Problem 24 - Essential Calculus Early Transcendentals

Question

\(21-32=\) Find an equation of the plane. The plane through the point \((2,0,1)\) and perpendicular to the line \(x=3 t, y=2-t, z=3+4 t\)

Step-by-Step Solution

Answer

The equation of the plane is \(3x - y + 4z = 10\).

Step 1: Determine the Normal Vector of the Plane

The equation of a plane can be represented as \(ax + by + cz = d\), where \((a, b, c)\) is the normal vector to the plane. Since the plane is perpendicular to the line \(x = 3t, y = 2 - t, z = 3 + 4t\), the direction vector of the line, \((3, -1, 4)\), will be the normal vector of the plane.

Step 2: Use the Point-Plane Form Equation

We have a point on the plane \((2, 0, 1)\), and the normal vector \((3, -1, 4)\) from the previous step. We can plug these values into the point-plane equation form: \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\). Substituting the normal vector and the point gives the equation: \(3(x - 2) - 1(y - 0) + 4(z - 1) = 0\).

Step 3: Simplify the Equation

Expand the equation from the previous step: \(3(x - 2) - 1(y - 0) + 4(z - 1) = 0\) becomes \(3x - 6 - y + 4z - 4 = 0\). Combine like terms to get the final equation of the plane: \(3x - y + 4z = 10\).

Key Concepts

Normal Vector

When dealing with planes, a normal vector plays a crucial role. You can think of this vector as an arrow pointing straight out of the surface of the plane. It's kind of like the plane's unique fingerprint in 3D space.
A normal vector is denoted by \((a, b, c)\), which directly relates to the plane equation \(ax + by + cz = d\).
For a given line equation of the form \(x = at, y = bt, z = ct\), the direction vector \((a, b, c)\) is used as the normal vector if the plane is perpendicular to this line. This is because the normal vector is, in essence, perpendicular to every line that lies within the plane.
In our exercise, the plane is perpendicular to a line with the direction vector \((3, -1, 4)\). Thus, this direction vector becomes the normal vector for the plane.

Point-Plane Form

The point-plane form of a plane is a powerful tool in plane geometry. It's an equation that helps to find the plane using a known point on the plane and a normal vector.
This form can be written as: \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\).
Here's a simple breakdown:

• \((x_0, y_0, z_0)\) is a point on the plane.
• \((a, b, c)\) is the normal vector to the plane.
• The equation describes a plane by using this geometry.

In the exercise, the plane passes through the point \((2, 0, 1)\), and the normal vector \((3, -1, 4)\) is already known.
By plugging these into the point-plane form equation, you can build the equation of that particular plane in a straightforward manner.

Direction Vector

Direction vectors provide guidance on how a line extends in space. They're powered by three components \((a, b, c)\), which show movement along the x, y, and z axes.
In a line equation like \(x = 3t, y = 2 - t, z = 3 + 4t\), the direction vector \((3, -1, 4)\) symbolizes its path in a three-dimensional world.
But what does the direction vector have to do with planes? When you're given that a plane is perpendicular to such a line, the direction vector of the line can actually serve as the normal vector for the plane. This is because of their perpendicular nature to each other.
So, without much fuss, identifying how the line moves (via its direction vector) directly assists in laying out the essentials of the plane that cuts across perpendicularly.