I know that to find the area of parallelogram, we have to either find the cross product of its adjacent sides or half the cross product of its diagonals. However I've encountered a question to find the area of a parallelogram through an analytical method. What exactly is analytical method? Simple algebra ? If then too, I'm stuck cause the question provides me the coordinates of the parallelogram and the simple method to find the area of a parallelogram is the multiplication of its base to its height. Again I'm stuck at finding the height
I then tried to cut it into two triangles but I can't find the area as the coordinates possess three dimensions and all areas of triangle formula I've learnt is of two coordinates .
What then is the solution ?
Best Answer
Using an analytical approach means you derive the formula for the area of the parallelogram, and use that formula to find the exact result.
Let's say you know the location of vertex $\vec{c}$, and two vertices adjacent to it, $\vec{a}$ and $\vec{b}$. You know $$\vec{a} = \left[ \begin{matrix} a_x \\ a_y \\ a_z \end{matrix} \right ], \quad \vec{b} = \left[ \begin{matrix} b_x \\ b_y \\ b_z \end{matrix} \right ], \quad \vec{c} = \left[ \begin{matrix} c_x \\ c_y \\ c_z \end{matrix} \right ]$$
To find an analytical solution to the area of the parallelogram, you need to find the function $$A(a_x, a_y, a_z, b_x, b_y, b_z, c_x, c_y, c_z)$$ If you are not given three vertices of the parallelogram, but say edge lengths and angles, you need to find the area of the parallelogram in terms of those variables you are given.
For example, the two edges meeting at vertex $\vec{c}$ can be described using vectors $\vec{u}$ and $\vec{v}$, $$\vec{u} = \left[ \begin{matrix} u_x \\ u_y \\ u_z \end{matrix} \right] = \left[ \begin{matrix} a_x - c_x \\ a_y - c_y \\ a_z - c_z \end{matrix} \right], \quad \vec{v} = \left[ \begin{matrix} v_x \\ v_y \\ v_z \end{matrix} \right] = \left[ \begin{matrix} b_x - c_x \\ b_y - c_y \\ b_z - c_z \end{matrix} \right]$$ and we know that $$\begin{aligned} A &= \left\lVert \vec{u} \times \vec{v} \right\rVert \\ ~ &= \sqrt{ (\vec{u} \times \vec{v}) \cdot (\vec{u} \times \vec{v}) } \\ ~ &= \sqrt{ (u_y v_z - u_z v_y)^2 + (u_z v_x - u_x v_z)^2 + (u_x v_y - u_y v_x)^2 } \\ \end{aligned}$$ You can simplify this to find the desired function $A$.
Another example: The part of $\vec{v}$ perpendicular to $\vec{u}$ is $\vec{v}_\perp$, $$\vec{v}_\perp = \vec{v} - \vec{u} \frac{\vec{u} \cdot \vec{v}}{\vec{u} \cdot \vec{u}}$$ and is the "height" of the parallelogram, so the area is $$\begin{aligned} A &= \left\lVert \vec{u} \right\rVert \left\lVert \vec{v}_\perp \right\rVert \\ ~ &= \sqrt{ \vec{u} \cdot \vec{u} } \sqrt{ \vec{v}_\perp \cdot \vec{v}_\perp } \\ ~ &= \sqrt{ u_x^2 + u_y^2 + u_z^2 } \sqrt{ \bigr( \vec{v} - \vec{u} \frac{\vec{u} \cdot \vec{v}}{\vec{u} \cdot \vec{u}} \bigr) \cdot \bigr( \vec{v} - \vec{u} \frac{\vec{u} \cdot \vec{v}}{\vec{u} \cdot \vec{u}} \bigr) } \\ ~ &= \sqrt{ u_x^2 + u_y^2 + u_z^2 } \sqrt{ \vec{v} \cdot \vec{v} - 2 \vec{u} \cdot \vec{v} \frac{\vec{u} \cdot \vec{v}}{\vec{u} \cdot \vec{u}} + \frac{ (\vec{u} \cdot \vec{v})^2}{\vec{u} \cdot \vec{u}} } \\ \end{aligned}$$ This, too, simplifies to the same desired parallelogram area function.