Equations of Planes — Question 7

Question 7

Find the plane in the family containing the intersection line of $x+y+z=2$ and $2x-y+3z=1$ that is perpendicular to the plane $x-2y+z=4$.

Question 7 – Solution

Strategy Apart from $2x-y+3z=1$ itself, every plane through the common line has equation $(x+y+z-2)+\lambda(2x-y+3z-1)=0$. The omitted plane has normal dot product $7\ne 0$ with the given normal, so it cannot qualify. Choose $\lambda$ so the family normal is perpendicular to the given normal.

Normal condition The family normal is $\left\langle 1+2\lambda,1-\lambda,1+3\lambda\right\rangle$. Dotting with $\left\langle 1,-2,1\right\rangle$ gives $7\lambda$, so $\lambda=0$.

Result The required plane is therefore simply $$\boxed{x+y+z=2}.$$

Verification Its normal $\left\langle 1,1,1\right\rangle$ has dot product $1-2+1=0$ with $\left\langle 1,-2,1\right\rangle$, and it contains the stated intersection line.