Math Help - 3D Vectors Parallelogram Area

1. 3D Vectors Parallelogram Area

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
thanks

2. For the area of the triangle, you can use the cross product.

$\overline{P_{1}P_{2}}=(-3,-4,-1)$....[1]

$\overline{P_{1}P_{3}}=(-3,2,-9)$...[2]

$\overline{P_{1}P_{2}}\times\overline{P_{1}P_{3}}=( 38,-24,-18)$....[3]

Consequently, from the area of a triangle, use the norm of [3], $\sqrt{(38)^{2}+(-24)^{2}+(-18)^{2}}=2\sqrt{586}$

$A=\frac{1}{2}||\overline{P_{1}P_{2}}\times\overlin e{P_{1}P_{3}}||=\sqrt{586}\approx{24.21}$

3. Originally Posted by Flash
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
thanks
A parallelogram
- 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):

$M(P,Q)=\left(-\frac12~,~ 1~,~-\frac12\right)$ and $M(R,S)=\left(\frac12~,~ -1~,~\frac32\right)$...... $\implies~ false$

$M(P,R)=\left(0~,~ \frac12~,~1\right)$ and $M(Q,S)=\left(0~,~ \frac12~,~0\right)$...... $\implies~ false$

Therefore PQRS is not a parallelogram

to #2:

Same procedure: Check twice. Finally you get

$M(P,R)=\left(2~,~ 2~,~-2\right)$ and $M(Q,S)=\left(2~,~2~,~-2\right)$...... $\implies~ right$

Check if $|\overrightarrow{PQ}|=|\overrightarrow{RS}|$

$\overrightarrow{PQ}=(4, 1, -3)$ and

$\overrightarrow{RS}=(-4, -1, 3)$

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:

$\overrightarrow{PQ} \times \overrightarrow{RS} = (4, 1, -3) \times (-4, -1, 3) = (6, -24, -8)~\implies~|(6, -24, -8)|=\sqrt{676} = 26$

4. Originally Posted by earboth
To calculate the area you are supposed to know that the absolute value of the cross-product yields the value of the area:

$\overrightarrow{PQ} \times \overrightarrow{RS} = (4, 1, -3) \times (-4, -1, 3) = (6, -24, -8)~\implies~|(6, -24, -8)|=\sqrt{676} = 26$
thanks for the answer but I have a question, I tried doing the cross product and it gives (0,0,0) what am I doing wrong? this is what I did (I dont know how to use matrices to calculate the Cross Product)

1 -3 4 1
-1 3 -4 -1

(1)(3) - (-1)(-3) =0
(-3)(-4) - (4)(3) =0
(4)(-1) - (1)(-4) =0

5. Hello,

I think this is because there is a typo... It should be the cross product $\overrightarrow{PQ}\times \overrightarrow{QR}$.

To get it in the form of a matrix :

$\begin{pmatrix} a \\ b \\ c \end{pmatrix} \times \begin{pmatrix} c \\ d \\e \end{pmatrix}$

Consider $\begin{pmatrix} a & c \\ b & d \\ c & e \end{pmatrix}$

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

6. Originally Posted by Flash
thanks for the answer but I have a question, I tried doing the cross product and it gives (0,0,0) what am I doing wrong? this is what I did (I dont know how to use matrices to calculate the Cross Product)

1 -3 4 1
-1 3 -4 -1

(1)(3) - (-1)(-3) =0
(-3)(-4) - (4)(3) =0
(4)(-1) - (1)(-4) =0
Sorry for the confusion: Since $\overline{PQ} \parallel \overline{RS}$ the cross-product must yield the null vector.

Probably you have done the calculations suggested by Moo and have got the value of the area as $A=\sqrt{40} \approx 6.32$

But check my results. As you have seen I'm the MoD (= master of desaster) if it comes to calculations.