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 find the equation of the plane that is perpendicular to the given line, we first extract the direction vector of the line from its parametric equations. The line can be expressed as \( (x,y,z) = (-4, 7, 0) + t(-1, -3, 5) \). This indicates that the direction vector of the line is \( \mathbf{d} = (-1, -3, 5) \). Since the plane is perpendicular to this line, the normal vector of the plane \( \mathbf{n} = (-1, -3, 5) \) can be used to construct the plane's equation. The general form of the plane's equation is given by: \[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \] Substituting \( (x_0, y_0, z_0) = (-4, 7, 2) \) and \( (n_1, n_2, n_3) = (-1, -3, 5) \): \[ -1(x + 4) - 3(y - 7) + 5(z - 2) = 0 \] Expanding this yields the equation of the plane: \[ -x - 4 + 3y - 21 + 5z - 10 = 0 \] Thus, simplifying gives: \[ -x + 3y + 5z - 35 = 0. \] Or, rearranging into the desired format with a coefficient of -1 for \( x \): \[ x - 3y - 5z + 35 = 0. \] This is the equation of the plane through point \( P_0(-4, 7, 2) \) and perpendicular to the specified line.