Problem 26 - Essential Calculus Early Transcendentals

Question

\(21-32=\) Find an equation of the plane. The plane through the origin and the points \((2,-4,6)\) and \((5,1,3)\)

Step-by-Step Solution

Answer

The plane's equation is \\(6x - 18y + 22z = 0\\).

Step 1: Identify the Given Points

We are given three points through which the plane passes: the origin \(0,0,0\), \(2, -4, 6\), and \(5, 1, 3\). From these, we will determine two direction vectors on the plane.

Step 2: Find Direction Vectors

Find two vectors in the plane formed by these points. Vector \mathbf{v_1}\ can be from \(0,0,0\) to \(2,-4,6\): \mathbf{v_1} = (2 - 0, -4 - 0, 6 - 0) = (2, -4, 6).\ Similarly, vector \mathbf{v_2}\ from \(0,0,0\) to \(5, 1, 3\) is \mathbf{v_2} = (5 - 0, 1 - 0, 3 - 0) = (5, 1, 3).\

Step 3: Compute the Cross Product

The normal vector to the plane, \mathbf{n}\, can be found using the cross product of \mathbf{v_1}\ and \mathbf{v_2}.\ Calculate \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{bmatrix} i & j & k \ \ 2 & -4 & 6 \ \ 5 & 1 & 3 \ \end{bmatrix}\.\ This evaluates to \(6i - 18j + 22k \), so \mathbf{n} = (6, -18, 22).\

Step 4: Write the Equation of the Plane

Having the normal vector \mathbf{n} = (6, -18, 22)\, the equation of the plane can be obtained by using the formula \(6(x-0) - 18(y-0) + 22(z-0) = 0\). Simplify to \(6x - 18y + 22z = 0\). This is the equation of the plane in standard form.

Key Concepts

Direction Vectors

Direction vectors are essential in defining the orientation of a plane in three-dimensional space. When given three points, such as the origin \((0, 0, 0)\), \((2, -4, 6)\),\ and \((5, 1, 3)\), you can construct direction vectors that lie within the plane.
Direction vectors, denoted as \(\mathbf{v_1}\) and \(\mathbf{v_2}\), are determined by calculating the difference between coordinates of these points.

• For \(\mathbf{v_1}\), calculate from the origin to \((2, -4, 6)\) resulting in \(\mathbf{v_1} = (2, -4, 6)\).
• For \(\mathbf{v_2}\), calculate from the origin to \((5, 1, 3)\) resulting in \(\mathbf{v_2} = (5, 1, 3)\).

Understanding direction vectors assists us in finding characteristics like the normal vector that further defines the plane's position.

Cross Product

The cross product is a mathematical operation that allows us to find a vector that is perpendicular to two given vectors. This is particularly useful when trying to find the normal vector to a plane.
For vectors \(\mathbf{v_1}\) = (2, -4, 6) and \(\mathbf{v_2}\) = (5, 1, 3), the cross product \(\mathbf{v_1} \times \mathbf{v_2}\) is calculated using the formula:\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -4 & 6 \ 5 & 1 & 3 \end{vmatrix} \]This expands to:

• \(\mathbf{i}[(-4)(3) - (6)(1)]\)
• \(-\mathbf{j}[(2)(3) - (6)(5)]\)
• \(+ \mathbf{k}[(2)(1) - (-4)(5)]\)

After calculating these, you'll find the normal vector \(\mathbf{n} = (6, -18, 22)\). The result is the vector orthogonal to both \(\mathbf{v_1}\) and \(\mathbf{v_2}\), crucial for defining our plane.

Normal Vector

A normal vector to a plane is one that is perpendicular to every direction vector within the plane. Once found, it plays a key role in determining the equation of the plane.
In our case, after performing the cross product of direction vectors \(\mathbf{v_1}\) and \(\mathbf{v_2}\), we obtained the normal vector \(\mathbf{n} = (6, -18, 22)\).
The components of this normal vector \((a, b, c) = (6, -18, 22)\) relate directly to the coefficients in the standard form equation of the plane.

• The normal vector can be considered as the vector \((a,b,c)\) in the equation of the plane \(ax + by + cz = d\).
• This means our normal vector's role in the equation is foundational, defining its orientation in 3D space.

Recognizing and utilizing the normal vector aids in converting vector information into the plane's standard equation.

Standard Form of a Plane Equation

The standard form of a plane equation is given by the formula \( ax + by + cz = d \). This equation succinctly represents the plane in three-dimensional space using a normal vector.
Given the normal vector \( \mathbf{n} = (6, -18, 22) \) and a point on the plane, such as the origin \((0, 0, 0)\), the plane equation becomes:\[ 6(x-0) - 18(y-0) + 22(z-0) = 0 \]Simplifying this, we get:
\(6x - 18y + 22z = 0\)

• The coefficients \(6, -18,\ and \22\) directly correlate with the normal vector components.
• This format allows us to see how any point \((x, y, z)\) satisfies the plane’s condition.

Converting vector information into standard form lets us apply the equation for analytical geometry and further calculations.