Proving Vector Dot Products: AC · BP = 0 | Step-by-Step Guide

[53Mark53]

Homework Statement

Homework Equations

let the point where the line through B and X intersects with AC be P

The Attempt at a Solution

[/B]
I know that

ACdotBP = 0

AC = AD+DC

BP = PC+ CD

Therefore (AD+DC)dot(PC+CD)=0

I also know that:

ECdotCE = 0

BCdotDA=0

However I am stuck on showing that the dot product of ACdotBP is equal to 0

If anyone can help me it would be much appreciated

thanks

 

[Samy_A]

Homework Statement

View attachment 100047

Homework Equations

let the point where the line through B and X intersects with AC be P

The Attempt at a Solution

[/B]
I know that

ACdotBP = 0

AC = AD+DC

BP = PC+ CD

Therefore (AD+DC)dot(PC+CD)=0

I also know that:

ECdotCE = 0

BCdotDA=0

However I am stuck on showing that the dot product of ACdotBP is equal to 0

If anyone can help me it would be much appreciated

thanks

How did you get BP = PC+ CD?
And isn't AC⋅BP = 0 what you have to prove?

 

[53Mark53]

thanks[/QUOTE]

How did you get BP = PC+ CD?
And isn't AC⋅BP = 0 what you have to prove?

Sorry that should say BP=BC+CP

And yes I am trying to prove AC⋅BP = 0

I am just unsure on the way that I should go about it

 

[Samy_A]

You know that AX⋅BC=0, CX⋅BA=0.

What you could do is pick an origin O (that could be any point, doesn't really matter), and express the vectors in terms of this origin.
For example: AX = OX - OA, BC = OC - OB, and similarly for the other vectors.
Now express AX⋅BC=0, CX⋅BA=0 in these terms, and compare the two expressions that you get.

Notice that you have to prove that BX⋅CA = 0.

 

[53Mark53]

You know that AX⋅BC=0, CX⋅BA=0.

What you could do is pick an origin O (that could be any point, doesn't really matter), and express the vectors in terms of this origin.
For example: AX = OX - OA, BC = OC - OB, and similarly for the other vectors.
Now express AX⋅BC=0, CX⋅BA=0 in these terms, and compare the two expressions that you get.

Notice that you have to prove that BX⋅CA = 0.

When I substitute AX = OX - OA, BC = OC - OB and BA = BO+OA, CX=CA+AB

I get (OX-OA)⋅(OC-OB)=0 and (CA+AE)⋅(BO+OA)=0

What would I do from here would I have to rearrange the equations?

 

[Samy_A]

When I substitute AX = OX - OA, BC = OC - OB and BA = BO+OA, CX=CA+AB

I get (OX-OA)⋅(OC-OB)=0 and (CA+AE)⋅(BO+OA)=0

What would I do from here would I have to rearrange the equations?

Actually, when you rewrite AX⋅BC=0 and CX⋅BA=0 in terms of vectors originating in O, what you get is:
AX⋅BC=0 indeed becomes (OX-OA)⋅(OC-OB)=0
But CX⋅BA=0 becomes (OX-OC)⋅(OA-OB)=0.

Now use the distributive property of the dot product to compute these two expressions.
Write the results one under the other and look at them for a moment. Combining them in the correct way will lead to the hoped for result.

 

[53Mark53]

Actually, when you rewrite AX⋅BC=0 and CX⋅BA=0 in terms of vectors originating in O, what you get is:
AX⋅BC=0 indeed becomes (OX-OA)⋅(OC-OB)=0
But CX⋅BA=0 becomes (OX-OC)⋅(OA-OB)=0.

Now use the distributive property of the dot product to compute these two expressions.
Write the results one under the other and look at them for a moment. Combining them in the correct way will lead to the hoped for result.

I am not sure if this is the distributive property of the dot product but this is what I got:

OXOC+OAOB=0
OXOA+OCOB=0

How would i got about combining these?

 

[Samy_A]

I am not sure if this is the distributive property of the dot product but this is what I got:

OXOC+OAOB=0
OXOA+OCOB=0

How would i got about combining these?

The distributive property of the dot product is:
(a + b)⋅c = a⋅c + b⋅c (where a,b and c are vectors).
From this it follows, for example, that (a + b)⋅(c + d) = a⋅c + b⋅c + a⋅d + b⋅d,
or (a - b)⋅(c - d) = a⋅c - b⋅c - a⋅d + b⋅d (where a,b c and d are vectors).

Apply this correctly to:
(OX-OA)⋅(OC-OB)=0
(OX-OC)⋅(OA-OB)=0

 

[53Mark53]

The distributive property of the dot product is:
(a + b)⋅c = a⋅c + b⋅c (where a,b and c are vectors).
From this it follows, for example, that (a + b)⋅(c + d) = a⋅c + b⋅c + a⋅d + b⋅d,
or (a - b)⋅(c - d) = a⋅c - b⋅c - a⋅d + b⋅d (where a,b c and d are vectors).

Apply this correctly to:
(OX-OA)⋅(OC-OB)=0
(OX-OC)⋅(OA-OB)=0

(OX-OA)⋅(OC-OB)=OX⋅OC-OA⋅OC-OX⋅OB+OA⋅OB

(OX-OC)⋅(OA-OB)=OX⋅OA-OC⋅OA-OX⋅OB+OC⋅OB

How would I combine this now?

 

[Samy_A]

(OX-OA)⋅(OC-OB)=OX⋅OC-OA⋅OC-OX⋅OB+OA⋅OB

(OX-OC)⋅(OA-OB)=OX⋅OA-OC⋅OA-OX⋅OB+OC⋅OB

How would I combine this now?

You left out that both expressions are 0.

Did you notice that two terms appear in both expressions? That should give you a clue about the next step: how to combine the two expressions.Remember that the final goal is to prove that BX⋅CA = 0.

 

[53Mark53]

You left out that both expressions are 0.

Did you notice that two terms appear in both expressions? That should give you a clue about the next step: how to combine the two expressions.Remember that the final goal is to prove that BX⋅CA = 0.

(OX-OA)⋅(OC-OB)=OX⋅OC-OA⋅OC-OX⋅OB+OA⋅OB=0

(OX-OC)⋅(OA-OB)=OX⋅OA-OC⋅OA-OX⋅OB+OC⋅OB=0

OX⋅OA-OC⋅OA-OX⋅OB+OC⋅OB=OX⋅OC-OA⋅OC-OX⋅OB+OA⋅OB

OX⋅OA+OC⋅OB=OX⋅OC+OA⋅OB

now how would I prove that BX⋅CA = 0?

 

[Samy_A]

(OX-OA)⋅(OC-OB)=OX⋅OC-OA⋅OC-OX⋅OB+OA⋅OB=0

(OX-OC)⋅(OA-OB)=OX⋅OA-OC⋅OA-OX⋅OB+OC⋅OB=0

OX⋅OA-OC⋅OA-OX⋅OB+OC⋅OB=OX⋅OC-OA⋅OC-OX⋅OB+OA⋅OB

OX⋅OA+OC⋅OB=OX⋅OC+OA⋅OB

now how would I prove that BX⋅CA = 0?

The distributive property of the dot product is still valid.
Two terms contain OX, two terms contain OB ...

 

[53Mark53]

The distributive property of the dot product is still valid.
Two terms contain OX, two terms contain OB ...

OX⋅OA+OC⋅OB=OX⋅OC+OA⋅OB

OX⋅OA-OX⋅OC=OA⋅OB+OC⋅OB

OX(OA-OC)=(OA+OC)OB

How would I prove it now?

 

[Samy_A]

OX⋅OA+OC⋅OB=OX⋅OC+OA⋅OB

OX⋅OA-OX⋅OC=OA⋅OB+OC⋅OB

OX(OA-OC)=(OA+OC)OB

How would I prove it now?

You have a sign error: what you should get is OX⋅OA-OX⋅OC=OA⋅OB-OC⋅OB

Now, look back at post #4: for any vector YZ, YZ = OZ - OY.

 

[53Mark53]

You have a sign error: what you should get is OX⋅OA-OX⋅OC=OA⋅OB-OC⋅OB

Now, look back at post #4: for any vector YZ, YZ = OZ - OY.

BX⋅CA=0

BX=OX-OB

CA=OA-OC

(OX-OB)⋅(OA-OC) = 0

OX⋅OA-OC⋅OX-OB⋅OA+OB⋅OC=0

OX⋅OA-OX⋅OC=OA⋅OB-OC⋅OB

Therefore: OX⋅OA=OA⋅OB-OC⋅OB+OX⋅OC

OA⋅OB-OC⋅OB+OX⋅OC-OC⋅OX-OB⋅OA+OB⋅OC=0

0=0

Is this now correct?

Thanks

 

[Samy_A]

BX⋅CA=0

BX=OX-OB

CA=OA-OC

(OX-OB)⋅(OA-OC) = 0

OX⋅OA-OC⋅OX-OB⋅OA+OB⋅OC=0

OX⋅OA-OX⋅OC=OA⋅OB-OC⋅OB

Therefore: OX⋅OA=OA⋅OB-OC⋅OB+OX⋅OC

OA⋅OB-OC⋅OB+OX⋅OC-OC⋅OX-OB⋅OA+OB⋅OC=0

0=0

Is this now correct?

Thanks

I'm not sure I follow the logic here. BX⋅CA = 0 is what you have to prove.

You had OX⋅OA - OX⋅OC = OA⋅OB - OC⋅OB.
That can be written as OX⋅(OA-OC) = OB⋅(OA - OC)
That becomes OX⋅CA = OB⋅CA.
(All these steps actually appear in your previous post.)

From that you easily get the value of BX⋅CA.

 

[53Mark53]

I'm not sure I follow the logic here. BX⋅CA = 0 is what you have to prove.

You had OX⋅OA - OX⋅OC = OA⋅OB - OC⋅OB.
That can be written as OX⋅(OA-OC) = OB⋅(OA - OC)
That becomes OX⋅CA = OB⋅CA.
(All these steps actually appear in your previous post.)

From that you easily get the value of BX⋅CA.

OX⋅CA-OB⋅CA. =0

(OX-OB)⋅CA=0

BX=(OX-OB)

therefore

BX⋅CA=0

Is this now correct?

 

[Samy_A]

OX⋅CA-OB⋅CA. =0

(OX-OB)⋅CA=0

BX=(OX-OB)

therefore

BX⋅CA=0

Is this now correct?

Yes.

 

[53Mark53]

Yes.

Thanks for your help I really appreciate it!