What is the area of triangle ABC in 3-D?

[ivybond]

Question:
Find the area of triangle ABC that has the coordinates A(1, 3, 0), B(0, 2, 5), and C(-1, 0, 2).
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One way to do that is to find the lengths of ABC sides and use Heron's formula. Not pretty nymbers!

You can find the distance h between C and AB.
Area A(ABC) = 1/2 AB h.

I noticed that ABC "split in half" by yz-plane.
Midpoint M(0,3/2,1) of AC is in yz-plane. ABC's median BM is in yz-plane.
This fact for some reason fascinated me, and, I guess, made me find area of AMB first and then double it to get area of ABC.
BM = sqrt(65)/2.
BM equation
(0)x + (8)y + (-1)z + (-11) = 0,
distance between A and BM is
H=|(0)(1)+(8)(3)+(-10)(0)+(-11)| /sqrt (0^2+8^2+(-1)^2))
H=13/sqrt(65)

A(ABC) = BM*H = (sqrt(65)/2)*(13/sqrt(65)) = 13 / 2 = 6.5

The answer seems to be too good to require such a long solution.

I also think that the fact that vertices A, B, and C are respectively in xy-plane, yz-plane, and xz-plane, should provide a shortcut.

Any suggestions?

 

[tony873004]

I'll take a guess
Can you just use Pothagorean to find the length of the sides?

length_{AB} = \sqrt{( |A_x-B_x|)^2 + (|A_y-B_y|)^2 +(|A_z-B_z|)^2}

Do the same for AC and BC, then use Heron's formula.

 

[robphy]

Use the cross-product. Calculate (say) \vec {AB} \times \vec{AC}.

 

[dicerandom]

I'm getting a different number for the area, but I'm not sure where your mistake is.

Have you learned about the cross product in your class? Given two vectors, such as the vectors AB and AC, if you take their cross product the magnitude of the resulting vector is the area of the parallelogram with sides AB, AC (the other two sides are parallel to AB, AC of course). The area of the triangle ABC is then half the area of the parallelogram.

 

[ivybond]

I am sorry I failed to mention that this question was given in the Algebra 2 class to somebody I know.
Using vectors is the fastest way to go (I did not check my answer though ), but I don't think they covered it in the class.

Thanks anyway!