Problem 13 - APEX Calculus

Question

Give the equation of the described plane in standard and general forms. Contains the parallel lines \(\vec{\ell}_{1}(t)=\langle 1,1,1\rangle+t\langle 1,2,3\rangle\) and $$ \vec{\ell}_{2}(t)=\langle 1,1,2\rangle+t\langle 1,2,3\rangle $$

Step-by-Step Solution

Answer

The plane's equations are in standard form: \(2x - y = 1\) and in general form: \(2x - y - 1 = 0\).

Step 1: Identify Direction Vector and Point on the Plane

Notice that both lines \(\vec{\ell}_{1}(t)\) and \(\vec{\ell}_{2}(t)\) have the same direction vector \(\langle 1, 2, 3 \rangle\). Hence, they are parallel. Take a point on each line, such as \(\langle 1, 1, 1 \rangle\) from \(\vec{\ell}_{1}(t)\) and \(\langle 1, 1, 2 \rangle\) from \(\vec{\ell}_{2}(t)\) to find a vector between them.

Step 2: Find the Vector Between the Two Points

The vector between the points \(\langle 1, 1, 1 \rangle\) (from \(\vec{\ell}_{1}\)) and \(\langle 1, 1, 2 \rangle\) (from \(\vec{\ell}_{2}\)) is found by subtracting the first point from the second: \(\langle 1, 1, 2 \rangle - \langle 1, 1, 1 \rangle = \langle 0, 0, 1 \rangle\).

Step 3: Determine the Normal Vector to the Plane

The normal vector \(\mathbf{n}\) to the plane can be found using the cross product of the direction vector \(\langle 1, 2, 3 \rangle\) and the vector \(\langle 0, 0, 1 \rangle\) between the points. Calculate the cross product: \(\langle 1, 2, 3 \rangle \times \langle 0, 0, 1 \rangle = \langle 2, -1, 0 \rangle\).

Step 4: Use a Point and Normal Vector for the Plane Equation

Using the point \(\langle 1, 1, 1 \rangle\) and the normal vector \(\langle 2, -1, 0 \rangle\), the equation for the plane is given by the formula \(2(x - 1) - 1(y - 1) + 0(z - 1) = 0\). Simplify to get the standard form: \(2x - y = 1\).

Step 5: Convert to General Form

The general form of the plane is simply the simplified standard form as \(2x - y - 1 = 0\).

Key Concepts

Direction Vectors

In the realm of geometry, direction vectors play a crucial role when dealing with lines and planes. A direction vector provides the direction in which a line extends. For the given lines \[ \vec{\ell}_{1}(t)=\langle 1,1,1\rangle+t\langle 1,2,3\rangle \] and \[ \vec{\ell}_{2}(t)=\langle 1,1,2\rangle+t\langle 1,2,3\rangle \], the direction vector is \(\langle 1, 2, 3 \rangle\).
Both lines share this direction vector, which means they are parallel and run alongside each other without intersecting.
Recognizing parallelism is crucial when working with planes, as it implies certain geometric relationships, including how vectors can define a plane.

Cross Product

The cross product is an operation that applies to two vectors in three-dimensional space, producing another vector that is orthogonal (or perpendicular) to both original vectors.
In this exercise, you take the cross product of the direction vector \(\langle 1, 2, 3 \rangle\) and the vector between the lines \(\langle 0, 0, 1 \rangle\), achieving the result \(\langle 2, -1, 0 \rangle\).
Here's what a cross product accomplishes:

• Helps in finding a normal vector to a plane.
• Provides the magnitude representing the area of the parallelogram formed by the two vectors.

In our scenario, the result \(\langle 2, -1, 0 \rangle\) not only identifies a unique normal vector but also serves as a crucial ingredient in the plane's equation.

Point on the Plane

When defining a plane, often we need a point that lies on it. This is a point through which the plane passes, a fixed element within the infinite expanse of a plane.
In our case, we can select \(\langle 1, 1, 1 \rangle\), which is merely an example point from one of the lines.
The selected point, together with a normal vector, helps in formulating the equation of the plane.
Knowing any additional point would also suffice, but keeping points that are easiest to identify makes calculations straightforward and prevents complexity.

Normal Vector

The normal vector is pivotal when describing a plane. It stands perpendicular to the plane's surface, determining the plane's orientation in space.
In this exercise, the normal vector is derived from the cross product giving \(\langle 2, -1, 0 \rangle\). This vector expresses every direction perpendicular to our plane and is essential when drafting the plane's equation.

• It ensures the plane's representation is uniquely defined.
• Acts as a guide to calculate distances or angles involving the plane.

Using this normal vector in the point-normal form formula \( \mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0 \) allows the creation of the standard equation for the plane. Here, \( \mathbf{r_0} \) is the selected point and \( \mathbf{n} \) is our normal vector, seamlessly guiding us to express the plane in mathematical form.