How do you find the area of the parallelogram with vertices
k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3)?

Question

Luke Alvarez · Accepted Answer

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.

Emily Ammerman · Answer

undefined To find the area of the parallelogram formed by the given vertices, you can use the cross product of two vectors formed by the sides of the parallelogram. Let's label the vectors as $\overrightarrow{LM}$ and $\overrightarrow{LN}$. Then, the cross product of these vectors will give us a vector perpendicular to the parallelogram, and its magnitude will give us the area of the parallelogram.

First, calculate the vectors:
$\overrightarrow{LM} = \overrightarrow{m} - \overrightarrow{l}$
$\overrightarrow{LN} = \overrightarrow{n} - \overrightarrow{l}$

Next, calculate the cross product of $\overrightarrow{LM}$ and $\overrightarrow{LN}$, denoted as $\overrightarrow{LM} \times \overrightarrow{LN}$. This will give you a vector perpendicular to the parallelogram.

Finally, calculate the magnitude of $\overrightarrow{LM} \times \overrightarrow{LN}$. This magnitude will give you the area of the parallelogram formed by the given vertices.

How do you find the area of the parallelogram with vertices k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3)?