Area of a Triable with three vertices

[ND3G]

Determine the area of a triangle with vertices A(1,0,-5), B(-2,2,3) and C(-3,1,-1).

1/2|a x b|

|a x b| = (1, 0, -5) x (-2, 2, 3)
= ((0)(3) - (2)(-5), (-5)(-2) - (3)(1), (1)(2) - (-2)(0))
= (10, 7, 4)

|(10, 7, 4)|
= sqr(10^2 + 7^2 + 4^2)
= sqr(165)

(1/2)(sqr(165)) = 6.42

Therefore the area of the triangle is approximately 6.42 square units.

 

[ChaoticLlama]

1. this isn't calculus
2. you should really explain where you got vectors a and b from.
3. why are you dotting when you should be crossing?
4. do you know the definition of area for a triangle with vertices A,B,C?

 

[HallsofIvy]

First you have the vector wrong: neither (-3,1,-2) nor (4,-5,2) is a vector between two of the vertices. Second, the area of a triangle where the vectors a, b form two sides is NOT (1/2)|a.b|, it is (1/2)|axb|.

 

[ND3G]

Sorry, my first post was a complete mess. Please see the revised post above.

Thanks, Chaotic. I am aware this is not Calculus. However, the geometry posts I made yesterday were moved here so I figured I would save the mods a step...

I am also aware that triangle is misspelled in the header but I can not correct it at this point.

 

[ND3G]

Does this look right? I keep thinking I should be using point C as well.

 

[ChaoticLlama]

Does this look right? I keep thinking I should be using point C as well.

yes, you would be right in thinking this.

look at points A,B,C; maybe try making vectors out of those points? then crossing some of them?

 

[Styx]

Ok, how about this:

|a| = (1, 0, -5) x (-2, 2, 3)
= ((0)(3) - (2)(-5), (-5)(-2) - (3)(1), (1)(2) - (-2)(0))
= (10, 7, 4)
=|(10, 7, 4)|

|b| = (-2,2,3) x (-3, 1, -1)
=((2)(-1) - (1)(3), (3)(-3) - (-1)(-2), (-2)(1) - (-3)(2))
=(-5,-11,4)
=|(-5, -11, 4)|

|a x b| = (10, 7, 4) x (-5,-11,4)
=((7)(4)-(-11)(4), (4)(-5)-(4)(10), (10)(-11)-(-5)(7))
=(72, -60, -75)
=(24, -20, -25)
=|(24, -20, -25)|

sqr(24^2 + (-20)^2 + (-25)^2) = 40
(1/2)(40) = 20