Equations of Planes — Question 2

Question 2 – Solution

Strategy Combine the line direction with a vector from a point on the line to $P$. Their cross product is a plane normal.

Step 1: Two directions in the plane Take $A=(1,-2,0)$ from the line. Its direction is $\vec d=\left\langle 2,1,-1\right\rangle$, while $$\overrightarrow{AP}=P-A=\left\langle 0-1,1-(-2),4-0\right\rangle=\left\langle -1,3,4\right\rangle.$$

Step 2: Find a normal Compute $$\vec d\times\overrightarrow{AP} =\left\langle 1(4)-(-1)(3),\ (-1)(-1)-2(4),\ 2(3)-1(-1)\right\rangle =\left\langle 7,-7,7\right\rangle.$$ Thus $\vec n=\left\langle 1,-1,1\right\rangle$. Using point $A$, $$(x-1)-(y+2)+(z-0)=0,$$ so the plane is $$\boxed{x-y+z=3}.$$

Uniqueness The vectors $\vec d$ and $\overrightarrow{AP}$ are not parallel, so they span exactly one plane.

Verification Point $P$ gives $0-1+4=3$. A general point on $L$ is $(1+2t,-2+t,-t)$, and $$(1+2t)-(-2+t)-t=3$$ for every $t$, proving that the whole line is contained in the plane.