A community for students.
Here's the question you clicked on:
 0 viewing
goldmahkot
 3 years ago
I was comparing the areas of a triangle that could be obtained via the conventional method, dot product, and cross product. Strangely, the latter did not agree with the former two.
Consider two vectors, A <1,1,1> and B<1,1,0>. Both these vectors form a right angled triangle with an angle of 45deg between both vectors, a hypotenuse of SQRT(3), and length of SQRT(2).
goldmahkot
 3 years ago
I was comparing the areas of a triangle that could be obtained via the conventional method, dot product, and cross product. Strangely, the latter did not agree with the former two. Consider two vectors, A <1,1,1> and B<1,1,0>. Both these vectors form a right angled triangle with an angle of 45deg between both vectors, a hypotenuse of SQRT(3), and length of SQRT(2).

This Question is Closed

goldmahkot
 3 years ago
Best ResponseYou've already chosen the best response.0Now, the conventional method for finding the area of a triangle is (height X length)/2. In this case, it would be (SQRT(3)(sin45deg) X SQRT(2))/2. The answer obtained is 0.866.

goldmahkot
 3 years ago
Best ResponseYou've already chosen the best response.0For the dot product, remember that we could find the area of a triangle with a "shortcut method" i.e. (A'Bsin(x))/2. Therefore, the area would be (SQRT(3)(sin45deg) X SQRT(2))/2 which gives an answer of 0.866.

goldmahkot
 3 years ago
Best ResponseYou've already chosen the best response.0But when it comes to the cross product, things get a little strange.\[A \times B = <1,1,0>\]To find its magnitude (which is equals to the area of a parallelogram): \[\sqrt{(1)^{2}+(1^{2}) + 0^{2}} = \sqrt{2}\]To find the area of the triangle, we divide the answer above by 2: \[\frac{ \sqrt2 }{2} = 0.7071\] The area obtained using the cross product is different from the other two methods!

Waynex
 3 years ago
Best ResponseYou've already chosen the best response.1The area you calculated via the cross product is the correct one. The problem you have on the first two is that the angle between A and B is not pi/4. The angle would be pi/4 if both vectors were in the xy plane, but one vector is shift upward in angle, and that changes the angle between them. The angle is actual arccos[2/sqrt(6)], or aprox. 0.61 radians. The first equation would be sqrt(3) * sqrt(2) * sin(0.615...) * 0.5 = 0.707....

goldmahkot
 3 years ago
Best ResponseYou've already chosen the best response.0Waynex, Thanks for you input on my question. You were right; I was assuming the angle between A and B to be pi/4 (a HUGE mistake). After recalculating the equations, all three areas agree with each other. I can now sleep in peace!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.