Area of triangle in 3D space

Once it occurred to me if there was a very quick way (split seconds) of computing the area of the triangle formed by the points of axis intercepts of the plane
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ Solution:
A half tetrahedron is formed by the planes $xy, yz, xz$ and $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Three of the sides of the tetrahedron have area $ab/2, bc/2, ac/2$. These sides are orthogonal
Hence the vector equivalent of the resultant is
$\frac{abi}{2} + \frac{bcj}{2} + \frac{ack}{2}$  (where $i,j,k$ are unit vectors)
Modulus of the resultant is (   $\sqrt{a^2b^2/4 + b^2c^2/4 + a^2c^2/4}$   )

Area of interest is

$$ \sqrt { a^2b^2/4 + b^2c^2/4 + a^2c^2/4 }
$$
This method could be applied to the generic case where we need to find the area of the
triangle formed by ($x1,y1,z1) (x2,y2,z2) (x3,y3,z3)$
Steps would be as below

Transform the coordinates in such a manner that the coordinates above turn into axes intercepts
$(0,0,a) (0,b,0) (c,0,0)$

Then apply the formula above

RP
info@softanalytics.net

Analyzing the differentiability of $\sqrt{x^2}$

Analyzing differentiability of $\sqrt{x^2}$

As we see a quick simplification of $\sqrt{x^2}$ yields $x$ which is a straight line.
Another simplification yields $|x|$

Clearly $|x|$ is not differentiable whereas $x$ is differentiable.

Some people might contest the $|x|$ simplification saying that is not true. I think both simplifications are correct and hence depending upon the simplification this may or may not be differentiable.

RP
info@softanalytics.net