The answer is: A=sqrt265.
There are two ways, the first one ie VERY LONG and complicate, the second one VERY SHORT and easy, but we have to use the vectorial product.
The first one:
First of all, let's check if the shape is really a parallelogram:
KL=sqrt((x_K-x_L)^2+(y_K-y_L)^2+(x_K-z_L)^2)=
=sqrt((1-1)^2+(2-3)^2+(3-6)^2)=sqrt(0+1+9)=sqrt10.
MN=sqrt((3-3)^2+(8-7)^2+(6-3)^2)=sqrt(0+1+9)=sqrt10.
So KL=MN
The direction of KL is the vector vecv such as:
vecv=(x_K-x_L,y_K-y_L,z_K-z_L)=(0,1,3).
The direction of MN is the vector vecw such as:
vecw=(x_M-x_N,y_M-y_N,z_M-z_N)=(0,1,3).
So vecv is parallel to vecw.
So, since KL=MN and KL is parallel to MN, the shape is a parallelogram.
The area of a parallelogram is: A=b*h.
We can assume that the base b is KL=sqrt10, but finding the height is more complicated, because it is the distance of the two line r, that contains K and L, and s, that contains M and N.
A plane, perpendicular to a line, can be written:
a(x-x_P)+b(y-y_P)+c(z-z_P)=0,
where vecd(a,b,c) is a whatever vector perpendicular to the plan, and P is a whaterver point that lies on the plan.
To find pi, that is a plan perpendicular to r, we can assume that vecd=vecv and P=K.
So:
pi: 0(x-1)+1(y-2)+3(z-3)=0rArry+3z-11=0.
A line can be written as the system of three equation in parametric form:
x=x_P+at
y=y_P+bt
z=z_P+ct
Where P is a whatever point of the line and vecd(a,b,c) is a whatever vector, direction of the line.
To find s, we can assume that P=M, and vecd=vecw.
So s:
x=3+0t
y=8+1t
z=6+3t
or:
x=3
y=8+t
z=6+3t.
Now, solving the system between pi and s we can find Q, foot of the height conducted from K to s.
y+3z-11=0
x=3
y=8+t
z=6+3t
8+t+3(6+3t)-11=0rArr10t=-15rArrt=-3/2.
So, to find the point Q, it is necessary to put t=-3/2 in the equation of s.
x=3
y=8-3/2
z=6+3(-3/2)
So:
x=3
y=13/2
z=3/2
Now, to find h, we can use the formula of the distance of two points, K and Q, just seen before:
h=sqrt((1-3)^2+(2-13/2)^2+(3-3/2)^2)=sqrt(2^2+(9/2)^2+(3/2)^2)=sqrt(4+81/4+9/4)=sqrt((16+81+9)/4)=sqrt106/2.
Finally the area is:
A=sqrt10sqrt106/2=sqrt1060/2=sqrt(4*265)/2=sqrt265.
The second one.
We can remember that the vectorial product between two vectors is a vector whose lenghts is the area of the parallelogram that has the two vector as two sides.
The vector: vec(KL)=(0,1,3),
the vector vec(KM)=(2,6,3).
And now we have to do: vec(KL)xxvec(KM)
We can build the matrix:
first row: [i,j,k],
second row [0,1,3],
third row[2,6,3].
The determinant is the vector: -15veci+6vecj-2veck, and his lenghth is: sqrt(225+36+4)=sqrt265 that is the area requested.
