Problem 39 - University Calculus: Early Transcendentals

Question

If \(\overrightarrow{A B}=\mathbf{i}+4 \mathbf{j}-2 \mathbf{k}\) and \(B\) is the point \((5,1,3),\) find \(A\)

Step-by-Step Solution

Answer

Point A is at \((4, -3, 5)\).

Step 1: Understand the Vector Representation

The vector \(\overrightarrow{A B}\) is given by \(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\). This represents the direction and magnitude from point A to point B.

Step 2: Define Points A and B in coordinates

Point B is given as \((5, 1, 3)\). Let's assume that point A has coordinates \((x, y, z)\). Then the vector \(\overrightarrow{A B}\) from A to B can be represented as \((5-x)\mathbf{i} + (1-y)\mathbf{j} + (3-z)\mathbf{k}\).

Step 3: Set up the Equation for Corresponding Components

For the vector equations to be equivalent, each component of vector \(\overrightarrow{A B}\) must match. Therefore, set the equations: \[ 5-x = 1 \] \[ 1-y = 4 \] \[ 3-z = -2 \]

Step 4: Solve for x, y, and z

Solve each equation to find the coordinates of point A:- From \(5-x = 1\), we get \(x = 4\).- From \(1-y = 4\), we get \(y = -3\).- From \(3-z = -2\), we get \(z = 5\).

Step 5: Write the Coordinates of Point A

The coordinates of point A are found to be \((4, -3, 5)\).

Key Concepts

Vector Representation

In vector mathematics, understanding vector representation is essential. Vectors are used to describe both magnitude and direction, making them a crucial concept in physics and engineering. A vector is expressed in terms of its components along a coordinate axis.
In the problem, vector \(\overrightarrow{AB}\) is given as \(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\). These are the components along the x, y, and z axes, respectively.

• The \(\mathbf{i}\) component (coefficient of \(\mathbf{i}\)) represents the movement in the x-direction.
• The \(\mathbf{j}\) component (coefficient of \(\mathbf{j}\)) shows the movement in the y-direction.
• The \(\mathbf{k}\) component (coefficient of \(\mathbf{k}\)) indicates movement in the z-direction.

Vectors can also be visualized as arrows pointing from the initial point to the terminal point. The length of the arrow shows the vector's magnitude, while the arrow's direction aligns with the vector's direction. Understanding these basics will simplify calculating vectors between points in different contexts.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves studying geometry using a coordinate system. In this context, we describe points in space using coordinates which help in defining vectors and scalar products.
When given a point \(B\) with coordinates \((5, 1, 3)\), it indicates the exact location along the x, y, and z axes. With coordinates, one can easily represent vectors by establishing points in space.
It's crucial to think of vectors as differences between points. For instance, assuming point \(A\) has the unknown coordinates \((x, y, z)\), vector \(\overrightarrow{AB}\) can be derived by calculating the difference between point \(B\) and \(A\). Hence, \((5-x)\mathbf{i} + (1-y)\mathbf{j} + (3-z)\mathbf{k}\) follows because it's this difference that creates the vector \(\overrightarrow{AB}\). Understanding how these coordinates interact provides the foundation for solving geometry-based problems using both algebra and geometry principles.

Solving Equations

Solving equations is a fundamental part of algebra and is widely applied in problems involving vectors. It's about finding the values of unknown variables that satisfy the given mathematical statements.
For vectors, like \(\overrightarrow{AB}\), to be equivalent, their corresponding components must match. In the exercise, equations are set up as:

• \(5-x = 1\)
• \(1-y = 4\)
• \(3-z = -2\)

These component-wise equations allow solving for each coordinate separately, resulting in unique solutions.
The solutions are found by manipulating the equations such that the unknowns, \(x\), \(y\), and \(z\), are isolated. For instance:

• Rearranging \(5-x = 1\) gives \(x = 5-1 = 4\).
• Solving \(1-y = 4\) gives \(y = 1-4 = -3\).
• Handling \(3-z = -2\) reveals \(z = 3+2 = 5\).

This technique ensures each component correlates correctly between the vector representation and the coordinates it connects. With practice, solving such equations becomes intuitive, greatly aiding in understanding vector operations.