Find AB X AC: Prove Triangle Area is 3/2

In summary, the conversation was about finding the area of a triangle given three points in three-dimensional space. The solution involved using the cross product of two vectors and dividing it by two to get the area. The conversation also touched on the relationship between the cross product and the area of a parallelogram, with the magnitude of the cross product being equal to the area of a parallelogram. Further clarification was provided on the difference between a cube and a parallelogram in terms of area.
  • #1
Scandal
4
0
Find: AB X AC (vectors). Prove that the area of the triangle is [tex]\frac{3}{2}[/tex]

Further information
We're given the points A(1,0,0) , B(0,2,2), C(1,1,2)
So considering how this is the same question there is no doubt that we have to use the normal vector to find the answer.

The attempt at a solution
AB X AC = [2,2,-1]

|AB X AC| = [tex]\sqrt{2^{2}+2^{2}+(-1)^{2}}[/tex] = 3

Dividing this by 2 gives us the answer we're looking for. However, why are the length of the normal vector equal to the area of a cube? Are there any other special characteristics of the normal vector I should know before my oral exam in two days? The theme is geometry considered around third-dimension vectors and planes. All help is good help as I can't really catch the essence of the abstract normal vector.
 
Physics news on Phys.org
  • #2
You are looking for the area of a triangle, not a cube.

How does the cross product relate to the area of a parallelogram?
 
  • #3
Might be I was unclear (perhaps of my poor English), but I found the answer, so this might not qualify for the homework help part of the forum, I don't know. The question is mostly why do the normal vector (cross product as many like to refer to it) have the properties of the area of a cube stretched by sides AB and AD (or in this case AC). In my perception it just a line which stretches outwards from the plane, 90 degrees in both 2D axises. I can't seem to comprehend why this line is so important, but somehow it is.
 
  • #4
I realize we have a language issue here. This is a cube:
rubiks-cube.jpg
A parallelogram is a plane figure:

http://www.coolmath.com/reference/images/parallelogram1.gif [Broken]

The area of a parallelogram is the product of the lengths of any two adjacent sides and the sine of the angle between those sides:

[tex]A=ab\sin\theta[/tex]

This is also the magnitude of the cross product of the displacement vectors corresponding to those adjacent sides.

685px-Cross_parallelogram.png


That the triangle has half the area of the parallelogram should be easy to visualize. A diagonal of a parallelogram splits the parallelogram into two identical triangles.
 
Last edited by a moderator:
  • #5
There it is! Thank you. I see now how cube was a wrong choice of words. What I meant to say was a square (3D vs. 2D), but as you say the correct would be a parallelogram. I think just the feeling of knowing somewhat what the normal vector is provides a lot when it comes to the presentation. I want to thank you again for the help.
 

1. What is the formula for finding the area of a triangle?

The formula for finding the area of a triangle is: A = 1/2 * base * height, where A is the area, base is the length of the base of the triangle, and height is the height of the triangle.

2. How do you find the length of AB and AC in a triangle?

To find the length of AB and AC, you can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. So, you can use this formula to find the length of AB and AC and then use the formula for finding the area of a triangle.

3. How do you prove that the area of a triangle is 3/2?

To prove that the area of a triangle is 3/2, you can use the formula A = 1/2 * base * height and plug in the values for base and height. Then, you can simplify the equation and show that it equals 3/2. You can also use other methods of proof, such as using the Pythagorean theorem or the properties of similar triangles.

4. What is the relationship between the length of AB and AC in a triangle?

The length of AB and AC in a triangle can vary depending on the type of triangle (e.g. equilateral, isosceles, scalene). However, in a right triangle, the length of AB and AC are related by the Pythagorean theorem, as mentioned before. In other triangles, the relationship between AB and AC may be determined by the angles and sides of the triangle.

5. Can the area of a triangle be greater than 3/2?

Yes, the area of a triangle can be greater than 3/2. The area of a triangle is determined by the length of its base and height, and these values can vary. Therefore, the area of a triangle can be greater or smaller than 3/2 depending on the specific measurements of the triangle.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
2K
Replies
1
Views
822
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Precalculus Mathematics Homework Help
Replies
11
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Precalculus Mathematics Homework Help
Replies
9
Views
1K
Replies
3
Views
6K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
3K
Back
Top