Problem 18 - Calculus Early Transcendentals

Question

Find an equation of the following planes. The plane passing through the points \((-1,1,1),(0,0,2),\) and (3,-1,-2)

Step-by-Step Solution

Answer

Question: Find the equation of a plane passing through the points \(A(-1, 1, 1)\), \(B(0, 0, 2)\), and \(C(3, -1, -2)\). Answer: The equation of the plane is \(5x - y + 2z + 4 = 0\).

Step 1: Find two vectors in the plane

Let's denote the given points as \(A(-1, 1, 1)\), \(B(0, 0, 2)\), and \(C(3, -1, -2)\). We can find two vectors in the plane by calculating the difference between each pair of points. Vector \(\vec{AB}\) = \(B - A\) and vector \(\vec{AC}\) = \(C - A\) \(\vec{AB} = (0 - (-1), 0 - 1, 2 - 1) = (1, -1, 1)\) \(\vec{AC} = (3 - (-1), -1 - 1, -2 - 1) = (4, -2, -3)\)

Step 2: Compute the cross product of the two vectors

Now we need to find the normal vector of the plane, \(\vec{n}\), which is the cross product of the two vectors \(\vec{AB}\) and \(\vec{AC}\): $\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 4 & -2 & -3 \\ \end{vmatrix}$ To calculate the cross product, we have: \(\vec{n} = \hat{i}((-1)(-3) - (1)(-2)) - \hat{j}((1)(-3) - (1)(4)) + \hat{k}((1)(-2) - (-1)(4))\) \(\vec{n} = (3 + 2)\hat{i} + (3 - 4)\hat{j} + (-2 + 4)\hat{k} = 5\hat{i} - \hat{j} + 2\hat{k}\) The normal vector of the plane is \(\vec{n} = (5, -1, 2)\).

Step 3: Find the equation of the plane

With a point in the plane, \(A(-1, 1, 1)\), and the normal vector, \(\vec{n} = (5, -1, 2)\), we can find the equation of the plane: \(5(x - (-1)) - 1(y - 1) + 2(z - 1) = 0\) Distribute the coefficients and simplify: \(5x + 5 - y + 1 + 2z - 2 = 0\) The final equation of the plane is: \(5x - y + 2z + 4 = 0\)

Key Concepts

Vectors in 3D

Vectors in 3D are vital for describing and analyzing objects and their movement in three-dimensional space. A vector in 3D is represented as an ordered triplet of numbers: \( x, y, z \). These represent the vector's components along the x, y, and z axes respectively. For example, the vector \( \vec{v} = (1, -1, 1) \) represents a movement of 1 unit in the x-direction, -1 unit in the y-direction, and 1 unit in the z-direction. Such vectors can express both direction and magnitude in 3D space.
When working with vectors, you can perform various operations:

• Addition: Combine the corresponding components to add vectors. For \( \vec{u} = (a, b, c) \) and \( \vec{v} = (d, e, f) \), the resultant vector \( \vec{u} + \vec{v} = (a+d, b+e, c+f) \).
• Subtraction: Similar to addition but you subtract the corresponding components.
• Scalar Multiplication: Multiply each component by a scalar value. For vector \( \vec{v} = (x, y, z) \) and scalar \( k \), \( k\vec{v} = (kx, ky, kz) \).

Understanding these operations is crucial for building more complex constructs like calculating surfaces and phenomena in physics or engineering.

Cross Product

The cross product is a significant operation in vector algebra, especially in 3D space, and is used primarily to find a vector that is perpendicular to two given vectors. This orthogonal vector is often referred to as the normal vector in contexts like finding plane equations.
To find the cross product of two vectors, say \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \), use the following determinant formula:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ \end{vmatrix} \]This calculates each component of the cross product by solving the determinant:

• i-component: \( (a_2 b_3 - a_3 b_2) \).
• j-component: \( -(a_1 b_3 - a_3 b_1) \).
• k-component: \( (a_1 b_2 - a_2 b_1) \).

The resulting vector is perpendicular to both input vectors. This property is especially useful when working with planes, as it helps derive a normal vector providing orientation to the plane.

Normal Vector

A normal vector is a vector that is perpendicular to a surface at a given point. In the context of planes in 3D, the normal vector is crucial for defining the plane's orientation. It tells us the direction perpendicularly away from the surface.
Suppose we have a plane in 3D space, defined by two lines through known vectors. The normal vector to this plane can be found using the cross product of these two vectors. This process involves:

• Identifying vectors: Choose vectors that lie on the plane, often between known points on the plane.
• Calculating their cross product: Use the formula of cross product to find a vector perpendicular to these vectors. This perpendicular vector is your normal vector.

The importance of a normal vector:

• Plane Equation: The normal vector \( \vec{n} = (A, B, C) \) is used in the general plane equation: \( Ax + By + Cz + D = 0 \).
• Uniqueness: It uniquely determines the orientation and slant of the plane. A change in the normal vector alters the entire plane's alignment.

Understanding the role and computation of the normal vector is essential in solving real-world problems in physics, engineering, and computer graphics, such as determining forces, collisions, and visual rendering of objects.