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

In summary, the student is trying to solve a problem involving the dot product of two vectors. He knows that the dot product is equal to zero if and only if the vectors originate from the same origin. He has found that if he uses the distributive property of the dot product, he can solve for the two expressions that must be equal to each other.
  • #1
53Mark53
52
0

Homework Statement


help.png


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
 
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  • #2
53Mark53 said:

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?
 
  • #3
thanks[/QUOTE]
Samy_A said:
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
 
  • #4
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.
 
  • #5
Samy_A said:
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?
 
  • #6
53Mark53 said:
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.
 
  • #7
Samy_A said:
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?
 
  • #8
53Mark53 said:
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
 
  • #9
Samy_A said:
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?
 
  • #10
53Mark53 said:
(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.
 
  • #11
Samy_A said:
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?
 
  • #12
53Mark53 said:
(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 ...
 
Last edited:
  • #13
Samy_A said:
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?
 
  • #14
53Mark53 said:
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.
 
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  • #15
Samy_A said:
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
 
  • #16
53Mark53 said:
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.
 
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  • #17
Samy_A said:
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?
 
  • #18
53Mark53 said:
OX⋅CA-OB⋅CA. =0

(OX-OB)⋅CA=0

BX=(OX-OB)

therefore

BX⋅CA=0

Is this now correct?
Yes.
 
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  • #19
Samy_A said:
Yes.

Thanks for your help I really appreciate it!
 

What is a vector dot product?

A vector dot product, also known as a scalar product, is a mathematical operation that takes two vectors and produces a scalar (a single number) as the result. It is a way to measure the similarity of two vectors and is often used in physics and engineering.

How do you prove the equation AC · BP = 0?

To prove this equation, you need to use the properties of the vector dot product. First, you can expand the left side of the equation using the distributive property. Then, you can use the fact that the dot product of perpendicular vectors is 0. Finally, you can use the commutative property to rearrange the order of the vectors. This will show that the left side of the equation is equal to 0, proving the equation.

What are the steps to proving the vector dot product equation?

The steps to proving the vector dot product equation are as follows:

  1. Expand the left side of the equation using the distributive property.
  2. Use the fact that the dot product of perpendicular vectors is 0.
  3. Rearrange the order of the vectors using the commutative property.
  4. Show that the left side is equal to 0, proving the equation.

Why is the vector dot product important in science?

The vector dot product is important in science because it allows us to measure the similarity or perpendicularity of vectors. This is useful in many fields such as physics, engineering, and computer graphics. It also has applications in calculating work, energy, and projections in three-dimensional space.

Can the vector dot product be applied to more than two vectors?

Yes, the vector dot product can be applied to any number of vectors. However, the resulting scalar will be a combination of all the vectors involved. For example, the dot product of three vectors would result in a scalar that is the sum of all three dot products. This is known as the generalized dot product.

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