Equations of Planes — Question 1

Question 1 – Solution

Strategy Form $\overrightarrow{AB}$ and $\overrightarrow{AC}$; their cross product is normal to the plane.

Step 1: In-plane vectors Subtract coordinates component by component: $$\overrightarrow{AB}=B-A=\left\langle 3-1,1-0,-1-2\right\rangle=\left\langle 2,1,-3\right\rangle,$$ $$\overrightarrow{AC}=C-A=\left\langle -1-1,4-0,0-2\right\rangle=\left\langle -2,4,-2\right\rangle.$$

Step 2: Normal vector Their cross product is $$\overrightarrow{AB}\times\overrightarrow{AC} =\left\langle 1(-2)-(-3)(4),\ (-3)(-2)-2(-2),\ 2(4)-1(-2)\right\rangle =\left\langle 10,10,10\right\rangle.$$ Divide by $10$ to use the primitive normal $\vec n=\left\langle 1,1,1\right\rangle$. The point-normal equation through $A$ is $$(x-1)+(y-0)+(z-2)=0.$$ Therefore $$\boxed{x+y+z=3},\qquad \boxed{\hat n=\frac 1{\sqrt 3}\left\langle 1,1,1\right\rangle}.$$

Step 3: Intercepts On the $x$-axis, $y=z=0$, so $x=3$. The other axes work similarly: $$\boxed{(3,0,0),\ (0,3,0),\ (0,0,3)}.$$

Verification Substitution of $A$, $B$, and $C$ gives $3$ each time. The two in-plane vectors dot to zero with $\left\langle 1,1,1\right\rangle$, independently confirming the orientation.