Area of a Triangle Using Vectors.

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  • #1
smakhtar
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Homework Statement


A triangle has verticies A(-2,1,3), B(7,8,-4), and C(5,0,2). Determine the area of the triange ABC.
The correct answer is 35.9 square units.

Homework Equations


Has to be done by using dot product and/or cross product.

Dot product: a(dot) b= |a||b|cos(theta)
Cross product: a x b= |a||b|sin(theta)
||- this is used to indicate magnitude

The Attempt at a Solution


AC=(5-(-2),0-1,2-3)
=(7,-1,1)
AB= (7-(-2),8-1,-4-3)
=(9,7,-7)
BC=(5-7,0-8,-1-4)
=(-2,-8,6)

I tried using this formula AB (dot) BC x AC, and I got 358 units squared, which is incorrect. I am trying to get to the correct answer mentioned above. What would be the formula that would give me the correct answer? Please reply as soon as possible
 

Answers and Replies

  • #2
SammyS
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Homework Statement


A triangle has verticies A(-2,1,3), B(7,8,-4), and C(5,0,2). Determine the area of the triangle ABC.
The correct answer is 35.9 square units.

Homework Equations


Has to be done by using dot product and/or cross product.

Dot product: a(dot) b= |a||b|cos(theta)
Cross product: a x b= |a||b|sin(theta)
||- this is used to indicate magnitude

The Attempt at a Solution


AC=(5-(-2),0-1,2-3)
=(7,-1,1)
AB= (7-(-2),8-1,-4-3)
=(9,7,-7)
BC=(5-7,0-8,-1-4)
=(-2,-8,6)

I tried using this formula AB (dot) BC x AC, and I got 358 units squared, which is incorrect. I am trying to get to the correct answer mentioned above. What would be the formula that would give me the correct answer? Please reply as soon as possible
The scalar triple product, AB (dot) BC x AC, gives a volume, so it's in cubed units.

What does CB x CA represent geometrically?
 
  • #3
smakhtar
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It has to be done using Cartesian vectors not geometric.
 
  • #4
SammyS
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It has to be done using Cartesian vectors not geometric.
But the point here is to help you understand how to answer the question, eventually using vectors. And the question asked does refer to vectors.

To repeat:

What does the cross-product (vector product), CB × CA, represent geometrically?
 
  • #5
smakhtar
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CA and CB are vectors. It geometrically represents the right hand rule where the thumb represents the z axis.
 
  • #6
smakhtar
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And the finders and palm would refer to the x and y axis.
 
  • #7
SammyS
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What does the magnitude of CB × CA represent ?
 
  • #8
smakhtar
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The length of the sides.
 
  • #9
SammyS
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The length of the sides.
No.

The magnitude of each vector is the length of each side.

What's the magnitude of the cross-product?
 
  • #10
smakhtar
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The magnitude of cross product equals the magnitude of |CB||CA|sin (theta).
 
  • #11
Nathanael
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The magnitude of cross product equals the magnitude of |CB||CA|sin (theta).

Yes, but there's a visual meaning of the cross product as well.

Perhaps you didn't learn (or don't remember) it so I'll just tell you, CA X CB is equal to the area of the parallelogram with sides being |CA| and |CB| and the angle between the sides is the angle between the CA and CB (what you called theta)
(From the definition of "parallelogram," the other two sides are parallel to CA and CB.)

So what would the area of triangle with points (A,B) (A,C) and (B,C) be?
 
  • #12
smakhtar
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I am assuming that you are referring to the coordinates mentioned above. I think it would be (AB x BC) (Dot) CA.
 
  • #13
Nathanael
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I think it would be (AB x BC) (Dot) CA.

That would be a volume. Draw the triangle and draw the parallelogram I described (AB X BC) and see if you can find a relationship.
 
  • #14
SammyS
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I am assuming that you are referring to the coordinates mentioned above. I think it would be (AB x BC) (Dot) CA.

Use the information Nathaniel gave you.

I already indicated that a scalar triple product, such as AB · (BC × AC) is a volume. By the way, your result for that was incorrect in your Original Post. For any three points A, B, C, that triple product is a vector of zero magnitude.


By the way, you could consider |CA|sin (θ) to be the altitude of a triangle and |CB| as the base.
 
  • #15
smakhtar
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So it would be A= |CB|(|CA|sin (theta) all over 2?
 
  • #16
SammyS
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So it would be Area= |CB|(|CA|sin (theta) all over 2?
(Don't use ' A ' for multiple purposes -- a point and an area .)

Yes that will work, but consider the following.

|CB|(|CA|sin (θ) is the magnitude of the cross-product of vectors CB and CA .

and

You can use any two of the three following vectors that you previously determined.
AC=(5-(-2),0-1,2-3)
=(7,-1,1)
AB= (7-(-2),8-1,-4-3)
=(9,7,-7)
BC=(5-7,0-8,-1-4)
=(-2,-8,6)​


Finally, CB x CA is a vector perpendicular to the plane in which points A, B, and C lie, having the magnitude of the parallelogram with adjacent sides formed by CB and CA .
 
  • #17
smakhtar
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Hey, I finally figured it out. Thanks for the help. I did BC x AC and I got it right, which is the same as |BC||AC|sin (theta) right?
 
  • #18
Nathanael
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Hey, I finally figured it out. Thanks for the help. I did BC x AC and I got it right, which is the same as |BC||AC|sin (theta) right?

Yes that is the same, but the correct answer is half of that.



Also you made a mistake right here:
AC=(5-(-2),0-1,2-3)
=(7,-1,1)
2-3 = -1, not 1 (could've been a typo, but still it could've affected your answer)
 
  • #19
SammyS
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Hey, I finally figured it out. Thanks for the help. I did BC x AC and I got it right, which is the same as |BC||AC|sin (theta) right?
That's the area of the parallelogram.

Divide by 2 to get the area of the triangle.
 
  • #20
smakhtar
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I did.
 

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