Find an equation of the line where the planes \( Q \) and \( R \) intersect. Q: \( -2 x-3 y+4 z=1 \); \( R: x+y+z=1 \) Choose the correct answer below. A. An equation of the line is given by \( x=-\frac{1}{3}-3 t, y=-2 t, z=\frac{1}{4}+4 t \), where \( -\infty

Ask by Owen Ingram. in the United States

Feb 03,2025

To find the equation of the line where the planes \( Q \) and \( R \) intersect, we'll follow these steps: 1. **Identify the Equations of the Planes:** - Plane \( Q \): \( -2x - 3y + 4z = 1 \) - Plane \( R \): \( x + y + z = 1 \) 2. **Find the Direction Vector of the Line:** The direction vector \( \mathbf{v} \) of the intersection line is perpendicular to the normals of both planes. The normal vectors are: - \( \mathbf{n}_Q = (-2, -3, 4) \) - \( \mathbf{n}_R = (1, 1, 1) \) The direction vector \( \mathbf{v} \) is given by the cross product \( \mathbf{n}_Q \times \mathbf{n}_R \): \[ \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & -3 & 4 \\ 1 & 1 & 1 \\ \end{vmatrix} = (-7, 6, 1) \] 3. **Find a Point on the Line:** Solve the system of equations to find a specific point on both planes. Let's set \( z = 0 \): \[ \begin{cases} -2x - 3y = 1 \\ x + y = 1 \\ \end{cases} \] Solving these equations: - From the second equation: \( y = 1 - x \) - Substitute into the first equation: \[ -2x - 3(1 - x) = 1 \implies -2x - 3 + 3x = 1 \implies x = 4 \] - Then \( y = 1 - 4 = -3 \) So, one point on the line is \( (4, -3, 0) \). 4. **Write the Parametric Equations of the Line:** Using the direction vector and the point: \[ \begin{cases} x = 4 - 7t \\ y = -3 + 6t \\ z = t \\ \end{cases} \quad \text{where} \quad t \in \mathbb{R} \] **Therefore, the correct answer is:** **C. An equation of the line is given by \( x = 4 - 7 t \), \( y = -3 + 6 t \), \( z = t \), where \( -\infty < t < \infty \).**