mathmate
  • mathmate
QUIZ TIME : easy question : area of triangle The plane 6x+4y+2z=12 intersects the coordinate planes to form three sides of a triangle. Find the area of the triangle. The challenge here is to show as many ways as you can find to get the correct answer. Give exact values if you can, eg. sqrt(14) instead of 3.742. NOTE to readers: Please award a medal to someone who presents a formula or method that you would not have used.
Mathematics
jamiebookeater
  • jamiebookeater
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mathmate
  • mathmate
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amistre64
  • amistre64
x=0, get the intercepts; put them in a distance formula .... di this for each variable to zero out. then you have the side measures and can do a heron on it
amistre64
  • amistre64
or... divide by 12 and divide off the tops to get the intercepts if we dont have time to do them one by one

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amistre64
  • amistre64
6x+4y+2z=12 12 12 12 /6 /4 /2 ------------- 2 3 6 are our intercepts
amistre64
  • amistre64
one way is we can take these and form vectors to get an angle to do a sin area formula with
anonymous
  • anonymous
3.164 squre unit approx using area = squrt{ s(s-a)(s-b)(s-c)) ;s=(a+b+c)/2
amistre64
  • amistre64
(2,0,0) (0,0,6) -(0,3,0) -(0,3,0) ------- ------- <2,-3,0> <0,-3,6> cos(yaxis) = <2,-3,0>.<0,-3,6> 9 ----------------- = --------------- |<2,-3,0>| |<0,-3,6>| sqrt(4+9+9+36) cos(y) = 9/sqrt(58) y = cos-1(9/sqrt(58)) |<2,-3,0>| |<0,-3,6>| sin(cos-1(9/sqrt(58))) / 2 9/sqrt(58) sin(cos-1(9/sqrt(58))) / 2 maybe :)
mathmate
  • mathmate
So far I see Heron's formula from Amistre64 and arijit, and vectors from Amistre64. Please provide detailed calculations and answers.
amistre64
  • amistre64
my error is in the sqrt(58) :) sqrt(4+9) * sqrt(9+36) sqrt(13*45) 3 sqrt(65)
amistre64
  • amistre64
Base * (..... ......height..........) /2 9/3sqrt(65) sin(cos-1(9/3sqrt(65))) / 2 3/sqrt(65) sin(cos-1(3/sqrt(65))) / 2 hopefully lol
anonymous
  • anonymous
PLEASE SEE THE DRAWING........
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amistre64
  • amistre64
i got a few more errors, but the concepts seems solid ;)
mathmate
  • mathmate
So 2,3,6 are the intercepts, sqrt(13), sqrt(45), sqrt(40) are the sides.
amistre64
  • amistre64
right, and you can heron that for sheer simplicity
mathmate
  • mathmate
Some "exact" answers would be nice! :)
mathmate
  • mathmate
Agree, we also want as many different ways as possible.
amistre64
  • amistre64
|dw:1324567464017:dw|
amistre64
  • amistre64
sqrt(45) sin(y) = height sqrt(13) = base
amistre64
  • amistre64
b*h/2 = 3sqrt(14) if i did it right that time
mathmate
  • mathmate
Wow! First exact answer! 3sqrt(14) (that's what I have too). So far we have intercepts: 2,3,6 sides : sqrt(13), sqrt(45), sqrt(40) Area : 3sqrt(14) Methods: Heron : Amistre64, arigit Vectors: Amistre64 (to be completed) bh/2 : Amistre64
amistre64
  • amistre64
the bh/2 is after the vectors give us a sin for an angle to determine the height
amistre64
  • amistre64
i wonder if its worth it to do a 3d integration :) or at least take the measures and translate them to a xy congruent triangle
phi
  • phi
\[A= \frac{1}{2}\sqrt{|x|^2|y|^2-(x \cdot y)^2}\] with (using amistre's two vectors) x=<2,-3,0> y= <0,-3,6> gives 0.5sqrt( 13*45-81)= 1.5sqrt(65-9)= 1.5sqrt(56)= 3sqrt(14)
amistre64
  • amistre64
work in progress here |dw:1324568575430:dw| \[\int_{0}^{6}distance.x.to.y\ dz\]
amistre64
  • amistre64
\[X_i=\left(\frac{z-6}{3},0,z\right)\] \[Y_i=\left(0,\frac{z-6}{2},z\right)\] \[X_i-Y_i=\left(\frac{z-6}{3},\frac{-z+6}{2},0\right)\] distance from X to Y:\[\sqrt{(\frac{z-6}{3})^2+(\frac{-z+6}{2})^2}\] \[\sqrt{\frac{z^2-6z+36}{9}+\frac{z^2-6z+36}{4}}\] \[\sqrt{\frac{4z^2-4.6z+4.36+9z^2-9.6z+9.36}{36}}\] \[\frac{\sqrt{13z^2-78z+468}}{6}\] ergo lol \[\int_{0}^{6} \frac{\sqrt{13z^2-78z+468}}{6}dz\]
amistre64
  • amistre64
hmm, wolf says that about 19, which is a bit off from the 3sqrt(14) i wonder why
amistre64
  • amistre64
348 not 468 ... but still thats off
amistre64
  • amistre64
oh well, it was a thought. guess i dont know how to freehand integrals yet :)
mathmate
  • mathmate
OK, here's what we have so far (let me know if I missed anything) Phi: does the formula come from cross product, it's neat, like a generalized form of bh/2. Related to that, I suggest the cross product of Amistre64's two vectors. (1/2)AxB=(1/2)<0,-3,6>x<2,-3,0>=(1/2)sqrt(504)=3sqrt(14) Methods: 1. Heron : Amistre64, arigit 2. (1/2)xysin(theta) + bh/2 : Amistre64 3. integration : Amistre64 (in progress) 4. (1/2)sqrt(|x|^2|y|^2-x.y) : phi 5. half magnitude of cross-product, (1/2)|PxQ|: mathmate

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