# Vector Algebra: Finding Resultant Forces at Optimal Angles | Expert Help

• IonizingJai
In summary, the conversation discusses the determination of the angles at which two forces, represented by vectors A+B and A-B, must act in order to result in a vector with a magnitude of sqrt(A^2+B^2). Different approaches are presented, including equating the magnitude of the resultant vector to the given value and using trigonometric identities. However, there is confusion and disagreement over the correct answer, leading to further discussion and clarification of the initial problem.
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.].

IonizingJai said:
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) ## .
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.

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.].

IonizingJai said:
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.].

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

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

IonizingJai said:
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)##
Simplify what's above!

You're making this much harder than it needs to be.
IonizingJai said:
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## .

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.

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)?

##A\sqrt{2}##

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

IonizingJai said:
##A\sqrt{2}##
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.

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?

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

IonizingJai said:
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)}##
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.

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 ?

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.

## 1. What is Vector Algebra and how is it useful in finding resultant forces?

Vector Algebra is a branch of mathematics that deals with the manipulation and application of vectors. In the context of finding resultant forces, Vector Algebra allows us to combine multiple forces acting on an object into a single, equivalent force known as the resultant force. This is useful in determining the overall effect of multiple forces on an object and predicting its motion.

## 2. What are the steps involved in finding the resultant force using Vector Algebra?

The first step is to identify all the individual forces acting on the object and represent them as vectors. Next, we use Vector Algebra operations such as addition and subtraction to combine these vectors and find the resultant force. Finally, we can use trigonometric functions to determine the magnitude and direction of the resultant force.

## 3. Can Vector Algebra be applied to three-dimensional systems?

Yes, Vector Algebra can be applied to both two-dimensional and three-dimensional systems. In three-dimensional systems, we use three-dimensional vectors to represent the forces and the resultant force is calculated using the same principles as in two-dimensional systems.

## 4. What is the significance of finding the optimal angle in Vector Algebra?

The optimal angle refers to the angle at which the resultant force is maximum or minimum, depending on the context. In other words, it is the angle that gives the most efficient result. Finding the optimal angle is important in situations where we want to minimize or maximize the effect of the resultant force, such as in structural design or optimizing the performance of a machine.

## 5. How can Vector Algebra help in real-life applications?

Vector Algebra is used in various real-life applications, including engineering, physics, and navigation. It is used to analyze and predict the motion of objects, design structures that can withstand different forces, and calculate the direction and speed of moving objects. It is also used in computer graphics and game development to simulate realistic movements and collisions.

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