# Vector Algebra help ?

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1. May 24, 2015

### IonizingJai

Question: At what angles must be the two forces $\vec A+\vec B$ and $\vec A-\vec B$ act so that the resultant may be :
$$\sqrt{ A^2+B^2}$$
Attempt at solution :

Let the given forces be $\vec F_1=\vec A+\vec B$ and $\vec F_2=\vec A-\vec B$ .
Now, Resultant vector : $\vec F_1 + \vec F_2 = (\vec A+\vec B) + (\vec A-\vec B)$ .
Magnitude of Resultant: $|\vec F_1 + \vec F_2| = \sqrt{(A+B)^2+(A-B)^2+2(A+B)(A-B)\cos(\theta)}$.
Here, $\theta$ being the angle between $\vec F_1 and \vec F_2$ .
Stuck here $=\sqrt{2(A^2+B^2)+2(A^2-B^2)\cos(\theta)}$ .
[ i'm noob with LaTeX too.]

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2. May 24, 2015

### HallsofIvy

Staff Emeritus
You realize that $(\vec{A}+ \vec{B})+ (\vec{A}- \vec{B})= 2\vec{A}$ don't you? And, since the magnitude of $2\vec{A}$ does not depend on $\vec{B}$, I have to ask "the resultant of what"?
You seem to have misunderstood the question.

3. May 24, 2015

### Ray Vickson

4. May 24, 2015

### Staff: Mentor

It's OK, Ray. I asked him to start a new thread, as I locked and will soon delete the old thread.

5. May 24, 2015

### IonizingJai

Yeah , i never approached the problem this way, sorry.
Also i found another way :
Let the given forces be $\vec F_1=\vec A+\vec B$ and $\vec F_2=\vec A-\vec B$ .
Now, Resultant vector $\vec F_1 + \vec F_2 = (\vec A+\vec B) + (\vec A-\vec B)$
Magnitude of Resultant:
$|\vec F_1 + \vec F_2| = \sqrt{(A+B)^2+(A-B)^2+2(A+B)(A-B)\cos(\theta)}$
...................1
But according to question the Resultant may be (but it should have been magnitude of the resultant, the question btw is correct as i read it.)
$= \sqrt{(A^2+B^2)}$
.....................2
Equating 1 and 2 we get .
and $2(A^2+B^2)+2(A^2-B^2)\cos\theta = (A^2+B^2)$ .
and $2(A^2-B^2)\cos\theta = - (A^2+B^2)$ .
and $\cos\theta= -1/2$.
That is $\theta = cos^{-1}(-0.5)$
$\theta = - 120 degrees$ between the two Forces $\vec F_1 and \vec F_2$ .

6. May 24, 2015

### Staff: Mentor

Simplify what's above!

You're making this much harder than it needs to be.

7. May 25, 2015

### IonizingJai

Alright , then that means i am wrong. So , if the question is correct , it ask for the value of the angle between the two forces so that the Resultant maybe
$\sqrt{(A^2+B^2)}$.
Mark44 : as you said i should simplify the but HallsofIvy already showed its not possible to arrive on the answer, since it will equal $2\vec A$ ?
Help ?
BTW , The question as asked in the First post is exactly as it is in language and i just copied it there.

8. May 25, 2015

### Staff: Mentor

So (A + B) + (A - B) = 2A, right?
What's the magnitude of the sum of the two vectors (which would be the magnitude of the resultant of A + B and A - B)?

9. May 25, 2015

### IonizingJai

$A\sqrt{2}$

10. May 25, 2015

### theodoros.mihos

Is this correct? $$\frac{0}{3} = \frac{0}{5} \Rightarrow \,\text{beacuse}\, 0=0 \,\text{then}\, 3=5$$
If you want:
$$|\mathbf{F}_1+\mathbf{F}_2| = 2A = \sqrt{A^2+B^2} \Rightarrow 3A^2 = B^2$$
that mean B2=3A2 and A is what you like or vs.

Last edited: May 25, 2015
11. May 25, 2015

### mooncrater

Will ( A vector) +(B vector) have a magnitude (|A|+|B|)2?

12. May 25, 2015

### HallsofIvy

Staff Emeritus
No, that is still wrong! Assuming that you have stated the problem correctly, so that A+ B+ A- B= 2A, the length of 2A is NOT $\sqrt{2}$ times the length of A.

13. May 25, 2015

### theodoros.mihos

This is the initial problem: the "resultant" of two vectors A and B must be $\sqrt{A^2+B^2}$.
The writer works on |resutlant|.
The other posibillity is that mean:
$$\sqrt{A^2+B^2} = (\mathbf{A}+\mathbf{B})\cdot(\mathbf{A}-\mathbf{B}) = A^2 + B^2 +2AB\cos{\theta} \Rightarrow \cos{\theta} = \frac{\sqrt{A^2+B^2}-(A^2+B^2)}{2AB}$$
what I can say???

14. May 25, 2015

### IonizingJai

Sorry , but i don't get what are you guys implying ?

15. May 25, 2015

### vela

Staff Emeritus
This isn't correct. For example, $\lvert \vec F_1 \rvert^2$ isn't equal to $A^2 + B^2$, and $\lvert \vec F_1 \rvert$ isn't equal to $A+B$. Rather, it should be
$$\lvert \vec F_1 \rvert^2 = \lvert \vec{A} + \vec{B} \rvert^2 = A^2 + B^2 + 2AB\cos\theta_\text{AB},$$ and $\lvert \vec F_1 \rvert$ is the square root of that. You might get the feeling this approach is going to be really messy, and you'd be right. Use Mark's suggestion to simplify first.

16. May 26, 2015

### IonizingJai

but if you check my solution at post NO , #5 , i have done exactly what you said , i have only left the steps in which i had to cancel many things out and expand and stuff. if you check it you will find that i have done.
and i don't get what Mark44 is implying , the question gets stuck that way since, $\vec A$ will not depend on $\vec B$ and thus there will be no angle between them ?

17. May 26, 2015

### vela

Staff Emeritus
Reread what I wrote a bit more carefully. You should see that your expression for the magnitude of the resultant is incorrect.

By the way, I suspect you used $\sqrt{A^2+B^2} = A+B$. That's wrong too.