Question

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.

Answer 1

I think the area for the parallelogram would be the length of the cross product.

|u x v|

Answer 2

Therefore, the area of the triangle would be the half of }u x v|

Answer 3

Thanks!

Answer 4

Could you write it out? I'm a little confused, I guess.

Answer 5

what have you some up with in the cross product so far?

Answer 6

I have (20, 0, 0) but I feel like I've done something wrong...

I'm not very good with vectors.

Answer 7

\[\begin{vmatrix}X&Y&Z\\U_x&U_y&U_z\\V_x&V_y&V_z\end{vmatrix}\to
X\begin{vmatrix}\\U_y&U_z\\V_y&V_z\end{vmatrix}
-Y\begin{vmatrix}\\U_x&U_z\\V_x&V_z\end{vmatrix}
+Z\begin{vmatrix}\\U_x&U_y\\V_x&V_y\end{vmatrix}\]

Answer 8

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

Answer 9

How does what you just did work?

Answer 10

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

Answer 11

turn the matrix on its side pretty much and expand down the first column; taking the determinant of each sub matrix along the way

Answer 12

<1,5,-4> ^2 = 1 + 25 + 16 = sqrt(42)

sqrt(42)/2 = area triangle

Answer 13

thank you!!

Answer 14

there are other ways to do it if you cant cross that well :)

Answer 15

but the cross bypasses any trig you might not know

Answer 16

Ok. I think this works pretty well for me:P