Problem 5 - Calculus: Early Transcendentals

Question

Find an equation of the plane containing (1,0,0) and the line \(\langle 1,0,2\rangle+t\langle 3,2,1\rangle .\)

Step-by-Step Solution

Answer

The equation of the plane is \(2x - 3y = 2\).

Step 1: Identify Points on the Line

To find points on the plane, we need at least three points. We are given the point \((1,0,0)\). Now let's find two points on the line \(\langle 1,0,2\rangle + t\langle 3,2,1\rangle\).For \(t=0\), the point is \((1,0,2)\) (by substituting \(t=0\) in the line equation).For \(t=1\), the point is \((4,2,3)\) (by substituting \(t=1\) in the line equation).

Step 2: Use the Points to Find Two Vectors in the Plane

The points \((1,0,0)\), \((1,0,2)\), and \((4,2,3)\) lie on the plane.Let's find two vectors from these points:\[ \overrightarrow{v_1} = (1,0,2) - (1,0,0) = (0,0,2) \]\[ \overrightarrow{v_2} = (4,2,3) - (1,0,0) = (3,2,3) \]

Step 3: Find the Normal Vector to the Plane

The normal vector \(\mathbf{n}\) of the plane can be found by taking the cross product of \(\overrightarrow{v_1} = (0,0,2)\) and \(\overrightarrow{v_2} = (3,2,3)\):\[ \mathbf{n} = \overrightarrow{v_1} \times \overrightarrow{v_2} \]\[ = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 0 & 2 \ 3 & 2 & 3 \end{vmatrix} \]\[ = (0(3) - 2(2))\mathbf{i} - (0(3) - 2(3))\mathbf{j} + (0(2) - 0(3))\mathbf{k} \]\[ = -4\mathbf{i} + 6\mathbf{j} + 0\mathbf{k} \]\[ = (-4, 6, 0) \]

Step 4: Form the Plane Equation

With the normal vector \((-4, 6, 0)\) and a point on the plane \((1,0,0)\), the equation of the plane is given by:\[ -4(x-1) + 6(y-0) + 0(z-0) = 0 \]Simplifying, we get:\[ -4x + 4 + 6y = 0 \]\[ -4x + 6y = -4 \]Therefore, the equation of the plane is:\[ 2x - 3y = 2 \]

Key Concepts

Understanding Vectors

Vectors are fundamental tools in mathematics and physics, representing both magnitude and direction. They are essential in describing movements and forces. In simpler terms, you can think of a vector as an arrow pointing from one place to another.

• It has a direction (where the arrow points).
• It has a magnitude (how long the arrow is).

In a three-dimensional space, vectors have three components, denoted as \((x, y, z)\). These components tell us about its movement in the x, y, and z directions. For instance, the vector \((3, 2, 1)\) moves 3 units in the x-direction, 2 in the y-direction, and 1 in the z-direction. Understanding vectors is crucial because they help in calculating distances, directions, and other quantities in various problems like the one we are discussing here, involving the plane equation.

The Cross Product Explained

The cross product is a mathematical operation performed on two vectors. It results in a third vector, which is orthogonal (meaning perpendicular) to the original two. The cross product is primarily used in three-dimensional space.

• It's represented as \(\mathbf{a} \times \mathbf{b}\).
• The resulting vector is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).
• The magnitude of the cross product vector is related to the area of the parallelogram formed by the two vectors.

In our problem, we used the cross product to find the normal vector of the plane. This normal vector helps us determine the plane's orientation in space. By taking the cross product of two vectors in the plane, we find a vector that points directly out of the plane, crucial for establishing the plane equation.

Importance of the Normal Vector

The normal vector is a key concept in understanding the geometry of planes. It's a vector that is perpendicular to the surface of the plane. Once you have the normal vector, you can easily describe the orientation of the plane in three-dimensional space.

• It helps in defining the angle of the plane with respect to the coordinate axes.
• It's used to formulate the plane's equation.

In the exercise, the normal vector was found using the cross product. It turned out to be \((-4, 6, 0)\), which tells us how the plane is tilted in space. Having the normal vector allows us to write down the plane equation, enabling us to understand where the plane lies in relation to other objects in 3D space.

Line in 3D Space

A line in 3D space represents the set of points extending indefinitely in both directions. It can be described using a direction vector and a point the line passes through.

• Usually written in the form \((x_0, y_0, z_0) + t(a, b, c)\).
• Where \(t\) is a parameter that represents how far along the line you are.

In our problem, the line is represented by \<1, 0, 2> + t\<3, 2, 1>\, meaning it passes through the point \((1, 0, 2)\) and has a direction given by the vector \((3, 2, 1)\). By plugging in different values of \(t\), you get different points along the line, which helps in determining the geometry of the plane as well.