For the area of the triangle, you can use the cross product.
Consequently, from the area of a triangle, use the norm of ,
Hi can anybody help me I have to determine if PQRS is a parallelogram and then find the area
1- P (1,1,1) Q (-2,3,-2) R (-1,0,1) S (2,-2,2)
2- P (-1,1,1) Q(3,2,-2) R(5,3,-5) S (1,2,-2)
What is the procedure to do this?
Also i have to determine the area of the triangle P(4,-1,3) Q(1,-5,2) R(1,1,-6)
One last thing what is the formula to calculate the angle between 2 vectors when is a 3d vector?
any help will be appreciated it
- has 2 pairs of parallel sides
- the parallel sides have the same length
- the mean value of the coordinates of opposite vertices is the midpoint of both diagonals.
to #1: Calculate the midpoint of the diagonals (because you don't know which vertices are opposite you have to check in 2 steps):
Therefore PQRS is not a parallelogram
Same procedure: Check twice. Finally you get
Obviously both vectors have the same length.
To calculate the area you are supposed to know that the absolute value of the cross-product yields the value of the area:
1 -3 4 1
-1 3 -4 -1
(1)(3) - (-1)(-3) =0
(-3)(-4) - (4)(3) =0
(4)(-1) - (1)(-4) =0
I think this is because there is a typo... It should be the cross product .
Your method is correct
To get it in the form of a matrix :
The cross product will be :
+ determinant of the remaining after removing the first line
- determinant of the remaining after removing the second line
+ determinant of the remaining after removing the third line
Probably you have done the calculations suggested by Moo and have got the value of the area as
But check my results. As you have seen I'm the MoD (= master of desaster) if it comes to calculations.