Question

A triangle has vertices A(1,3,4) B (3,-1,1) C (5,1,1). What is the area of ABC

Answer 1

I believe its 9.1. @amistre64 Am I correct?

Answer 2

this has something to do with a dot product .... or a cross product

Answer 3

assume 2d for the moment to help my brain recycle these

Answer 4

half magnitude of the cross product vector sounds familiar

Answer 5

A(1,3,4) B (3,-1,1) C (5,1,1)
-1-3-4 -1-3-4 -1-3-4
-----------------------------
2,-4,-3 4,-2,-3

x 2 4
y -4 -2
z -3 -3

x = 12+6 = 18
y = -(-6+12) = -6
z = -4+16 = -12

sqrt(18^2+6^2+12^2)/2

i get about 11.22 if thats right

Answer 6

6,-6,12 had an error

Answer 7

sqrt(36+36+144)/2 = 7.35

Answer 8

Thanks bud, but 9.1 was correct. Thanks a ton

Answer 9

hmm, il have to rechk my vectors then :)

Answer 10

vectors are good

cross is 6, -6, 12

mag is 6sqrt(6); divided by 2 is 3sqrt(6) is 7.35

hmmm

Answer 11

\[cos(\alpha)=\frac{a.b}{|a||b|}\]
\[sin(\alpha)=sin(cos^{-1}(\frac{a.b}{|a||b|}))\]
\[|a||b|\frac12sin(\alpha)=|a||b|\frac12sin(cos^{-1}(\frac{a.b}{|a||b|}))\]
would work too