anonymous
  • anonymous
Let u=(4,0,1), v=(5,-1,0), and w=(-3,1,-2). Find the area of the triangle determined by u and v.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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slaaibak
  • slaaibak
I think the area for the parallelogram would be the length of the cross product. |u x v|
slaaibak
  • slaaibak
Therefore, the area of the triangle would be the half of }u x v|
anonymous
  • anonymous
Thanks!

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anonymous
  • anonymous
Could you write it out? I'm a little confused, I guess.
amistre64
  • amistre64
what have you some up with in the cross product so far?
anonymous
  • anonymous
I have (20, 0, 0) but I feel like I've done something wrong... I'm not very good with vectors.
amistre64
  • amistre64
\[\begin{vmatrix}X&Y&Z\\U_x&U_y&U_z\\V_x&V_y&V_z\end{vmatrix}\to X\begin{vmatrix}\\U_y&U_z\\V_y&V_z\end{vmatrix} -Y\begin{vmatrix}\\U_x&U_z\\V_x&V_z\end{vmatrix} +Z\begin{vmatrix}\\U_x&U_y\\V_x&V_y\end{vmatrix}\]
amistre64
  • amistre64
thats the "formal" confusion for it: me, i just do this x 4 5 x = 1 y 0 -1 y = 5 z 1 0 z = -4
anonymous
  • anonymous
How does what you just did work?
amistre64
  • amistre64
so, the length of the vector <1,5,-4> its the same process as the fancy typing; but in a vertical format that I can keep track of better
amistre64
  • amistre64
turn the matrix on its side pretty much and expand down the first column; taking the determinant of each sub matrix along the way
amistre64
  • amistre64
<1,5,-4> ^2 = 1 + 25 + 16 = sqrt(42) sqrt(42)/2 = area triangle
anonymous
  • anonymous
thank you!!
amistre64
  • amistre64
there are other ways to do it if you cant cross that well :)
amistre64
  • amistre64
but the cross bypasses any trig you might not know
anonymous
  • anonymous
Ok. I think this works pretty well for me:P

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