Let's solve the problem step by step: ### Given Planes: \[ \pi_1: 2x - y + z - 1 = 0 \] \[ \pi_2: x + y + z - 6 = 0 \] We are to **describe how these planes intersect** and give the answer in **parametric equations form**. --- ## Step 1: Find the Direction Vector of the Line of Intersection The line of intersection of two planes is parallel to the cross product of their normal vectors. - The normal vector of \(\pi_1\), \( \vec{n}_1 = \langle 2, -1, 1 \rangle \) - The normal vector of \(\pi_2\), \( \vec{n}_2 = \langle 1, 1, 1 \rangle \) Find the cross product: \[ \vec{d} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \] \[ = \mathbf{i}((-1)\cdot1 - 1\cdot1) - \mathbf{j}(2\cdot1 - 1\cdot1) + \mathbf{k}(2\cdot1 - (-1)\cdot1) \] \[ = \mathbf{i}(-1 - 1) - \mathbf{j}(2 - 1) + \mathbf{k}(2 + 1) \] \[ = \mathbf{i}(-2) - \mathbf{j}(1) + \mathbf{k}(3) \] \[ = \langle -2, -1, 3 \rangle \] So, the **direction vector** of the line of intersection is \( \langle -2, -1, 3 \rangle \). --- ## Step 2: Find a Point on the Line To get a particular point, solve the system of equations: \[ \begin{cases} 2x - y + z = 1 \\ x + y + z = 6 \end{cases} \] Let’s choose \( z = t \) (a parameter). Substitute \( z = t \): From the second equation: \[ x + y + t = 6 \implies y = 6 - x - t \] Substitute \( y \) and \( z \) into the first equation: \[ 2x - (6 - x - t) + t = 1 \\ 2x - 6 + x + t + t = 1 \\ 2x + x + 2t - 6 = 1 \\ 3x + 2t = 7 \\ 3x = 7 - 2t \\ x = \frac{7 - 2t}{3} \] Then, \[ y = 6 - x - t = 6 - \frac{7 - 2t}{3} - t = \frac{18 - (7 - 2t) - 3t}{3} = \frac{18 - 7 + 2t - 3t}{3} = \frac{11 - t}{3} \] --- ## Step 3: Write the Parametric Equations Let \( t \) be the parameter: \[ x = \frac{7 - 2t}{3} \] \[ y = \frac{11 - t}{3} \] \[ z = t \] Or, in vector form: \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{7}{3} \\ \frac{11}{3} \\ 0 \end{pmatrix} + t \begin{pmatrix} -\frac{2}{3} \\ -\frac{1}{3} \\ 1 \end{pmatrix} \] Or, using the direction vector found earlier (\(\langle -2, -1, 3 \rangle\)), you can pick any specific point (for example, let \( t = 0 \)): When \( t = 0 \): \[ x = \frac{7}{3},\quad y = \frac{11}{3},\quad z = 0 \] So, \[ \vec{r} = \begin{pmatrix} \frac{7}{3} \\ \frac{11}{3} \\ 0 \end{pmatrix} + s \begin{pmatrix} -2 \\ -1 \\ 3 \end{pmatrix} \] where \( s \) is a parameter. --- ### **Final Answer (Parametric Equations Form):** \[ \boxed{ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{7}{3} \\ \frac{11}{3} \\ 0 \end{pmatrix} + s \begin{pmatrix} -2 \\ -1 \\ 3 \end{pmatrix} } \] where \( s \in \mathbb{R} \). **Explanation:** - The planes intersect in a line. - The direction of the line is \(\langle -2, -1, 3 \rangle\). - A point on the line is \(\left(\frac{7}{3}, \frac{11}{3}, 0\right)\). - The parametric equations above describe all the points of intersection.