Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

goldmahkot

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).

  • one year ago
  • one year ago

  • This Question is Closed
  1. goldmahkot
    Best Response
    You've already chosen the best response.
    Medals 0

    Now, 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.

    • one year ago
  2. goldmahkot
    Best Response
    You've already chosen the best response.
    Medals 0

    For the dot product, remember that we could find the area of a triangle with a "shortcut method" i.e. (|A'||B|sin(x))/2. Therefore, the area would be (SQRT(3)(sin45deg) X SQRT(2))/2 which gives an answer of 0.866.

    • one year ago
  3. goldmahkot
    Best Response
    You've already chosen the best response.
    Medals 0

    But 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!

    • one year ago
  4. Waynex
    Best Response
    You've already chosen the best response.
    Medals 1

    The 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....

    • one year ago
  5. goldmahkot
    Best Response
    You've already chosen the best response.
    Medals 0

    Waynex, 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!

    • one year ago
    • Attachments:

See more questions >>>

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.