Finding area of triangle with vector product (1 Viewer)

[prime-factor]

Hello. I need some help with a vector product question which involves finding the area of a triangle (i, j and k). The problem I am having is determining the i, j and k of the third side of the triangle. Attached is the picture.

I need vectors AB, and AC. I can get AB, but it is AC which I am not sure what to do.

BTW, the answer is 43.5 according to the book.

Area = (1/2) l a x b l ----------> 'x' = cross(vector) product

Thanks in Advance.

 

[Iruka]

We use the fact that the feet of the poles form an equilateral triangle. We can centre the coordinate system whereever we like, so I am going to put the origin halfway between the feet of two of the poles. Then the coordinates of the feet of the poles are (5,0,0), (-5,0,0) and (0, 5sqrt(3)/2,0), and the tops of the poles are at (5,0,2), (-5,0,3) and (0, 5sqrt(3)/2,2.5).

I think you should be able to work it out from there.

 

[Drongoski]

Hello. I need some help with a vector product question which involves finding the area of a triangle (i, j and k). The problem I am having is determining the i, j and k of the third side of the triangle. Attached is the picture.

I need vectors AB, and AC. I can get AB, but it is AC which I am not sure what to do.

BTW, the answer is 43.5 according to the book.

Area = (1/2) l a x b l ----------> 'x' = cross(vector) product

Thanks in Advance.

Solution outline:

Let the bases of the 3 poles be P, Q and R resp, with P at the origin of the xyz-axes, with Q on the x-axis. Let A, B and C be the top of the 3 poles so that PA is along the positive z-axis. Let M be mid=pt of PQ. Therefore MR = 5sqr(3)

Let V(PA) stand for vector PA etc.

Then: V(PA) = 2.5 k V(PQ) = 10 i V(PR) = 5 i + 5sqr(3) j V(RC) = 3 k

V(PC) = V(PR) + V(RC) = 5i + 5sqr(3)j + 3k

Therefore V(AC) = V(PC) - V(PA) = 5i +5sqr(3)j + 0.5k

V(PB) = V(PQ) + V(QB) = 10i + 2k
V(AB) = V(PB) - V(PA) = 10i + 0j - 0.5k

So we now have vectors AC and AB

Take the cross product AC x AB = -2.5 sqr(3) + 7.5j - 50sqr(3) k

Take 1/2 x Abs value = 43.5172 ..

Trust this helps!

Which Examination jurisdiction are you in ? Certainly not NSW HSC !

 

[prime-factor]

Thanks very much for your help guys. Sorry for the late reply. I'm in Queensland; it is very different to the HSC