Vector dot product question- due tomorrow, please help!!

Nov 2019
9
1
london
Have you tried the hints in that attachment? That gives you rather clear instructions.
Can you express all in terms of \(\displaystyle x,\;y\;\;\&\;z\;?\)
So you can’t help then?
 

topsquark

Forum Staff
Jan 2006
11,567
3,453
Wellsville, NY
Is that okay?
What Plato is trying to do is find out what you already know. That way he can help you better. We are a help site...We don't usually just post the answers.

If you can write everything in terms of x, y, and z please post what you have as well as any other work you've done.

-Dan
 
Nov 2019
9
1
london
What Plato is trying to do is find out what you already know. That way he can help you better. We are a help site...We don't usually just post the answers.

If you can write everything in terms of x, y, and z please post what you have as well as any other work you've done.

-Dan
Yes I understand that and I’ve explained to him that I can’t provide anything as I’m totally stuck. Like if he doesn’t want to help me then tell me, instead of wasting my time. What’s your suggestion? I’m literally clueless on what to do hence why I want to learn from his answer.
 

topsquark

Forum Staff
Jan 2006
11,567
3,453
Wellsville, NY
What Plato has been waiting for is to see your attempt at writing out the equation in x, y, and z. Without that we don't know what you might be doing wrong. We can't help if we don't know where the error might be.

Here are the highlights:

triangle.jpg
Your equation, in terms of x, y, and z is
\(\displaystyle | \vec{x} + \vec{y} |^2 z + | \vec{y} + \vec{z} |^2 x - y^2 | \vec{x} - \vec{z} | = | \vec{x} - \vec{z} | xz\)

Writing this out:
\(\displaystyle (x^2 + y^2 + 2xy~cos( \theta ) ) z + (y^2 + z^2 + 2yz~cos( \pi - \theta ) )x - y^2(x + z) = (x + z)xz\)
(where \(\displaystyle \theta\) is the angle ADC.)

Expanding these out and noting that \(\displaystyle cos( \pi - \theta ) = - cos( \theta )\) the y terms drop out, leaving
\(\displaystyle x^2z + xz^2 = (x + z)xz\), which is true for all x, z.

-Dan
 
Last edited:
  • Like
Reactions: chip and romsek
Nov 2019
9
1
london
Here are the highlights:

View attachment 39594
Your equation, in terms of x, y, and z is
\(\displaystyle | \vec{x} + \vec{y} |^2 z + | \vec{y} + \vec{z} |^2 x - y^2 | \vec{x} - \vec{z} | = | \vec{x} - \vec{z} | xz\)

Writing this out:
\(\displaystyle (x^2 + z^2 + 2xz~cos( \theta ) ) z + (y^2 + z^2 + 2yz~cos( \pi - \theta ) )x - y^2(x + z) = (x + z)xz\)
(where \(\displaystyle \theta\) is the angle ADC.)

Expanding these out and noting that \(\displaystyle cos( \pi - \theta ) = - cos( \theta )\) the y terms drop out, leaving
\(\displaystyle x^2z + xz^2 = (x + z)xz\), which is true for all x, z.

-Dan
Wow thank you for the help Dan. That’s much clearer now, all thanks to you.