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 ?

  • $\begingroup$ I think it is with the use of coordinates. In 3D case, area equals the magnitude of the cross product of the vectors that represent adjacent sides. If you are in 2D, set the third coordinate equal $0$ and follow the 3D method $\endgroup$
    – user376343
    Commented Jan 9, 2021 at 14:54
  • $\begingroup$ But the 3D method involves vectors, right? Can you explain a bit more if you can ? $\endgroup$
    – Parvathy
    Commented Jan 9, 2021 at 14:58
  • $\begingroup$ Depending on the source of the problem, "analytical method" may just mean without a calculator and without just measuring a diagram or something; then your cross product approach would be fine. If the source doesn't say what they mean by "analytical method" (and it's not mentioned in an official syllabus either), then I would use something like the cross product method and hope for the best. $\endgroup$
    – Mark S.
    Commented Jan 9, 2021 at 15:45

1 Answer 1


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.


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