Math Problem Statement

In each part, find the vector component of ~v along ~b and the vector component of ~v orthogonal to ~b. Then sketch the vectors ~v, proj~b ~v, and ~v − proj~b ~v. (a) ~v = 2ˆi −ˆj, ~b = 3ˆi +ˆj

Solution

We are tasked with finding two vector components of v\vec{v} with respect to b\vec{b}:

  1. The vector component of v\vec{v} along b\vec{b} (denoted as projbv\text{proj}_{\vec{b}} \vec{v}).
  2. The vector component of v\vec{v} orthogonal to b\vec{b}, which is vprojbv\vec{v} - \text{proj}_{\vec{b}} \vec{v}.

Given:

v=2i^j^=(21),b=3i^+j^=(31)\vec{v} = 2\hat{i} - \hat{j} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad \vec{b} = 3\hat{i} + \hat{j} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}

Step 1: Calculate the projection of v\vec{v} along b\vec{b}

The formula for the projection of a vector v\vec{v} along another vector b\vec{b} is: projbv=vbbbb\text{proj}_{\vec{b}} \vec{v} = \frac{\vec{v} \cdot \vec{b}}{\vec{b} \cdot \vec{b}} \vec{b}

Find vb\vec{v} \cdot \vec{b} (dot product):

vb=(2)(3)+(1)(1)=61=5\vec{v} \cdot \vec{b} = (2)(3) + (-1)(1) = 6 - 1 = 5

Find bb\vec{b} \cdot \vec{b}:

bb=(3)2+(1)2=9+1=10\vec{b} \cdot \vec{b} = (3)^2 + (1)^2 = 9 + 1 = 10

Now, calculate projbv\text{proj}_{\vec{b}} \vec{v}:

projbv=510b=12(31)=(3212)\text{proj}_{\vec{b}} \vec{v} = \frac{5}{10} \vec{b} = \frac{1}{2} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} \\ \frac{1}{2} \end{pmatrix}

Step 2: Calculate the vector component of v\vec{v} orthogonal to b\vec{b}

This component is simply vprojbv\vec{v} - \text{proj}_{\vec{b}} \vec{v}: vprojbv=(21)(3212)=(232112)=(1232)\vec{v} - \text{proj}_{\vec{b}} \vec{v} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} - \begin{pmatrix} \frac{3}{2} \\ \frac{1}{2} \end{pmatrix} = \begin{pmatrix} 2 - \frac{3}{2} \\ -1 - \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ -\frac{3}{2} \end{pmatrix}

Step 3: Sketch the vectors

The vectors to be sketched are:

  • v=(21)\vec{v} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}
  • projbv=(3212)\text{proj}_{\vec{b}} \vec{v} = \begin{pmatrix} \frac{3}{2} \\ \frac{1}{2} \end{pmatrix}
  • vprojbv=(1232)\vec{v} - \text{proj}_{\vec{b}} \vec{v} = \begin{pmatrix} \frac{1}{2} \\ -\frac{3}{2} \end{pmatrix}

I'll prepare the sketch, which shows v\vec{v}, projbv\text{proj}_{\vec{b}} \vec{v}, and the orthogonal component vprojbv\vec{v} - \text{proj}_{\vec{b}} \vec{v}.

Would you like a more detailed explanation of the process?

Meanwhile, here are 5 questions to expand on the topic:

  1. How is the projection of a vector geometrically interpreted?
  2. What is the significance of the dot product in determining the projection?
  3. Can you explain why vprojbv\vec{v} - \text{proj}_{\vec{b}} \vec{v} is orthogonal to b\vec{b}?
  4. How does the projection formula change when vectors are not in standard form?
  5. What are some real-world applications of vector projection?

Tip: The vector projection represents the "shadow" of one vector onto another, giving the component that aligns with the second vector.

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Math Problem Analysis

Mathematical Concepts

Vector Projection
Vector Orthogonality
Dot Product

Formulas

proj_b v = (v • b / b • b) b
v - proj_b v

Theorems

Projection Theorem
Orthogonality of Vector Components

Suitable Grade Level

Grades 10-12