Question

Find the point of intersection of the plane \(3 x-y+4 z=7\) and the line through (5,4,-3) that is perpendicular to this plane.

Short answer

To find the intersection point of the plane and the line, derive the equation of the line that is perpendicular to the plane and passes through the given point, then solve for the line-plane intersection point. Insert the obtained value of \(t\) back into the line equation to obtain the coordinates of the intersection point.

Step-by-step solution

Determine the Equation of the Line

The line's direction is the same as the plane's normal vector, which is given by the coefficients (3, -1, 4). So the directional vector of the line is (3, -1, 4). Given that this line also passes through the point (5, 4, -3), the parametric equation of this line can be written as: \(x = 5 + 3t\), \(y = 4 - t\), \(z = -3 + 4t\)

Find the Intersection Point

Substitute the parametric equation of the line into the equation for the plane to solve for \(t\). From the plane equation, we get \(3x - y+ 4z = 7\). Substituting \(x\), \(y\) and \(z\) we get \(3(5+3t) - (4-t) + 4(-3+4t) = 7\). This single equation can be solved to find the value of \(t\).

Solve for \(t\) and Substitute Back

Solve the equation you obtained in the previous step for \(t\). This will yield the accurate value of \(t\). Substitute this value of \(t\) back into the line equation to determine coordinates of the point of intersection.

Key concepts

Parametric Equations

When studying the intersection of a plane and a line, it's essential to understand parametric equations. These equations express the coordinates of the points on a line as functions of a single parameter, commonly denoted as 't'. In our example, the line's equations were given by:

• \(x = 5 + 3t\)
• \(y = 4 - t\)
• \(z = -3 + 4t\)

Using t allows us to generate any point on the line by merely substituting the value of t into these equations. The power of parametric equations lies in their simplicity and flexibility in representing lines in three dimensions, which can be a complex task when attempting to describe geometric figures in multivariable calculus.

Normal Vector

Understanding the concept of a normal vector is crucial in three-dimensional geometry and calculus. The normal vector to a plane is a vector that is perpendicular to every line lying in the plane itself. For our plane equation \(3x - y + 4z = 7\), the coefficients of x, y, and z (3, -1, 4) form the normal vector of the plane. This vector is vital because it lets us find lines either parallel or perpendicular to the plane, and acts as a directional beacon for calculating intersections. In the given exercise, the normal vector is directly used to define the direction of the line that is to intersect with the plane.

Solving Systems of Equations

The problem of finding the intersection between a plane and a line boils down to solving systems of equations. This process involves finding the values that satisfy all equations simultaneously. In our exercise, we combine the parametric equations of the line and the equation of the plane to form a system. However, because we have one equation and one unknown (the parameter t), it simplifies to a single-variable problem. By substituting the parametric equations into the plane equation, we establish a relationship that only holds true where the line and plane intersect, which leads us to the correct value of t and thus the specific point of intersection.

Plane Equation

The foundation of our problem is rooted in the plane equation, which describes a two-dimensional surface within three-dimensional space. A plane can be defined using the equation \(Ax + By + Cz + D = 0\), where A, B, and C are the components of the normal vector, and D is the constant that positions the plane relative to the origin. In our case, the equation \(3x - y + 4z = 7\) signifies a plane in space. When we seek the intersection with a line, we essentially look for a point that satisfies this equation while also lying on the line defined by the parametric equations.

Geometry in Calculus

The intersection of a plane and a line is a fundamental application of geometry in calculus. Problems like this integrate concepts from both algebra and calculus, allowing us to visualize and solve geometric issues in a calculus framework. It highlights how different elements, such as vectors, equations, and parameters, work together to depict and solve real-world problems involving shapes and movements in space. Understanding this integration is critical for advancing in calculus, as it lays the groundwork for more complex topics such as multivariable functions, vector fields, and differential geometry.