three-dimensional planes

[mathstresser]

Find the area of the parallelograpm with vertices K(1,2,3), L(1,3,6), M(3,8,6), and N(3,7,3).

I don't really know where to start. To find area, don't we need to use the cross product? But I don't know what to do for the area. Do we need to do the dot product in there? I know how to do both the dot product and the cross product, I just don't know where to do it.

Please help!

 

[ChaoticLlama]

only crossproduct is required here... but you should understand why before you begin.

draw youself a parallelogram, pick one vertex. That vertex is then the tail of two vectors \(\displaystyle \L\vec a\& \vec b\).
label the angle between vector a & b as theta at the vertex.
draw a line perpendicular from the head of vector a to vector b.

Area of parallelogram = \(\displaystyle \L\ h\left| {\vec b} \right|\)

where.. \(\displaystyle \L\sin \theta = \frac{h}{{\left| {\vec a} \right|}}\)

Therefore...
Area of parallelogram = \(\displaystyle \L\left| {\left| {\vec a} \right|\left| {\vec b} \right|\sin \theta } \right|\)

If you can't go from there... then... get help from your teacher. You should have seen that exact lesson in class. if not, find another person to teach you Algebra.

 

[soroban]

Hello, mathstresser!

Find the area of the parallelograpm with vertices: K(1,2,3), L(1,3,6), M(3,8,6), N(3,7,3).

You're right . . . we want the cross product of two vectors.
\(\displaystyle \;\;\)Actually, we want the magnitude of the cross product.
But first, we have two determine the two vectors to use.

Try to make a sketch of the four points to see how they are oriented.
[Yes, I know ... it's hard to make a 3-dimensional sketch on 2-dimensional paper.]

I found that they are arranged like this:

        L(1,3,6)        M(3,8,6) 
            * - - - - - - - *
           /               /
          /               /
         /               /
        /               /
       /               /
      * - - - - - - - *
  K(1,2,3)        N(3,7,3)

We need two adjcent vectors.

Let \(\displaystyle \vec{u}\,=\,\vec{KL}\,=\,\langle0,1,3\rangle\,\) and \(\displaystyle \,\vec{v}\,=\,\vec{KN}\,=\,\langle2,5,0\rangle\)

Then: \(\displaystyle \,\vec{u}\,\times\,\vec{v}\;=\;\begin{vmatrix}i & j & k \\ 0 & 1 & 3 \\ 2 & 5 & 0\end{vmatrix}\;=\;-15i\,+\,6j\,-\,2k\;=\;\langle-15,6,-2\rangle\)

\(\displaystyle \text{Area} \;=\;|\vec{u}\,\times\,\vec{v}|\;=\;\sqrt{(-15)^2\,+\,6^2\,+\,(-2)^2}\;=\;\sqrt{265}\)

 

[mathstresser]

Thanks!