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)?

Feb 4, 2015

The answer is: $A = \sqrt{265}$.

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:

$K L = \sqrt{{\left({x}_{K} - {x}_{L}\right)}^{2} + {\left({y}_{K} - {y}_{L}\right)}^{2} + {\left({x}_{K} - {z}_{L}\right)}^{2}} =$

$= \sqrt{{\left(1 - 1\right)}^{2} + {\left(2 - 3\right)}^{2} + {\left(3 - 6\right)}^{2}} = \sqrt{0 + 1 + 9} = \sqrt{10}$.

$M N = \sqrt{{\left(3 - 3\right)}^{2} + {\left(8 - 7\right)}^{2} + {\left(6 - 3\right)}^{2}} = \sqrt{0 + 1 + 9} = \sqrt{10}$.

So $K L = M N$

The direction of $K L$ is the vector $\vec{v}$ such as:

$\vec{v} = \left({x}_{K} - {x}_{L} , {y}_{K} - {y}_{L} , {z}_{K} - {z}_{L}\right) = \left(0 , 1 , 3\right)$.

The direction of $M N$ is the vector $\vec{w}$ such as:

$\vec{w} = \left({x}_{M} - {x}_{N} , {y}_{M} - {y}_{N} , {z}_{M} - {z}_{N}\right) = \left(0 , 1 , 3\right)$.

So $\vec{v}$ is parallel to $\vec{w}$.

So, since $K L = M N$ and $K L$ is parallel to $M N$, the shape is a parallelogram.

The area of a parallelogram is: $A = b \cdot h$.

We can assume that the base $b$ is $K L = \sqrt{10}$, but finding the height is more complicated, because it is the distance of the two line $r$, that contains $K \mathmr{and} L$, and $s$, that contains $M \mathmr{and} N$.

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

$a \left(x - {x}_{P}\right) + b \left(y - {y}_{P}\right) + c \left(z - {z}_{P}\right) = 0$,

where $\vec{d} \left(a , b , c\right)$ 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 $\vec{d} = \vec{v}$ and $P = K$.

So:

$\pi : 0 \left(x - 1\right) + 1 \left(y - 2\right) + 3 \left(z - 3\right) = 0 \Rightarrow y + 3 z - 11 = 0$.

A line can be written as the system of three equation in parametric form:

$x = {x}_{P} + a t$
$y = {y}_{P} + b t$
$z = {z}_{P} + c t$

Where $P$ is a whatever point of the line and $\vec{d} \left(a , b , c\right)$ is a whatever vector, direction of the line.

To find $s$, we can assume that $P = M$, and $\vec{d} = \vec{w}$.

So $s$:

$x = 3 + 0 t$
$y = 8 + 1 t$
$z = 6 + 3 t$

or:

$x = 3$
$y = 8 + t$
$z = 6 + 3 t$.

Now, solving the system between $\pi$ and $s$ we can find $Q$, foot of the height conducted from $K$ to $s$.

$y + 3 z - 11 = 0$
$x = 3$
$y = 8 + t$
$z = 6 + 3 t$

$8 + t + 3 \left(6 + 3 t\right) - 11 = 0 \Rightarrow 10 t = - 15 \Rightarrow t = - \frac{3}{2}$.

So, to find the point $Q$, it is necessary to put $t = - \frac{3}{2}$ in the equation of $s$.

$x = 3$
$y = 8 - \frac{3}{2}$
$z = 6 + 3 \left(- \frac{3}{2}\right)$

So:

$x = 3$

$y = \frac{13}{2}$

$z = \frac{3}{2}$

Now, to find $h$, we can use the formula of the distance of two points, $K \mathmr{and} Q$, just seen before:

$h = \sqrt{{\left(1 - 3\right)}^{2} + {\left(2 - \frac{13}{2}\right)}^{2} + {\left(3 - \frac{3}{2}\right)}^{2}} = \sqrt{{2}^{2} + {\left(\frac{9}{2}\right)}^{2} + {\left(\frac{3}{2}\right)}^{2}} = \sqrt{4 + \frac{81}{4} + \frac{9}{4}} = \sqrt{\frac{16 + 81 + 9}{4}} = \frac{\sqrt{106}}{2}$.

Finally the area is:

$A = \sqrt{10} \frac{\sqrt{106}}{2} = \frac{\sqrt{1060}}{2} = \frac{\sqrt{4 \cdot 265}}{2} = \sqrt{265}$.

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: $\vec{K L} = \left(0 , 1 , 3\right)$,
the vector $\vec{K M} = \left(2 , 6 , 3\right)$.

And now we have to do: $\vec{K L} \times \vec{K M}$

We can build the matrix:

first row: $\left[i , j , k\right]$,
second row $\left[0 , 1 , 3\right]$,
third row$\left[2 , 6 , 3\right]$.

The determinant is the vector: $- 15 \vec{i} + 6 \vec{j} - 2 \vec{k}$, and his lenghth is: $\sqrt{225 + 36 + 4} = \sqrt{265}$ that is the area requested.