Why does the dot product in this solution equal zero?

In summary: Otherwise, I don't know what would have caused the error.-In summary, the dot product of a and b can equal zero even if a and b are not at right angles to each other.
  • #1
Darkmisc
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Homework Statement
Prove that the medians bisecting the equal sides of an isosceles triangle are equal.
Relevant Equations
a · b = |a| × |b| × cos(θ)
a · b = ax × bx + ay × by
Hi everyone

I have the solutions for the problem. It makes sense except for one particular step.

Why does the dot product of a and b equal zero? I thought this would only be the case if a and b were at right angles to each other. The solutions seem to be a general proof and should work for all isosceles triangles (not just right angle triangles).

The solution would still make sense even if a · b didn't equal zero. Why does it equal zero here?

Thanks

image_2022-09-21_131238624.png

image_2022-09-21_131315774.png
 
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  • #2
Darkmisc said:
Homework Statement:: Prove that the medians bisecting the equal sides of an isosceles triangle are equal.
Relevant Equations:: a · b = |a| × |b| × cos(θ)
a · b = ax × bx + ay × by

Hi everyone

I have the solutions for the problem. It makes sense except for one particular step.

Why does the dot product of a and b equal zero? I thought this would only be the case if a and b were at right angles to each other. The solutions seem to be a general proof and should work for all isosceles triangles (not just right angle triangles).

The solution would still make sense even if a · b didn't equal zero. Why does it equal zero here?

Thanks

View attachment 314417
View attachment 314418
I don't think you need a b = 0 for the proof to work.
Edit: Now I just saw that you wrote just that.
Maybe they just forgot the missing terms.
 
Last edited:
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  • #3
Darkmisc said:
Homework Statement:: Prove that the medians bisecting the equal sides of an isosceles triangle are equal.
Relevant Equations:: a · b = |a| × |b| × cos(θ)
a · b = ax × bx + ay × by

Hi everyone

I have the solutions for the problem. It makes sense except for one particular step.

Why does the dot product of a and b equal zero? I thought this would only be the case if a and b were at right angles to each other. The solutions seem to be a general proof and should work for all isosceles triangles (not just right angle triangles).

The solution would still make sense even if a · b didn't equal zero. Why does it equal zero here?

Thanks

View attachment 314417
View attachment 314418
I'll echo Philip Koeck above, it does not matter if ##a \cdot b = 0##. As you can see, if |a| = |b| then
##a \cdot a - a \cdot b + \dfrac{1}{4} b \cdot b = \dfrac{1}{4} a \cdot a - a \cdot b + b \cdot b##
whether ##a \cdot b = 0## or not. But (never trust a diagram!) it almost looks like the diagram has a and b perpendicular. It seems that this is the whole problem statement but is there somewhere that says a and b are perpendicular?

-Dan
 
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  • #4
topsquark said:
I'll echo Philip Koeck above, it does not matter if ##a \cdot b = 0##. As you can see, if |a| = |b| then
##a \cdot a - a \cdot b + \dfrac{1}{4} b \cdot b = \dfrac{1}{4} a \cdot a - a \cdot b + b \cdot b##
whether ##a \cdot b = 0## or not. But (never trust a diagram!) it almost looks like the diagram has a and b perpendicular. It seems that this is the whole problem statement but is there somewhere that says a and b are perpendicular?

-Dan
No, it wasn't specified that a and b are perpendicular. I also think Philip Koeck was right about them forgetting to delete a · b.

There was a previous problem in which a and b were perpendicular. Maybe that's what caused the authors to mistakenly say that a · b = 0 in the present problem.
 
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Related to Why does the dot product in this solution equal zero?

1. Why is the dot product in this solution equal to zero?

The dot product is equal to zero when the two vectors being multiplied are perpendicular to each other. This means that the angle between the two vectors is 90 degrees and there is no component of one vector in the direction of the other vector.

2. How is the dot product calculated?

The dot product is calculated by multiplying the corresponding components of two vectors and then adding those products together. For example, if vector A is (a1, a2, a3) and vector B is (b1, b2, b3), the dot product would be a1*b1 + a2*b2 + a3*b3.

3. Can the dot product ever be negative?

Yes, the dot product can be negative if the angle between the two vectors is greater than 90 degrees. This means that one vector is pointing in the opposite direction of the other vector.

4. What is the significance of the dot product being zero?

When the dot product is zero, it means that the two vectors are orthogonal or perpendicular to each other. This can be useful in many mathematical and scientific applications, such as finding the angle between two vectors or determining if two lines intersect.

5. How is the dot product used in physics?

In physics, the dot product is used to calculate the work done by a force on an object. It is also used in calculating the magnitude of a vector and determining the direction of a vector in a specific coordinate system. Additionally, the dot product is used in physics equations involving displacement, velocity, and acceleration.

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