Solving Vector Problems in Physics

[mrssmth]

i just started physics and i missed the first day. I'm trying to do these vector problems but i can't quite understand it and my book isn't helping.

given that A+2B = x1j +y1j and 2A-B = x2i + y2j what is A?

given that A + B =x1i + y1j and A-B = x2i + y2j what is A?

given that A +B =x1i + y1j and A-B = x2i + y2j what is B?

 

[Simon Bridge]

Welcome to PF;
There's a bunch of people with these problems coming in today ... so having attended the class does not seem to have helped ;)

If you look at those equations, they are in pairs with A-B on one side and A+B on the other ... to get the A out, just add them. To get the B out, subtract them.

I'll show you:

Lets say A+B=xi+yj and A-B=wi+zj

then I can put v1= A+B and v2=A-B
then v1+v2 = (A+B)+(A-B) = (xi+yj)+(wi+zj) do you see now?v1+v2 = A+B+A-B = 2A = (x+w)i+(y+z)j

therefore A = (1/2)[(x+w)i+(y+z)j]

 

[Simon Bridge]

What we are doing is lining the relations up and adding the columns just like for a sum of two regular numbers.

<br /> \begin{array}{ccccccccc}<br /> &amp; A &amp; + &amp; B &amp; = &amp; x\hat{\imath} &amp; + &amp; y\hat{\jmath}&amp;\\<br /> +(&amp; A &amp; - &amp; B &amp; = &amp; w \hat{\imath}&amp; + &amp; z\hat{\jmath}&amp;)\\ \hline<br /> &amp; 2A &amp; &amp; &amp; = &amp; (x+w) \hat{\imath}&amp; + &amp; (y+z)\hat{\jmath}&amp;<br /> \end{array}<br />

 

[mrssmth]

That makes so much more sense!
Is this the right answer for "given that A + B =x1i + y1j and A-B = x2i + y2j what is A?"
I got A= (x1+x2)/2i +(y1+y2)/2j

 

[mrssmth]

Also for "A+2B = x1j +y1j and 2A-B = x2i + y2j what is A?" I got to this point and then got confused:

3A+B=(x1+x2)i +(y1+y2)j and now i don't know how to get the A all alone

 

[Simon Bridge]

When there is some constant multiplying the vector we want to get rid of, we have to add a multiple of the vectors together like this:

<br /> \begin{array}{rcccccccl}<br /> &amp; A &amp; + &amp; 2B &amp; = &amp; x\hat{\imath} &amp; + &amp; y\hat{\jmath}&amp;\\<br /> +2\times(&amp; 2A &amp; - &amp; B &amp; = &amp; w \hat{\imath}&amp; + &amp; z\hat{\jmath}&amp;)\\ \hline<br /> &amp; 5A &amp; &amp; &amp; = &amp; (x+2w) \hat{\imath}&amp; + &amp; (y+2z)\hat{\jmath}&amp;<br /> \end{array}<br />

notice how the +2x(... applies to the entire second row. I'm multiplying the second vector by 2 so that the second column will have a -2B in it.

You won't always be asked to find A, you may be asked to extract the B instead. In that case you have to subtract instead of add:

<br /> \begin{array}{rcccccccl}<br /> 2\times( &amp; A &amp; + &amp; 2B &amp; = &amp; x\hat{\imath} &amp; + &amp; y\hat{\jmath}&amp;)\\<br /> -(&amp; 2A &amp; - &amp; B &amp; = &amp; w \hat{\imath}&amp; + &amp; z\hat{\jmath}&amp;)\\ \hline<br /> &amp; &amp; &amp; 3B &amp; = &amp; (2x-w) \hat{\imath}&amp; + &amp; (2y-z)\hat{\jmath}&amp;<br /> \end{array}<br />

... in this case I multiplied the top row by 2 so the first column would be 2A-2A=0.