- #1

mrssmth

- 4

- 0

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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter mrssmth
- Start date

- #1

mrssmth

- 4

- 0

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?

- #2

Simon Bridge

Science Advisor

Homework Helper

- 17,874

- 1,657

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]

- #3

Simon Bridge

Science Advisor

Homework Helper

- 17,874

- 1,657

[tex]

\begin{array}{ccccccccc}

& A & + & B & = & x\hat{\imath} & + & y\hat{\jmath}&\\

+(& A & - & B & = & w \hat{\imath}& + & z\hat{\jmath}&)\\ \hline

& 2A & & & = & (x+w) \hat{\imath}& + & (y+z)\hat{\jmath}&

\end{array}

[/tex]

- #4

mrssmth

- 4

- 0

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

- #5

mrssmth

- 4

- 0

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

- #6

Simon Bridge

Science Advisor

Homework Helper

- 17,874

- 1,657

[tex]

\begin{array}{rcccccccl}

& A & + & 2B & = & x\hat{\imath} & + & y\hat{\jmath}&\\

+2\times(& 2A & - & B & = & w \hat{\imath}& + & z\hat{\jmath}&)\\ \hline

& 5A & & & = & (x+2w) \hat{\imath}& + & (y+2z)\hat{\jmath}&

\end{array}

[/tex]

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:

[tex]

\begin{array}{rcccccccl}

2\times( & A & + & 2B & = & x\hat{\imath} & + & y\hat{\jmath}&)\\

-(& 2A & - & B & = & w \hat{\imath}& + & z\hat{\jmath}&)\\ \hline

& & & 3B & = & (2x-w) \hat{\imath}& + & (2y-z)\hat{\jmath}&

\end{array}

[/tex]

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

Share:

- Last Post

- Replies
- 7

- Views
- 324

- Last Post

- Replies
- 10

- Views
- 200

- Last Post

- Replies
- 4

- Views
- 434

- Replies
- 20

- Views
- 495

- Last Post

- Replies
- 2

- Views
- 285

- Last Post

- Replies
- 4

- Views
- 432

- Last Post

- Replies
- 8

- Views
- 389

- Last Post

- Replies
- 6

- Views
- 347

- Last Post

- Replies
- 2

- Views
- 233

- Replies
- 4

- Views
- 335