Find the angle between two vectors

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Homework Help Overview

The problem involves determining the angle between two vectors A and B, both with a magnitude of 5.25, given that their sum results in a third vector equal to 6.73j. The context is rooted in vector addition and properties of triangles formed by vectors.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find the components of the vectors A and B, questioning the feasibility of doing so with the given information. There is mention of using the cosine rule to relate the vectors and their resultant. Some participants explore the relationships between vector components and consider simultaneous equations to solve for unknowns.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants have suggested using geometric interpretations and relationships between components, while others have expressed confusion about the directionality of the vectors and the implications of their magnitudes. There is no explicit consensus on the correct method or interpretation yet.

Contextual Notes

Participants note the absence of directional information for the vectors, which complicates the formation of a triangle. The problem constraints include the equal magnitudes of the vectors and the specific resultant vector provided.

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Homework Statement



Two vectors A and B have the same magnitude of 5.25. If the sum of these two vectors gives a third vector equal to 6.73j, determine the angle between A and B.

Homework Equations



For some vector \vec{R}: |\vec{R}| = \sqrt{R_x^2 + R_y^2}

The Attempt at a Solution



I feel like I need to find the components of A and B, but I can't figure out how to do that with only the information given -- is it even possible?

I think the magnitude of the third vector is 6.73. Since there was no \hat{i} term given for \vec{R}, I would say that |\vec{R}| = \sqrt{R_x^2 + R_Y^2} = \sqrt{6.73^2} = 6.73.

Given the components of A and B, this would be easy to solve. Without them I am utterly lost. What am I missing?
 
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jamesbrewer said:

Homework Statement



Two vectors A and B have the same magnitude of 5.25. If the sum of these two vectors gives a third vector equal to 6.73j, determine the angle between A and B.

Homework Equations



For some vector \vec{R}: |\vec{R}| = \sqrt{R_x^2 + R_y^2}

The Attempt at a Solution



I feel like I need to find the components of A and B, but I can't figure out how to do that with only the information given -- is it even possible?

I think the magnitude of the third vector is 6.73. Since there was no \hat{i} term given for \vec{R}, I would say that |\vec{R}| = \sqrt{R_x^2 + R_Y^2} = \sqrt{6.73^2} = 6.73.

Given the components of A and B, this would be easy to solve. Without them I am utterly lost. What am I missing?

The vectors A and B, along with the resultant will form a triangle - probably not right angled - so you can use the cosine Rule to solve.

a2 = b2 + c2 - 2.b.c.cos(A)
 
PeterO said:
The vectors A and B, along with the resultant will form a triangle - probably not right angled - so you can use the cosine Rule to solve.

a2 = b2 + c2 - 2.b.c.cos(A)

What are a, b, and c? The lengths of the sides of the triangle? If so, would that mean that a = |A|, b = |B|, and c = |C| (the resultant)?
 
jamesbrewer said:
What are a, b, and c? The lengths of the sides of the triangle? If so, would that mean that a = |A|, b = |B|, and c = |C| (the resultant)?

a, b & c are the sides,
A is the angle opposite side a

[so a, b & c could represent vectors A, B and C but not necessarily, depends which angle you are looking for]
 
Last edited:
How can I form a triangle if I don't know anything other than the vector's magnitude? I have nothing to tell me what direction it points in.
 
jamesbrewer said:
How can I form a triangle if I don't know anything other than the vector's magnitude? I have nothing to tell me what direction it points in.

You can stand a triangle up any way you like - its sides are still the same length and its angles are still the same size. A name like "the base" might apply to a different side, depending which way you arrange it, but that should not be a problem .. what's in a name?
 
Here's what I've got:

|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|cos\theta

6.73^2 = 5.25^2 + 5.25^2 - 2(5.25)^2 cos \theta

45.29 = 27.56 + 27.56 - 2(27.56) cos \theta

45.29 = 55.12 - 55.12 cos \theta

45.29 - 55.12 = - 55.12 cos \theta

-9.83 = -55.12 cos \theta

\frac{-9.83}{-55.12} = cos \theta

\theta = cos^-1 0.178

\theta = 79.75^o

My answer wasn't correct though, where did I go wrong?
 
Last edited:
One approach which you might want to consider is to look at the relationships between the vector components in terms of simultaneous equations. Suppose that the two initial vectors are A and B and the resultant is C.

Since C contains no x-component and has only a y-component, the sum Ax + Bx must be zero, or in other words, Bx = -Ax. Similarly, Ay + By must be Cy. You also have |A| = |B| = 5.25, so that's another pair of relationships. You have four unknowns (really three when you consider that the x-components of A and B are equal and opposite) and plenty of interrelationships to use to solve for them.
 
jamesbrewer said:
Here's what I've got:

|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|cos\theta

6.73^2 = 5.25^2 + 5.25^2 - 2(5.25)^2 cos \theta

45.29 = 27.56 + 27.56 - 2(27.56) cos \theta

45.29 = 55.12 - 55.12 cos \theta

45.29 - 55.12 = - 55.12 cos \theta

-9.83 = -55.12 cos \theta

\frac{-9.83}{-55.12} = cos \theta

\theta = cos^-1 0.178

\theta = 79.75^o

My answer wasn't correct though, where did I go wrong?

How incorrect was your answer? You have rounded off all the way through which could make a small difference
 
  • #10
PeterO said:
How incorrect was your answer? You have rounded off all the way through which could make a small difference

Unfortunately I have no idea. All I was told was "Incorrect answer."
 
  • #11
Let's step back from the problem and see if we can make additional simplifying deductions before invoking formulas.

Since the resultant of vectors A and B is a vector with only a y-component (6.73j), then A and B must have equal and opposite x-components. Further, since A and B have equal magnitudes (5.25), this then forces their y-components to be equal also. Why not equal and opposite you say? Because then their sum would be zero rather that +6.73.

attachment.php?attachmentid=39204&stc=1&d=1316810186.gif


So A and B "straddle" the positive y-axis, and make equal angles with that axis. The sum of their y-components is 6.73. Now it's time to write formulas. If you let \theta be the angle between either A or B and the Y-axis, what is an expression for the y-component of A or B? If double that is 6.73, can you solve for \theta? What then is the angle between A and B?
 

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  • #12
jamesbrewer said:
Unfortunately I have no idea. All I was told was "Incorrect answer."

Draw a diagram of the vectors, like gneil's and you will see that they actually want (180 - 79) degrees - with the decimal bits.
Rounding off needs to be addressed too.
gneil's method is excellent also - even preferable!
 
  • #13
Three vectors sum to zero. The magnitude of 2 vectors is equal and the third one is root 2 times the magnitude of the equal vectors. Find the angle between the three vectors. Plzz help me on dis one..:)
 
  • #14
Karthik k said:
Three vectors sum to zero. The magnitude of 2 vectors is equal and the third one is root 2 times the magnitude of the equal vectors. Find the angle between the three vectors. Plzz help me on dis one..:)

Since the three vectors will make up a triangle - so you end up where you started with a vector sum of zero - you should recognise the magnitudes 1,1,√2 as the sides of a very common triangle in trigonometry, from which we derive the exact vale of sin, cos and tan of a particular angle.
The other triangle used in trigonometry is known as the 2,1,√3 triangle [not that it relates directly to this question].
 

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