There space two ways, the an initial one ie really LONG and complicate, the second one very SHORT and also easy, yet we have to use the vectorial product.

You are watching: Find the area of the parallelogram with vertices a(−3, 0), b(−1, 7), c(9, 6), and d(7, −1).

The an initial one:

First of all, let"s examine if the shape is yes, 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 the #MN# is the vector #vecw# together as:

#vecw=(x_M-x_N,y_M-y_N,z_M-z_N)=(0,1,3)#.

So #vecv# is parallel come #vecw#.

So, since #KL=MN# and also #KL# is parallel come #MN#, the form is a parallelogram.

The area of a parallelogram is: #A=b*h#.

We can assume that the basic #b# is #KL=sqrt10#, however finding the height is an ext complicated, since it is the street of the 2 line #r#, that contains #K and L#, and also #s#, that contains #M and also N#.

A plane, perpendicular to a line, deserve to be written:

#a(x-x_P)+b(y-y_P)+c(z-z_P)=0#,

where #vecd(a,b,c)# is a every little thing vector perpendicular to the plan, and #P# is a whaterver suggest that lies ~ above the plan.

To uncover #pi#, the is a plan perpendicular come #r#, we have the right to assume the #vecd=vecv# and #P=K#.

So:

#pi: 0(x-1)+1(y-2)+3(z-3)=0rArry+3z-11=0#.

A line deserve to be written as the mechanism of 3 equation in parametric form:

#x=x_P+at##y=y_P+bt##z=z_P+ct#

Where #P# is a whatever suggest of the line and also #vecd(a,b,c)# is a everything vector, direction the the line.

To discover #s#, we can assume that #P=M#, and also #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# us can find #Q#, foot the the height carried out 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 uncover the point #Q#, the is vital to placed #t=-3/2# in the equation that #s#.

#x=3##y=8-3/2# #z=6+3(-3/2)#

So:

#x=3#

#y=13/2#

#z=3/2#

Now, to discover #h#, we have the right to use the formula of the street 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 2nd one.

We deserve to remember the the vectorial product in between two vectors is a vector whose lenghts is the area the the parallel that has the two vector as 2 sides.

The vector: #vec(KL)=(0,1,3)#,the vector #vec(KM)=(2,6,3)#.

And currently we need to do: #vec(KL)xxvec(KM)#

We can construct the matrix:

first row: #*#,second heat #<0,1,3>#,third row#<2,6,3>#.See more: How To Use Ref In Oil Painting, Linseed Oil Acts As:, Art Appreciation Chapter 7*

The determinant is the vector: #-15veci+6vecj-2veck#, and his lenghth is: #sqrt(225+36+4)=sqrt265# that is the area requested.