought that $|b|(1-cos\theta)=|b| sin \theta$ I have no idea why throught this at the time but I know this to be wrong. Can someone please give some advice on that matter thank you in advance.

 

[FactChecker]

I'm not sure about your notation, but it is true that a_||b + a_⊥b = a, where a_||b is the projection of vector a onto b and a_⊥b is the vector rejection of a on b. So if you can calculate a_||b, then you can use that to calculate a_⊥b and prove what the exercise asked for.

 

[Taylor_1989]

That what I oringinally throught but how could you, I can't seem to figure how to. I mean I know how the projection vector, $a_||b$ using dot product but how I convert to sine is what confusing me at the moment. Erm am I on the right lines with this:$|a_{||b}+a_{\perp b}|=|a|$ $\rightarrow$ $|a_{||b}|+|a_{\perp b}|=|a|$,

If this correct the $a_{||b}$=$|a|cos\theta$, but taking the mod of this is where it get confusing, what is the mod of this?

 

[FactChecker]

That what I oringinally throught but how could you, I can't seem to figure how to. I mean I know how the projection vector, $a_||b$ using dot product but how I convert to sine is what confusing me at the moment.

If you can calculate cos(θ) then you can get θ and then sin(θ).

Erm am I on the right lines with this:$|a_{||b}+a_{\perp b}|=|a|$ $\rightarrow$ $|a_{||b}|+|a_{\perp b}|=|a|$,

No. Use the Pythagorean theorem instead. a_||b and a_⊥b are at right angles.

If this correct the $a_{||b}$=$|a|cos\theta$, but taking the mod of this is where it get confusing, what is the mod of this?

It is not true. The mod is already there. |a_||b| = |a|cos(θ).

 

[Taylor_1989]

Im sorry, but I am now more confuse to how $arccos(\frac{|a_{||b}|}{|a|})$ to get the desired result. I really not seeing somthing here.

I know this not the way you are saying but can you use the face that $tan x= sinx/cosx$ & $cosx=a_{||b}/|a|$?So now I am thinking this, when you said pythagours

$(1) |B_{||}|^2+|B_{\perp}|^2=|B|^2$& $(2) B_{||}=Bcos\theta$

sub 1 into to 2 and rearrange for $|B_{\perp}|^2$

I then get: $B{\perp}=\sqrt(|B|^2)\sqrt(1-cos\theta^2)$

 

[FactChecker]

Im sorry, but I am now more confuse to how $arccos(\frac{|a_{||b}|}{|a|})$ to get the desired result. I really not seeing somthing here.

θ = arccos( |a_||b| / |a| )

I know this not the way you are saying but can you use the face that $tan x= sinx/cosx$ & $cosx=a_{||b}/|a|$?

a_||b is a vector. If you used |a_||b| that would be correct, but sort of a more complicated way to do things. The goal is to find θ, which would be the same as x. So it is simpler to just use θ = arccos( |a_||b| / |a| )

So now I am thinking this, when you said pythagours

$(1) |B_{||}|^2+|B_{\perp}|^2=|B|^2$

yes. They form a right triangle.

& $(2) B_{||}=Bcos\theta$

no. Look at figure 3 in post #1. B_|| and B are vectors that point in different directions. B_|| points in the direction of A. That's why I think the notation B_||A is better than B_||. Likewise B_⊥A is better notation than B_⊥. That notation would help you keep track of what vector they are parallel or perpendicular to (which is A).

sub 1 into to 2 and rearrange for $|B_{\perp}|^2$ I I have assumed that $B_{||}$ & $B_{\perp}$ are scalr projection of $A_B$ and not vector projections

They are the vectors shown in figure 3 of post #1.

 

[Taylor_1989]

Ok I am still a bit hazy. So the method i described is incorrect? Ok i cn see that B has been projected on to A, so the sclar projection is $B_{|| A}=|B|cos\theta$ so that why I throught I could use that in pythagours as the sclar projection would be the length so then i could you that with the length of the vect $B$ to show the result.

 

[FactChecker]

Ok I am still a bit hazy. So the method i described is incorrect? Ok i cn see that B has been projected on to A, so the sclar projection is $B_{|| A}=|B|cos\theta$ .

I'm still not sure that we are talking about the same thing. $B_{|| A}$ is a vector and $|B|cos\theta$ is not. So they are not equal. The scalar projection of B onto A is | $B_{|| A}$ | =$|B|cos\theta$. The modulus symbols are necessary. Please don't omit them.

 

[Taylor_1989]

First I apologise for my poor latex and also thank you being patient with me. I have finally figure out the problem. Thank you for your help it was much apprecited.