Prove c bisects angle between a and b | Vector Algebra

  • Thread starter Calpalned
  • Start date
  • Tags
    Angle
In summary: You should be able to see that the dot product of a and c will be proportional to cos(θ/2), where θ is the angle between a and b.
  • #1
Calpalned
297
6

Homework Statement


If c = |a|b + |b|a, where a, b, and c are all nonzero vectors, show that c bisects the angle between a and b.

Homework Equations


Angle between a & b is cos-1(a dot b)/(|a||b|)
Angle between a & c is cos-1(a dot c)/(|a||c|)
Angle AC is half of angle AB

The Attempt at a Solution


Given c = |a|b + |b|a, I plug that into the equation for the angle between a and c. I eventually get (|a||b|)(|a|b + |b|a2 = (2|a||c|)(a dot b). Is this right? I would also like to confirm:

- if (a dot a) is always equal to |a|^2
- how to differentiate absolute value and magnitude as they use the same symbol
- When I got (2|a||c|)(a dot b), do I do regular multiplication or use the dot product?
 
Physics news on Phys.org
  • #2
Calpalned said:

Homework Statement


If c = |a|b + |b|a, where a, b, and c are all nonzero vectors, show that c bisects the angle between a and b.

Homework Equations


Angle between a & b is cos-1(a dot b)/(|a||b|)
Angle between a & c is cos-1(a dot c)/(|a||c|)
Angle AC is half of angle AB

The Attempt at a Solution


Given c = |a|b + |b|a, I plug that into the equation for the angle between a and c. I eventually get (|a||b|)(|a|b + |b|a2 = (2|a||c|)(a dot b). Is this right?
This doesn't look right. If θ is the angle between a and c, you should have ##cos(θ) = \frac{a \dot c}{|a||c|}##. When you substitute in c and |c|, there should be some stuff in the denominator. I don't see that in what you have.
Calpalned said:
I would also like to confirm:

- if (a dot a) is always equal to |a|^2
Yes.
Calpalned said:
- how to differentiate absolute value and magnitude as they use the same symbol
Absolute values apply to real numbers. For a vector, |v| means the magnitude of the vector.
Calpalned said:
- When I got (2|a||c|)(a dot b), do I do regular multiplication or use the dot product?
Regular multiplication, if I'm understanding what you are asking. The two quantities in parentheses are real numbers.
 
  • #3
A simpler approach than you're taking is to calculate the cosines of the two angles; i.e., the angle between a and c, and the angle between c and b. As expected, these turn out to be equal.
 

FAQ: Prove c bisects angle between a and b | Vector Algebra

1. How do you prove that c bisects the angle between a and b?

To prove that c bisects the angle between a and b, we need to show that c divides the angle into two equal parts. This can be done by showing that the two angles formed by c are congruent, meaning they have the same measure.

2. What is the definition of angle bisector?

An angle bisector is a line or ray that divides an angle into two equal parts. It passes through the angle's vertex and divides the angle into two congruent angles.

3. What is the formula for finding the angle bisector of two vectors?

The formula for finding the angle bisector of two vectors a and b is given by c = (a + b) / ||a + b||, where ||a + b|| represents the magnitude of the vector sum of a and b. This formula can be derived using vector algebra and trigonometric identities.

4. Can you prove that c bisects the angle using vector algebra?

Yes, we can prove that c bisects the angle between a and b using vector algebra. We can use the dot product of the two vectors to show that the angles formed by c are congruent, thus proving that c bisects the angle between a and b.

5. What are some real-life applications of angle bisectors?

Angle bisectors have various applications in different fields such as architecture, engineering, and physics. In architecture, they are used to create symmetrical designs and layouts. In engineering, they are used for constructing structures with equal angles and for solving force-related problems. In physics, angle bisectors are used to calculate resultant forces and to analyze motion in two or three dimensions.

Back
Top