Let's solve each part step by step: --- ### **8.1: Parametric equation of the line through** \( P = (1, 0, -1) \) **parallel to** \( \vec{n} = \langle 2, 1, -1 \rangle \) **Step 1: General Form** The parametric equation of a line passing through point \( P_0(x_0, y_0, z_0) \) and parallel to vector \( \vec{d} = \langle a, b, c \rangle \) is: \[ \begin{align*} x &= x_0 + at \\ y &= y_0 + bt \\ z &= z_0 + ct \end{align*} \] **Step 2: Substitute the values** \[ \begin{align*} x &= 1 + 2t \\ y &= 0 + 1t = t \\ z &= -1 - t \end{align*} \] **Final Answer (8.1):** \[ \boxed{ \begin{align*} x &= 1 + 2t \\ y &= t \\ z &= -1 - t \end{align*} } \] where \( t \in \mathbb{R} \). --- ### **8.2: Line through points \( A = (1, 2, -1) \) and \( B = (-1, 2, -1) \)** **Step 1: Direction vector** \[ \vec{AB} = B - A = (-1 - 1, 2 - 2, -1 - (-1)) = (-2, 0, 0) \] **Step 2: Parametric Form** Using point \( A \) and direction vector \( \vec{d} = (-2, 0, 0) \): \[ \begin{align*} x &= 1 - 2t \\ y &= 2 \\ z &= -1 \end{align*} \] **Final Answer (8.2):** \[ \boxed{ \begin{align*} x &= 1 - 2t \\ y &= 2 \\ z &= -1 \end{align*} } \] where \( t \in \mathbb{R} \). --- ### **8.3: Intersection of Planes** Planes: 1. \( -5x + y - 2z = 3 \) 2. \( 2x - 3y + 5z = -7 \) **Step 1: Direction vector of intersection** The direction vector is the cross product of normals: - Normal to plane 1: \( \vec{n_1} = (-5, 1, -2) \) - Normal to plane 2: \( \vec{n_2} = (2, -3, 5) \) \[ \vec{d} = \vec{n_1} \times \vec{n_2} \] Calculate: \[ \begin{align*} \vec{d}_x &= (1)(5) - (-2)(-3) = 5 - 6 = -1 \\ \vec{d}_y &= -[(-5)(5) - (-2)(2)] = -[-25 + 4] = -[-21] = 21 \\ \vec{d}_z &= (-5)(-3) - (1)(2) = 15 - 2 = 13 \\ \end{align*} \] So \( \vec{d} = \langle -1, 21, 13 \rangle \). **Step 2: Find a point on the line** Let \( z = 0 \): - Plane 1: \( -5x + y = 3 \) - Plane 2: \( 2x - 3y = -7 \) Solve these simultaneously: Multiply plane 1 by 3: - \( -15x + 3y = 9 \) Add to plane 2: - \( 2x - 3y = -7 \) - \( -15x + 3y = 9 \) Add both: \[ (2x - 3y) + (-15x + 3y) = -7 + 9 \implies -13x = 2 \implies x = -\frac{2}{13} \] Substitute \( x \) into plane 1: \[ -5x + y = 3 \implies -5(-\frac{2}{13}) + y = 3 \implies \frac{10}{13} + y = 3 \implies y = 3 - \frac{10}{13} = \frac{29}{13} \] So, when \( z = 0 \), \( x = -\frac{2}{13} \), \( y = \frac{29}{13} \). So point \( P = \left(-\frac{2}{13}, \frac{29}{13}, 0\right) \). **Step 3: Parametric Equations** \[ \begin{align*} x &= -\frac{2}{13} - t \\ y &= \frac{29}{13} + 21t \\ z &= 0 + 13t \end{align*} \] **Final Answer (8.3):** \[ \boxed{ \begin{align*} x &= -\frac{2}{13} - t \\ y &= \frac{29}{13} + 21t \\ z &= 13t \end{align*} } \] where \( t \in \mathbb{R} \). --- **Let me know if you need any further explanation or details!**