1. **Problem statement:** We have two planes in $\mathbb{R}^3$: $$\Pi_1: x + y + z = 1$$ $$\Pi_2: x - y + z = 2$$ (a) Show their intersection is a line and find its parametric equation. (b) Find the plane(s) containing all points equidistant from $\Pi_1$ and $\Pi_2$. --- 2. **Part (a): Intersection of two planes** - The intersection of two planes is either empty, a line, or a plane. - Since the planes are not parallel (their normal vectors differ), their intersection is a line. 3. **Find the direction vector of the line:** - Normal vector of $\Pi_1$ is $\mathbf{n}_1 = (1,1,1)$. - Normal vector of $\Pi_2$ is $\mathbf{n}_2 = (1,-1,1)$. - The direction vector $\mathbf{d}$ of the line is perpendicular to both normals, so: $$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} = (1\cdot1 - 1\cdot(-1), -(1\cdot1 - 1\cdot1), 1\cdot(-1) - 1\cdot1) = (2, 0, -2)$$ - Simplify direction vector: $$\mathbf{d} = (1, 0, -1)$$ 4. **Find a point on the line:** - Solve the system: $$\begin{cases} x + y + z = 1 \\ x - y + z = 2 \end{cases}$$ - Add equations: $$2x + 2z = 3 \implies x + z = \frac{3}{2}$$ - Subtract second from first: $$2y = -1 \implies y = -\frac{1}{2}$$ - Let parameter $t = z$, then: $$x = \frac{3}{2} - t$$ - So parametric equations: $$x = \frac{3}{2} - t, \quad y = -\frac{1}{2}, \quad z = t$$ 5. **Final parametric form of the line:** $$\mathbf{r}(t) = \left(\frac{3}{2}, -\frac{1}{2}, 0\right) + t( -1, 0, 1)$$ --- 6. **Part (b): Plane(s) equidistant from $\Pi_1$ and $\Pi_2$** - Distance from point $(x,y,z)$ to plane $Ax+By+Cz+D=0$ is: $$d = \frac{|Ax + By + Cz + D|}{\sqrt{A^2 + B^2 + C^2}}$$ - Rewrite planes in standard form: $$\Pi_1: x + y + z - 1 = 0$$ $$\Pi_2: x - y + z - 2 = 0$$ - Norm of normals: $$\|\mathbf{n}_1\| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$$ $$\|\mathbf{n}_2\| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$$ - Points equidistant satisfy: $$\frac{|x + y + z - 1|}{\sqrt{3}} = \frac{|x - y + z - 2|}{\sqrt{3}} \implies |x + y + z - 1| = |x - y + z - 2|$$ - This gives two cases: Case 1: $$x + y + z - 1 = x - y + z - 2 \implies y = \frac{1}{2}$$ Case 2: $$x + y + z - 1 = -(x - y + z - 2) \implies x + y + z - 1 = -x + y - z + 2$$ Simplify: $$x + y + z - 1 = -x + y - z + 2$$ $$x + z + x + z = 2 + 1 - y + y$$ $$2x + 2z = 3$$ $$x + z = \frac{3}{2}$$ - So the two planes containing points equidistant from $\Pi_1$ and $\Pi_2$ are: $$y = \frac{1}{2}$$ $$x + z = \frac{3}{2}$$ --- **Final answers:** (a) The intersection is a line with parametric equations: $$x = \frac{3}{2} - t, \quad y = -\frac{1}{2}, \quad z = t$$ (b) The planes equidistant from $\Pi_1$ and $\Pi_2$ are: $$y = \frac{1}{2}$$ $$x + z = \frac{3}{2}$$