Unit Vector Determination for Vector Bisecting Angle

In summary: I don't see how it gives you that. How about taking the dot product of c with itself?In summary, the vector -i+j+k bisects the angle between the vector c and 3i+4j. The equation for the angle bisector is \frac{-1}{\lambda}. Using the dot product of c with itself, x=y=z=0.
  • #1
utkarshakash
Gold Member
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Homework Statement


The vector -i+j+k bisects the angle between the vector c and 3i+4j. Determine a unit vector along c.

Homework Equations



The Attempt at a Solution



Taking the dot product of the two vectors (other than c) gives me the cosine of the angle = 1/5√3.
This is also equal to the angle between the angle bisector and c. But now I can't take the dot product anymore as I don't know anything about c.
 
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  • #2
If you have two vectors, u and v, can you write down in terms of u and v a vector that bisects the angle between them?
 
  • #3
haruspex said:
If you have two vectors, u and v, can you write down in terms of u and v a vector that bisects the angle between them?

lu+mv where l and m are constants.
 
  • #4
utkarshakash said:
lu+mv where l and m are constants.

That can give you any angle in the plane formed by the vectors. To get the bisector there will be a relationship between l, m, |u| and |v|.
 
Last edited:
  • #5
haruspex said:
That can give you any angle in the plane formed by the vectors. To get the bisector there will be a relationship between l, m, |u| and |v|.
OK So here's what I did:
I assumed c=xi+yj+zk

[itex]-i+j+k= \lambda \left( \dfrac{3i+4j}{5} + \dfrac{ \vec{c}}{|\vec{c}|} \right) [/itex]

Equating the respective components of both sides I get three equations

[itex] \frac{3}{5} + \frac{x}{|\vec{c}|} = \frac{-1}{\lambda} \\
\frac{4}{5} + \frac{y}{|\vec{c}|} = \frac{1}{\lambda} \\
\lambda \frac{z}{|\vec{c}|} = 1 [/itex]

Now taking the dot product of angle bisector with c gives me the value of |c|=5(-x+y+z). Using this in the above three equations gives me x=y=z=0 !
 
  • #6
utkarshakash said:
Now taking the dot product of angle bisector with c gives me the value of |c|=5(-x+y+z).
I don't see how it gives you that. How about taking the dot product of c with itself?
 

FAQ: Unit Vector Determination for Vector Bisecting Angle

1. What is a unit vector?

A unit vector is a vector with a magnitude of 1 and is used to represent direction in a specific coordinate system. It is often denoted as a lowercase letter with a hat on top (such as ˆi, ˆj, ˆk).

2. How do you determine the unit vector of a vector?

To determine the unit vector of a vector, you first need to find the magnitude of the vector. Then, divide each component of the vector by its magnitude to get the unit vector. For example, if the vector is v = (3, 4), the magnitude is √(3² + 4²) = 5. Therefore, the unit vector would be u = (3/5, 4/5).

3. Why is it important to find the unit vector of a vector?

Finding the unit vector of a vector is important because it allows us to represent the direction of a vector without being affected by its magnitude. This is useful in a variety of applications, such as physics, engineering, and computer graphics.

4. Can the unit vector have a negative magnitude?

No, by definition, a unit vector must have a magnitude of 1. This means that it cannot have a negative magnitude, as the magnitude of a vector is always a positive value.

5. How is the unit vector related to the concept of normalization?

The unit vector is closely related to the concept of normalization. Normalization is the process of scaling a vector to have a magnitude of 1, which is essentially what a unit vector is. Therefore, finding the unit vector is a form of normalization.

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