q711.5 ### Finding an Equation of a PlaneTo determine the equation of a plane, we need a point through which the plane passes and a normal vector perpendicular to the plane. Here, we are given:1. **Point**: (2, 5, -6)2. **Normal Vector (n)**: \( \mathbf{n} = \mathbf{j} \)The normal vector \( \mathbf{j} \) corresponds to the unit vector in the y-direction, often written as (0, 1, 0) in coordinate form.#### Step-by-Step Solution:1. **Identifying the Normal Vector**: The normal vector \( \mathbf{n} \) points in the y-direction: \( (0, 1, 0) \).2. **Using the General Plane Equation**: The general form of the equation of a plane with a normal vector \((A, B, C)\) passing through a point \((x_0, y_0, z_0)\) is: \[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \] For our problem: \[ \mathbf{n} = (0, 1, 0) \quad \text{and} \quad \text{Point} = (2, 5, -6) \] Substituting into the equation: \[ 0(x - 2) + 1(y - 5) + 0(z + 6) = 0 \]3. **Simplification**: Simplifying the above equation, we get: \[ y - 5 = 0 \]4. **Final Equation**: Solving for \( y \), we get: \[ y = 5 \]Therefore, the equation of the plane that passes through the point (2, 5, -6) and is perpendicular to the vector \( \mathbf{j} \) is:\[y = 5\]This is a horizontal plane perpendicular to the y-axis, passing through all points where \( y = 5 \).

Name: q711.5 ### Finding an Equation of a PlaneTo determine the equation of
Uploaded: 2024-03-20T06:38:44.000Z
Channel: Bartleby
Description: q711.5 ### Finding an Equation of a PlaneTo determine the equation of a plane, we need a point through which the plane passes and a normal vector perpendicular to the plane. Here, we are given:1. **Point**: (2, 5, -6)2. **Normal Vector (n)**: \( \mat