Problem 32 - Thomas Calculus

Question

Find a plane through the points \(P_{1}(1,2,3), P_{2}(3,2,1)\) and perpendicular to the plane \(4 x-y+2 z=7\)

Step-by-Step Solution

Answer

The equation of the plane is \(2x + 12y - 2z = 20\).

Step 1: Find the Normal Vector of the Given Plane

The equation of the given plane is \(4x-y+2z=7\). The normal vector \(\mathbf{n}\) of this plane can be directly read from its coefficients: \(\mathbf{n} = \langle 4, -1, 2 \rangle\).

Step 2: Calculate the Direction Vector of the Line through Two Points

We need to find a direction vector for a line passing through \(P_1(1, 2, 3)\) and \(P_2(3, 2, 1)\). This can be found by subtracting the coordinates of \(P_1\) from \(P_2\): \(\mathbf{d} = (3 - 1, 2 - 2, 1 - 3) = \langle 2, 0, -2 \rangle\).

Step 3: Determine a Vector Perpendicular to the Direction Vector and Normal Vector

To find a vector perpendicular to both the normal vector \(\mathbf{n}\) and the direction vector \(\mathbf{d}\), we compute their cross product: \(\mathbf{n} \times \mathbf{d} = \langle 4, -1, 2 \rangle \times \langle 2, 0, -2 \rangle\). Calculating this gives us: \[\mathbf{n} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & -1 & 2 \ 2 & 0 & -2 \end{vmatrix} = \mathbf{i}(-1 \cdot -2 - 2 \cdot 0) - \mathbf{j}(4 \cdot -2 - 2 \cdot 2) + \mathbf{k}(4 \cdot 0 - -1 \cdot 2)\]Simplifying, we get: \(\langle -2, -12, 2 \rangle \).

Step 4: Find the Equation of the Plane

Once we have a normal vector to the desired plane, which is \(\mathbf{n}_{perp} = \langle -2, -12, 2 \rangle \), and a point on the plane, \(P_1(1, 2, 3)\), we can establish the equation of the plane as follows:\[-2(x - 1) - 12(y - 2) + 2(z - 3) = 0\]Expanding, we get:\[-2x + 2 - 12y + 24 + 2z - 6 = 0\]Simplifying, we have:\[-2x - 12y + 2z + 20 = 0\] or \[2x + 12y - 2z = 20\].

Key Concepts

Normal Vector

Understanding the normal vector is key to mastering planes in three-dimensional geometry. It is a vector that is perpendicular to the plane and plays a crucial role in defining the plane's orientation in space. In any plane equation given in the format of \(ax + by + cz = d\), the coefficients \(a, b,\) and \(c\) give us the components of the normal vector. For example, if you have a plane represented as \(4x - y + 2z = 7\), its normal vector is expressed as \(\langle 4, -1, 2 \rangle\).
Here are a few points to remember about normal vectors:

• They dictate the plane's orientation in a 3D space.
• A normal vector is essential for finding perpendicular planes or lines.
• Always consists of three components corresponding to the \(x, y,\) and \(z\) directions.

Grasping these concepts assures a clearer understanding of geometric relationships in 3D space.

Cross Product

The cross product is a mathematical operation on two vectors in three-dimensional space. It results in another vector that is perpendicular to both input vectors. This property makes it incredibly useful for finding normal vectors to surfaces or lines in geometry.
To compute the cross product of two vectors, such as \(\langle a, b, c \rangle\) and \(\langle d, e, f \rangle\), the formula is:\[\mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a & b & c \ d & e & f \end{vmatrix}\]This involves:

• Taking the determinant of a matrix composed of unit vectors \(\mathbf{i, j, k}\) and the components of each vector.
• Calculating the determinant results in a new vector with components derived through the formula.

For instance, given \(\langle 4, -1, 2 \rangle\) and \(\langle 2, 0, -2 \rangle\),the cross product is \(\langle -2, -12, 2 \rangle\).This vector is perpendicular to both starting vectors.

Plane Equation

Formulating the equation of a plane is accomplished once a normal vector and a point on the plane are known. This makes defining the position and orientation of a plane possible through mathematical equations.
Let's explore the general process to derive a plane's equation. Suppose you have the normal vector \(\mathbf{n} = \langle a, b, c \rangle\) and a point \(P(x_0, y_0, z_0)\) on the plane. The plane's equation is given by:\[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]This relies on:

• Incorporating the point on the plane to fix its position in space.
• Using the normal vector's components to ensure the correct orientation.

Following these steps, let's create a plane through point \(P_1(1, 2, 3)\). If the normal vector \(\langle -2, -12, 2 \rangle\) was given, substitute into the equation:\[-2(x - 1) - 12(y - 2) + 2(z - 3) = 0\]After rearranging, simplify to \[2x + 12y - 2z = 20\]. This simplified form allows easy classification and interpretation of the plane in space.