Problem 82 - Precalculus with Calculus Previews

Question

Find (a) \(\operatorname{proj}_{\mathbf{v}} \mathbf{u},\) and \(\mathbf{( b )}\) \(\operatorname{proj}_{\mathbf{u}} \mathbf{v}\). $$ \mathbf{u}=4 \mathbf{i}+2 \mathbf{j}, \mathbf{v}=-3 \mathbf{i}+\mathbf{j} $$

Step-by-Step Solution

Answer

(a) \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} = 3 \mathbf{i} - \mathbf{j} \), (b) \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} = -2 \mathbf{i} - \mathbf{j} \).

Step 1: Understand the formula for vector projection

The vector projection of one vector \( \mathbf{a} \) onto another vector \( \mathbf{b} \) is given by the formula: \( \operatorname{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b} \). This formula represents how much of \( \mathbf{a} \) is in the direction of \( \mathbf{b} \).

Step 2: Calculate the dot products

Calculate the dot products needed for the projection formulas.1. \( \mathbf{u} \cdot \mathbf{v} = (4)(-3) + (2)(1) = -12 + 2 = -10 \).2. \( \mathbf{v} \cdot \mathbf{v} = (-3)^2 + (1)^2 = 9 + 1 = 10 \).3. \( \mathbf{u} \cdot \mathbf{u} = (4)^2 + (2)^2 = 16 + 4 = 20 \).

Step 3: Find \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} \)

Now use the formula for projection to find \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} \):\[ \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} = \frac{-10}{10} \mathbf{v} = -1(-3 \mathbf{i} + \mathbf{j}) = 3 \mathbf{i} - \mathbf{j} \].

Step 4: Find \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} \)

Similarly, use the formula for projection to find \( \operatorname{proj}_{\mathbf{u}} \mathbf{v} \):\[ \operatorname{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} = \frac{-10}{20} \mathbf{u} = -\frac{1}{2}(4 \mathbf{i} + 2 \mathbf{j}) = -2 \mathbf{i} - \mathbf{j} \].

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector mathematics. It is used to determine the angle or relation between two vectors. Mathematically, the dot product is calculated by multiplying corresponding components of the vectors and then summing these products. If we have two vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} \), the dot product is given by:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

This result is a scalar, meaning it has magnitude but no direction. In physical terms, the dot product measures how much one vector extends or contributes in the direction of another.
Applications of the dot product include finding angles between vectors and projections, which are vital for various engineering and physics calculations.

Vector Operations

Vector operations involve calculations with vectors to reveal various properties and insights about their orientation or magnitude in space. Key operations include addition, subtraction, and multiplication of vectors.

• **Addition:** For vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} \), the sum is \( \mathbf{a} + \mathbf{b} = (a_1 + b_1)\mathbf{i} + (a_2 + b_2)\mathbf{j} \).


• **Subtraction:** The difference is \( \mathbf{a} - \mathbf{b} = (a_1 - b_1)\mathbf{i} + (a_2 - b_2)\mathbf{j} \).


• **Scalar Multiplication:** Multiplying a vector by a scalar changes its magnitude but not its direction. If \( k \) is a scalar, \( k\mathbf{a} = ka_1 \mathbf{i} + ka_2 \mathbf{j} \).

These operations allow you to manipulate vectors to study forces, velocity, and other vector-based phenomena.
Understanding these operations leads to better intuition when dealing with physical systems and helps in deriving complex relationships like the vector projection utilized in the original exercise.

Coordinate Geometry

Coordinate geometry uses algebraic expressions to describe geometric aspects of shapes and the positions of vectors in a plane. In two dimensions, vectors have an \( \, x\) (horizontal) component and a \( \, y\) (vertical) component, typically expressed as \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \).
Working with coordinate geometry allows for:

• Determining the position and orientation of vectors.


• Calculating geometric properties like distances and angles using the Pythagorean theorem and trigonometry.
• Solving vector-based problems in physics and engineering like determining paths and trajectories.

The beauty of coordinate geometry is in translating complex geometric problems into manageable algebraic equations.
In exercise calculations, applying the principles of coordinate geometry helps to break down vectors like \( \mathbf{u} = 4\mathbf{i} + 2\mathbf{j} \) and \( \mathbf{v} = -3\mathbf{i} + \mathbf{j} \).
By representing them in this manner, it becomes straightforward to carry out calculations like finding vector projections or combining vector components.