Equations of Planes — Question 4

Question 4 – Solution

Strategy The intersection direction is perpendicular to both normals. Solve the two plane equations for one point and one free parameter.

Step 1: Solve simultaneously Add the equations: $$(x+y-z)+(2x-y+z)=2+1\Longrightarrow 3x=3\Longrightarrow x=1.$$ Substitute $x=1$ into $x+y-z=2$: $$1+y-z=2\Longrightarrow y=1+z.$$ Let the free coordinate be $z=t$. Then $y=1+t$, so $$\boxed{\vec r=\left\langle 1,1,0\right\rangle+t\left\langle 0,1,1\right\rangle}.$$

Step 2: Confirm the direction The normals are $\vec n_1=\left\langle 1,1,-1\right\rangle$ and $\vec n_2=\left\langle 2,-1,1\right\rangle$. Their cross product is $\left\langle 0,-3,-3\right\rangle$, parallel to $\left\langle 0,1,1\right\rangle$.

Step 3: Angle Compute $\vec n_1\cdot\vec n_2=2-1-1=0$. Therefore $$\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{\|\vec n_1\|\|\vec n_2\|}=0, \qquad \boxed{\theta=90^\circ}.$$

Verification Substitution of $(1,1+t,t)$ gives $2$ in $\Pi_1$ and $1$ in $\Pi_2$ for every $t$.