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.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I think the area for the parallelogram would be the length of the cross product. |u x v|
Therefore, the area of the triangle would be the half of }u x v|
Thanks!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question