Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(11-14\), find the equation of the plane through the given points. \((1,1,2),(0,0,1)\), and \((-2,-3,0)\)

Step-by-Step Solution

Answer

The equation of the plane is \(-2x + y + z = 1\).

Step 1: Understanding the problem

We are asked to find the equation of a plane that passes through three points: \((1,1,2), (0,0,1)\), and \((-2,-3,0)\). The equation of a plane can be expressed in the form \(ax + by + cz = d\), where \(a, b, c\) are the components of the normal vector to the plane and \(d\) is the scalar constant.

Step 2: Find two direction vectors on the plane

Using the given points, find two vectors that lie on the plane. Consider vectors \( \mathbf{v_1} \) from \((1,1,2)\) to \((0,0,1)\) and \( \mathbf{v_2} \) from \((1,1,2)\) to \((-2,-3,0)\). Calculate \( \mathbf{v_1} = (0-1, 0-1, 1-2) = (-1, -1, -1)\) and \( \mathbf{v_2} = (-2-1, -3-1, 0-2) = (-3, -4, -2)\).

Step 3: Compute the normal vector to the plane

The normal vector \( \mathbf{n} \) to the plane can be found using the cross product of \( \mathbf{v_1} \) and \( \mathbf{v_2} \). Compute the cross product:\(\mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & -1 & -1 \ -3 & -4 & -2 \end{vmatrix}\)\(= \mathbf{i}((-1)(-2) - (-1)(-4)) - \mathbf{j}((-1)(-2) - (-1)(-3)) + \mathbf{k}((-1)(-4) - (-1)(-3))\)\(= \mathbf{i}(2 - 4) - \mathbf{j}(2 - 3) + \mathbf{k}(4 - 3)\)\(= -2\mathbf{i} + 1\mathbf{j} + 1\mathbf{k}\)Thus, \( \mathbf{n} = (-2, 1, 1)\).

Step 4: Write the equation of the plane

Use the point-normal form equation of the plane with point \((1,1,2)\) and normal vector \((-2, 1, 1)\): \(-2(x - 1) + 1(y - 1) + 1(z - 2) = 0\)Simplify to find the plane equation:\(-2x + 2 + y - 1 + z - 2 = 0\)\(-2x + y + z - 1 = 0\)Therefore, the equation of the plane is \(-2x + y + z = 1\).

Step 5: Verify the solution

Check whether all given points satisfy the equation \(-2x + y + z = 1\).1. For \((1,1,2)\): \(-2(1) + 1 + 2 = 1 \) which is true.2. For \((0,0,1)\): \(-2(0) + 0 + 1 = 1 \) which is true.3. For \((-2,-3,0)\): \(-2(-2) + (-3) + 0 = 1 \) which is true.All points satisfy the plane equation, so the solution is verified.

Key Concepts

Normal Vector

A plane in three-dimensional space is uniquely defined by its normal vector, which is a vector that is perpendicular to every line in the plane. Imagine it as a stick poking straight up from a flat surface. The direction and length of this stick determine the plane's orientation. When working with a plane equation, the components of the normal vector are incorporated as coefficients in the equation. Specifically, if the normal vector is \(\mathbf{n} = (a, b, c)\), then the plane equation can be represented in the form:

• \(ax + by + cz = d\).

In this equation, the components \(a\), \(b\), and \(c\) represent the orientation of the plane. These components are essential in defining how the plane is tilted or rotated in space.The normal vector is crucial because it determines whether two planes are parallel, if they share the same normal vector, or if they might intersect.

Cross Product

To find the normal vector to a plane using two direction vectors in the plane, we rely on the cross product. The cross product of two vectors results in a third vector that is perpendicular to both. This property makes it perfect for finding the normal vector of a plane.For example, let's consider the vectors \(\mathbf{v_1} = (-1, -1, -1)\) and \(\mathbf{v_2} = (-3, -4, -2)\), which lie on a plane. To find the normal vector \(\mathbf{n}\), we calculate \(\mathbf{v_1} \times \mathbf{v_2}\) using the determinant method:

• The formula for the cross product can be written using the determinant of a matrix:\[\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-1 & -1 & -1 \-3 & -4 & -2 \\end{vmatrix}\].
• Expanding the determinant gives us: \[ \mathbf{n} = -2\mathbf{i} + 1\mathbf{j} + 1\mathbf{k} \].
• Thus, the normal vector is \(\mathbf{n} = (-2, 1, 1)\).

In essence, the cross product allows a straightforward computation of a normal vector, which is vital for establishing a plane's equation.

Direction Vectors

Direction vectors provide pivot points for determining the geometry of the plane. A direction vector is simply a vector that shows a direction along the plane. If you have two distinct direction vectors in the plane, you can better grasp the plane's orientation and area.Using two points on the plane, you can derive a direction vector. For example, given points \((1,1,2)\) and \((0,0,1)\), you can form a vector \(\mathbf{v_1} = (-1,-1,-1)\). Subtract the coordinates:

• \((0 - 1, 0 - 1, 1 - 2) = (-1, -1, -1)\).

Similarly, another vector \(\mathbf{v_2}\) can be formed from points \((1,1,2)\) and \((-2,-3,0)\) by calculating:

• \((-2-1, -3-1, 0-2) = (-3, -4, -2)\).

These vectors represent lines on the plane and are used to compute the normal vector through their cross product.

Point-Normal Form

The point-normal form is a remarkably useful way to express the equation of a plane. This format directly incorporates a point on the plane and its normal vector. For the plane passing through a point \((x_0, y_0, z_0)\) with normal vector \(\mathbf{n} = (a, b, c)\), the point-normal form of the equation is:

• \[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\].

This equation highlights the role of the normal vector and a specific point on the plane in crafting the complete equation.By substituting the normal vector \((-2, 1, 1)\) and point \((1,1,2)\), the equation derived becomes:

• \[-2(x - 1) + 1(y - 1) + 1(z - 2) = 0\].
• After simplifying, it results in \(-2x + y + z = 1\).

This representation is not only neat, but it allows easy verification with any additional points. If a point satisfies this equation, it indeed lies on the plane.