Problem 40 - University Calculus: Early Transcendentals

Question

If \(\overrightarrow{A B}=-7 \mathbf{i}+3 \mathbf{j}+8 \mathbf{k}\) and \(A\) is the point \((-2,-3,6),\) find \(B .\)

Step-by-Step Solution

Answer

Point B is at coordinates \((-9, 0, 14)\).

Step 1: Understand the Vector Representation

The vector \(\overrightarrow{AB}\) represents the direction and magnitude from point \(A\) to point \(B\). In components, \(\overrightarrow{AB} = (x_B - x_A)\mathbf{i} + (y_B - y_A)\mathbf{j} + (z_B - z_A)\mathbf{k}\).

Step 2: Identify the Known Quantities

We know \(\overrightarrow{AB} = -7\mathbf{i} + 3\mathbf{j} + 8\mathbf{k}\), and point \(A\) is given as \((-2, -3, 6)\). Therefore, \((x_A, y_A, z_A) = (-2, -3, 6)\).

Step 3: Set Up the Equation for Each Component

From the vector equation \(\overrightarrow{AB} = (x_B - x_A)\mathbf{i} + (y_B - y_A)\mathbf{j} + (z_B - z_A)\mathbf{k}\), set up equations:\(x_B - (-2) = -7\),\(y_B - (-3) = 3\),\(z_B - 6 = 8\).

Step 4: Solve for Each Component Value

To find \(x_B\), solve \(x_B + 2 = -7\) which gives \(x_B = -9\).To find \(y_B\), solve \(y_B + 3 = 3\) which gives \(y_B = 0\).To find \(z_B\), solve \(z_B - 6 = 8\) which gives \(z_B = 14\).

Step 5: Compile the Coordinates of Point B

Combine the solutions for each component to determine point \(B\), which has coordinates \((-9, 0, 14)\).

Key Concepts

Vector Components

In the world of vectors, understanding vector components is essential. Vectors are often represented in terms of their components along the axes in a coordinate system. This allows us to describe a vector by its horizontal and vertical displacements.

In the context of the exercise, the vector between points A and B, denoted as \(\overrightarrow{AB}\), is expressed in terms of its components:

• \(-7\mathbf{i} \) – the component along the \(x\)-axis.
• \(3\mathbf{j} \) – the component along the \(y\)-axis.
• \(8\mathbf{k} \) – the component along the \(z\)-axis.

To fully describe a vector, each axis's component tells us how far and in what direction the vector moves. For example, a component of \(-7\mathbf{i} \) indicates a movement of 7 units in the negative \(x\)-direction. Vectors can thus be understood as a sum of these directional components.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to find the position of points using coordinates and algebraic representations. In our exercise, point A is given as \((-2, -3, 6)\). Each of these numbers corresponds to a position along the \(x\), \(y\), and \(z\) axes.

Finding point B involves calculating each coordinate based on a known vector. For instance, moving in the \(x\)-direction by \(-7\) from \(-2\) results in \(x_B = -9\). Similarly, moving \(3\) units upward from \(y_A = -3\) finds us at \(y_B = 0\), and moving \(8\) units forward from \(z_A = 6\) finds us at \(z_B = 14\).

This step-by-step movement along each axis showcases the power of coordinate geometry to visualize and solve problems in three-dimensional space. It enables us to pinpoint exact locations by relying on simple rules of movement along coordinates.

Problem-solving Steps

To solve problems involving vectors and coordinates, it's helpful to follow structured problem-solving steps. This method ensures that each part of the problem is tackled efficiently. Here's how to apply these steps to our problem:

• Identify Known Quantities: Begin by recognizing what information is given. Here, we have the vector \(\overrightarrow{AB}\) and point A.
• Set Up Equations: Use the standard vector component equation: \(\overrightarrow{AB} = (x_B - x_A)\mathbf{i} + (y_B - y_A)\mathbf{j} + (z_B - z_A)\mathbf{k}\) to form equations for each coordinate.
• Solve for Unknowns: Tackle each equation one at a time. Solve for \(x_B\), \(y_B\), and \(z_B\) individually, ensuring each step involves clear calculation.
• Compile the Solution: Combine the results of the individual equations to find the coordinates of point B.

By breaking the problem into these understandable steps, solutions become more approachable. This method can be extended to various problems dealing with vectors and points in space, making learning efficient and effective.