3
$\begingroup$

I know the geometrical meaning of the determinant of a matrix, and I know that, for example, the area of a parallelogram defined by two vectors $$ v=\begin{bmatrix}a \\b \end{bmatrix},\quad w=\begin{bmatrix}c \\d \end{bmatrix}, $$ is equal to $$ \det \begin{pmatrix} a & c \\ b & d \end{pmatrix}. $$

I know that the area of the triangle $ABC$ is equal to $$\frac{1}{2} \det\begin{pmatrix} 1 & 1 & 1\\ x_A & x_B & x_C \\ y_A & y_B & y_C \end{pmatrix}. $$

I would like to find a geometrical proof of the last formula. Why is the area of a triangle (numerically) equal to the volume of the solid generated by the three vectors $[1, x_A, y_A]$, $[1, x_B, y_B]$, $[1, x_C, y_C]$? I can verify it, but I can't see it.

$\endgroup$
  • $\begingroup$ Sorry, I forgot a fraction... thanks $\endgroup$ – zar Feb 26 at 11:00
2
$\begingroup$

Subtract the second and the third columns by the first (this is geometrically a shear). Now you get a block lower triangular matrix and the determinant becomes the area of a parallelogram (or strictly speaking, a slanted cylinder with a parallelogram base and unit height) with two adjacent sides $\pmatrix{x_B-x_A\\ y_B-y_A}$ and $\pmatrix{x_C-x_A\\ y_C-y_A}$. Multiply it by one half, you get the area of the triangle.

$\endgroup$
  • $\begingroup$ Thanks. I tried with a GeoGebra construction, and now I "see" it. $\endgroup$ – zar Feb 26 at 17:20
1
$\begingroup$

It might be easier to visualize if you move the row of $1$s to the bottom so that the triangle lies on the plane $z=1$. (This doesn’t change the value of the determinant.)

enter image description here

Together with the origin, this triangle forms a tetrahedron with altitude $1$, so its volume is numerically equal to the area of $\triangle{ABC}$. However, you can take any of the other faces as the base, say $\triangle{OBC}$. The area of this triangle is $\frac12\|B\times C\|$. (This is fundamentally the same as using a determinant in 2-D.) The altitude to $A$ from this face is the length of the projection of $A$ onto a normal $\mathbf n$ to the face, which can be computed via a dot product: $A\cdot{\mathbf n\over\|\mathbf n\|}$. For the normal, we can take $\mathbf n=B\times C$, so the volume of the tetrahedron, and hence the area of $\triangle{ABC}$, is $$\frac12\|B\times C\| \left(A\cdot{B\times C \over \|B\times C\|}\right) = \frac12A\cdot B\times C = \frac12\begin{vmatrix}x_A&x_B&x_C\\y_A&y_B&y_C\\1&1&1 \end{vmatrix}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.