Find the area of a triangle with vector sides

[thomas49th]

Homework Statement

Find the area of the triangle with vertices at the points with coordinates (1,2,3), (4,-3,2) and (8,1,1)

Homework Equations

The Attempt at a Solution

well area of a triangle is 0.5 x base x height . We can get the length of each side by taking the modulus of each vector, but the problem we have is finding the height. We need to find a the point where a line between 2 vectors cuts the other vector at 90° (the base)

How do we find this line?

Thanks
Thomas

 

[tiny-tim]

Hi Thomas!

Hint: area = absinθ = a x b

 

[thomas49th]

you mean it's simply the "cross product" or "vector product". Of course
axb = |a||b|sinx e

(where e is direction +/-)

do the modulus make any difference?

I would simply the determinant rule for finding it

i j k
1 2 3
8 1 1

-i + 23j - 15k

then take the modulus

sqrt(755)

Right?

Thanks
Thomas

 

[tiny-tim]

do the modulus make any difference?

I would simply the determinant rule for finding it

i j k
1 2 3
8 1 1

-i + 23j - 15k

then take the modulus

ooh yes, I left out the | | symbols!

yes, that's the right method, but you've only found the area of the triangle formed by points 1 and 3 and the origin, haven't you?

 

[Altabeh]

Homework Statement

Find the area of the triangle with vertices at the points with coordinates (1,2,3), (4,-3,2) and (8,1,1)

Homework Equations

The Attempt at a Solution

well area of a triangle is 0.5 x base x height . We can get the length of each side by taking the modulus of each vector, but the problem we have is finding the height. We need to find a the point where a line between 2 vectors cuts the other vector at 90° (the base)

How do we find this line?

Thanks
Thomas

Well, this may not be what your exercise is all about, but you can try "[URL formula[/URL].

 

[thomas49th]

ooh yes, I left out the | | symbols!

yes, that's the right method, but you've only found the area of the triangle formed by points 1 and 3 and the origin, haven't you?

Tim, havn't I got to half it to get that area :) Right? How does that help though.

I can remember from my A levels that there are position vectors, which are relative to the centre, direction vectors and I think they have a parameter in them telling you how far "along" the vector to go. I'm getting myself confused over the basics. :sigh:

Anyway thinking about it again it's the area = 0.5 |AB x AC|

a = (1,2,3), b = (4,-3,2) and c = (8,1,1)

AB = b - a
= 3,-5,-1
AC = c - a
= 4, 4, -1

but what does |AB x AC| mean? |AB| x |AC|?

Am I getting there?

@ Altabeh - I saw this mentioned on the WIKI page for triangles, though useful as it looks I want I'm having conflicting thoughts and confusing myself with the fundamentals which I want to sort out.

Thanks

 

[tiny-tim]

Tim, havn't I got to half it to get that area :) Right?

oooh, I'm not doing very well on this … yes, half!

I can remember from my A levels that there are position vectors, which are relative to the centre, direction vectors and I think they have a parameter in them telling you how far "along" the vector to go. I'm getting myself confused over the basics. :sigh:

Yes, that's right …

the formula for a general point on the line AB is a + k(b - a), for any number k.

(though of course you don't need that in this case)

Anyway thinking about it again it's the area = 0.5 |AB x AC|

a = (1,2,3), b = (4,-3,2) and c = (8,1,1)

AB = b - a
= 3,-5,-1
AC = c - a
= 4, 4, -1

That's right (except the last one is BC , but it doesn't matter anyway).

but what does |AB x AC| mean? |AB| x |AC|?

No, it means the modulus of the vector AB x AC (or AB x BC or AC x BC).

So just do (3,-5,-1) x (4, 4, -1), and take half the modulus.

 

[Altabeh]

No, it means the modulus of the vector AB x AC (or AB x BC or AC x BC).

So just do (3,-5,-1) x (4, 4, -1), and take half the modulus.

And at the end of your calculation, check your result to see if it matches the one I've prepared through Herron's formula here:

[tex](1/2)*sqrt(2\sqrt(14)+2\sqrt(66)+2\sqrt(29)) \approx 2.936908289[/tex].

AB

 

[thomas49th]

No, it means the modulus of the vector AB x AC (or AB x BC or AC x BC).

So just do (3,-5,-1) x (4, 4, -1), and take half the modulus.

So (3,-5,-1) x (4, 4, -1),

= (12, -20, 1)

= sqrt(144 + 400 + 1)

= sqrt(545)

= 5sqrt(109)

That's not what Altabeh got. Guessing I am the one who is wrong

 

[tiny-tim]

So (3,-5,-1) x (4, 4, -1),

= (12, -20, 1)

Nooo , you need the cross product (you did it in your post #3, except you used OA x OB instead of AB x BC)

 

[thomas49th]

i j k
3 - 5 -1
4 4 -1

(5 -- 4)i - (-3 - -4)j + (12 - - 20)k

9i -j + 32k

sqrt(81 + 1 +1024) = sqrt(1106) ~= 33.3

i take it I'm wrong, as Altabeh's answer is right?

Thanks

 

[thomas49th]

and take half of that

= 0.5sqrt(1106)

= 16.6

 

[vela]

The cross product method is the most straightforward, but you could also use the dot product to find the angle between two sides and use it to calculate the height of the triangle.

[tex]\cos\theta = \frac{(3, -5, -1)\cdot(4, 4, -1)}{|(3, -5, -1)| |(4, 4, -1)|} = \frac{3\times4+(-5)\times4+(-1)\times(-1)}{\sqrt{3^2+(-5)^2+(-1)^2}\sqrt{4^2+4^2+(-1)^2}}=\frac{-7}{\sqrt{33}\sqrt{35}}[/tex]

So

[tex]\sin\theta = \sqrt{1-\cos^2\theta}=\sqrt{1-\frac{(-7)^2}{33\times35}}=\frac{\sqrt{1106}}{\sqrt{33}\sqrt{35}}[/tex]

You can arbitrarily choose one side to be the base of the triangle. The height will then be the length of the other side multiplied by the sine of the angle between them, so you get

[tex]A=\frac{1}{2}bh = \frac{1}{2}\sqrt{33}\sqrt{35}\sin\theta = 0.5\sqrt{1106} = 16.6[/tex]

 

[tiny-tim]

Looks like it's 16.6.

 

[Altabeh]

So (3,-5,-1) x (4, 4, -1),

= (12, -20, 1)

= sqrt(144 + 400 + 1)

= sqrt(545)

= 5sqrt(109)

That's not what Altabeh got. Guessing I am the one who is wrong

No! I was wrong! The correct calculation is:

[tex]sqrt(s*(s-\sqrt{33})*(s-\sqrt{35})*(s-\sqrt{54}) \approx 16.62828916[/tex],

where [tex] s=(\sqrt{54}+\sqrt{35}+\sqrt{33})/2[/tex].

AB

 

[tiny-tim]

Thomas, you'd better use either my method or vela's!