Why does the dot product in this solution equal zero?

[Darkmisc]

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

 

[Philip Koeck]

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.

 

[topsquark]

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

 

[Darkmisc]

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.