1. **Problem statement:** We have a line \(\mathcal{L}\) (from part 2, assumed parametric form \(x=1+t, y=2-2t, z=3t\)) and a plane \(\mathcal{P}\) given by the equation \(x + 3y - z = -4\). (a) Show that \(\mathcal{L}\) intersects \(\mathcal{P}\) at a unique point and find its coordinates. (b) Find the line \(\mathcal{M}\) lying in \(\mathcal{P}\), perpendicular to \(\mathcal{L}\), passing through the intersection point. --- 2. **Recall formulas and rules:** - To find intersection, substitute parametric equations of \(\mathcal{L}\) into the plane equation. - A line \(\mathcal{M}\) perpendicular to \(\mathcal{L}\) means its direction vector is orthogonal to \(\mathcal{L}\)'s direction vector. - \(\mathcal{M}\) lies in \(\mathcal{P}\), so its direction vector is perpendicular to the plane's normal vector. --- 3. **Part (a) Intersection:** - Parametric equations of \(\mathcal{L}\): \[ x = 1 + t, \quad y = 2 - 2t, \quad z = 3t \] - Substitute into plane equation: \[ (1 + t) + 3(2 - 2t) - 3t = -4 \] - Simplify: \[ 1 + t + 6 - 6t - 3t = -4 \] \[ 7 - 8t = -4 \] - Solve for \(t\): \[ -8t = -11 \implies t = \frac{11}{8} \] - Find coordinates: \[ x = 1 + \frac{11}{8} = \frac{19}{8}, \quad y = 2 - 2 \times \frac{11}{8} = 2 - \frac{22}{8} = \frac{16}{8} - \frac{22}{8} = -\frac{6}{8} = -\frac{3}{4}, \quad z = 3 \times \frac{11}{8} = \frac{33}{8} \] - So the intersection point is \(\left(\frac{19}{8}, -\frac{3}{4}, \frac{33}{8}\right)\). --- 4. **Part (b) Line \(\mathcal{M}\):** - Direction vector of \(\mathcal{L}\) is \(\vec{d}_L = (1, -2, 3)\). - Plane normal vector is \(\vec{n} = (1, 3, -1)\). - \(\mathcal{M}\) lies in \(\mathcal{P}\), so its direction vector \(\vec{d}_M\) is perpendicular to \(\vec{n}\): \[ \vec{d}_M \cdot \vec{n} = 0 \] - \(\mathcal{M}\) is perpendicular to \(\mathcal{L}\), so: \[ \vec{d}_M \cdot \vec{d}_L = 0 \] - Find \(\vec{d}_M = (a,b,c)\) satisfying: \[ a + 3b - c = 0 \] \[ a - 2b + 3c = 0 \] - Solve system: From first: \(c = a + 3b\) Substitute into second: \[ a - 2b + 3(a + 3b) = 0 \implies a - 2b + 3a + 9b = 0 \implies 4a + 7b = 0 \] - Express \(a\) in terms of \(b\): \[ a = -\frac{7}{4}b \] - Then \(c = a + 3b = -\frac{7}{4}b + 3b = -\frac{7}{4}b + \frac{12}{4}b = \frac{5}{4}b\) - Choose \(b=4\) for simplicity: \[ a = -7, \quad b=4, \quad c=5 \] - Direction vector \(\vec{d}_M = (-7, 4, 5)\). - Parametric equations of \(\mathcal{M}\) passing through intersection point \(P_0 = \left(\frac{19}{8}, -\frac{3}{4}, \frac{33}{8}\right)\): \[ x = \frac{19}{8} - 7t, \quad y = -\frac{3}{4} + 4t, \quad z = \frac{33}{8} + 5t \] --- **Final answers:** - (a) Intersection point: \(\left(\frac{19}{8}, -\frac{3}{4}, \frac{33}{8}\right)\) - (b) Line \(\mathcal{M}\) parametric form: \[ x = \frac{19}{8} - 7t, \quad y = -\frac{3}{4} + 4t, \quad z = \frac{33}{8} + 5t \]