Find the equation for the plane through \( P_{0}(-4,7,2) \) perpendicular to the following line. \( x=-4-t, y=7-3 t, z=5 t,-\infty

Ask by Turnbull Jimenez. in the United States

Feb 03,2025

To determine the equation of the plane that passes through the point \( P_{0}(-4, 7, 2) \) and is perpendicular to the given line, follow these steps: 1. **Identify the Direction Vector of the Line:** The parametric equations of the line are: \[ \begin{cases} x = -4 - t \\ y = 7 - 3t \\ z = 5t \end{cases} \] From these, the direction vector \( \mathbf{d} \) of the line is: \[ \mathbf{d} = \langle -1, -3, 5 \rangle \] 2. **Determine the Normal Vector of the Plane:** Since the plane is perpendicular to the line, the normal vector \( \mathbf{n} \) of the plane is parallel to the direction vector of the line: \[ \mathbf{n} = \langle -1, -3, 5 \rangle \] 3. **Use the Point-Normal Form of the Plane Equation:** The point-normal form of the equation of a plane is: \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \] Substituting \( \mathbf{n} = \langle -1, -3, 5 \rangle \) and \( P_0(-4, 7, 2) \): \[ -1(x + 4) - 3(y - 7) + 5(z - 2) = 0 \] 4. **Simplify the Equation:** Expanding and simplifying the equation: \[ -x - 4 - 3y + 21 + 5z - 10 = 0 \] \[ -x - 3y + 5z + 7 = 0 \] To match the requirement of having a coefficient of -1 for \( x \), you can rearrange it as: \[ -x - 3y + 5z + 15 = 0 \] **Final Equation of the Plane:** \[ -x - 3y + 5z + 15 = 0 \] Answer: After simplifying, an equation for the plane is – x – 3 y + 5 z + 15 = 0