Equations of Planes — Question 10

Question 10

A plane cuts the positive coordinate axes at $A$, $B$, and $C$. The centroid of the intercept triangle $ABC$ is $$G=(2,3,4).$$ Tasks

1. Determine $A$, $B$, and $C$.

2. Find the plane in intercept and Cartesian forms.

3. Find the area of triangle $ABC$.

4. Find the volume of the tetrahedron bounded by the plane and the coordinate planes.

Question 10 – Solution

Strategy Write $A=(a,0,0)$, $B=(0,b,0)$, and $C=(0,0,c)$. Their centroid directly determines the three intercepts.

Step 1: Recover the intercepts Write $A=(a,0,0)$, $B=(0,b,0)$, and $C=(0,0,c)$. Then $$G=\frac{A+B+C}{3}=\left(\frac a3,\frac b3,\frac c3\right),$$ so $a/3=2$, $b/3=3$, and $c/3=4$. Hence $$\boxed{A=(6,0,0),\quad B=(0,9,0),\quad C=(0,0,12)}.$$

Step 2: Plane equation A plane with intercepts $a,b,c$ has equation $x/a+y/b+z/c=1$. Therefore $$\boxed{\frac x6+\frac y9+\frac z{12}=1},\qquad \boxed{6x+4y+3z=36}.$$

Step 3: Triangle area Compute $$\overrightarrow{AB}=\left\langle -6,9,0\right\rangle,\qquad \overrightarrow{AC}=\left\langle -6,0,12\right\rangle,$$ $$\overrightarrow{AB}\times\overrightarrow{AC} =\left\langle 9(12),\ 0(-6)-(-6)(12),\ (-6)(0)-9(-6)\right\rangle =\left\langle 108,72,54\right\rangle.$$ Its length is $$\sqrt{108^2+72^2+54^2}=\sqrt{19764}=18\sqrt{61}.$$ Hence $\boxed{\operatorname{Area}(ABC)=9\sqrt{61}}$.

Step 4: Tetrahedron volume The coordinate-axis edges from the origin are mutually perpendicular with lengths $6,9,12$. A tri-rectangular tetrahedron has one-sixth the corresponding box volume, so $$\boxed{V=\frac 16(6)(9)(12)=108}.$$

Verification Substitution of each intercept into $6x+4y+3z=36$ succeeds, and their coordinate average is $(2,3,4)$.