Find area of 3D triangle with vertexes $A(1,0,0),\>B(0,2,0),\>C(0,0,\frac23)$

Question

I have a question that asks me to calculate the area of triangle ABC in 3D space,

• The equation for plan is $2x+y+3z = 2$, and the plan intersects each axes and we have $A(1,0,0);B(0,2,0);C(0,0,2/3)$ as its intersection points. Please calculate the area of the triangle ABC.

I have no idea what should I do in order to obtain this result!

Thanks

Answer 1

You can find the length of each side of the triangle by distance formula. Then Heron's formula will give you the area.

Answer 2

The side vectors $AB = (-1,2,0)$ and $AC = (-1,0,2/3)$, between which their angle is given by

$$\cos \theta = \frac{AB\cdot AC}{|AB||AC|}$$

Thus, the area is

$$\frac12 |AB||AC|\sin\theta = \frac12 \sqrt{|AB|^2|AC|^2-(AB\cdot AC)^2}=\frac12 \sqrt{5\cdot \frac{13}9-1}=\frac{\sqrt{14}}3$$

Answer 3

Finding length of each of the three sides $AB,BC,CA$ and using Area formula is straightforward:

$$ c^2=1^+2^2=45/9;$$ $$ a^2=2^2+\left(\frac23\right)^2={40}/{9};$$ $$b^2=1^2+\left(\frac23\right)^2=13/9;$$

Apply formula of Brahmagupta/Heron

$$ 2s= (a+b+c); Area=\sqrt{s(s-a)(s-b)(s-c)} $$

Simplifying $$ Area= \dfrac{\sqrt{14}}{3}$$

Another way is to find (one half of) cross product any two of vectors $AB,BC,CA$ and evaluate $(3 \,\times ,3)$ area matrix.

Yet another way to find matrix is evaluating (one half of the) matrix:

$$ \left( (1,x1,y1),(1,x2,y2),(1,x3,y3) \right)$$

This method can be also used later if desired to find Volume in 3-space of enclosing tetrahedron by $(4\times 4)$ matrix with a different constant coefficient($=\dfrac16$) in front.

Answer 4

We can simply use De Gua’s Theorem: $$A=\sqrt{\left(\frac{1\times2}{2}\right)^2+\left(\frac{2\times\frac{2}{3}}{2}\right)^2+\left(\frac{\frac{2}{3}\times1}{2}\right)^2}=\boxed{\frac{\sqrt{14}}{3}}.$$

Answer 5

Observe that the origin and the three vertices of the triangle form a tetrahedron. Its volume is equal to $\frac16\lvert A\times B\cdot C\rvert$. The triple product in this expression is equal to the determinant of the matrix $\begin{bmatrix}A&B&C\end{bmatrix}$, but this matrix is diagonal, so it’s simply the product of the three intercepts.

On the other hand, the volume is also equal to $\frac13A_0h$, where $A_0$ is the triangle area that you’re trying to find, and $h$ is the height of the tetrahedron. In this case this height is the distance of the origin to the plane. Use the point-plane distance formula, which I’m sure you know already, to find $h$, and solve for $A_0$.