Math Problem Statement

Given the vectors A⃗ = 2.00ˆi + 3.00ˆj and B⃗ = 3.00ˆi − 5.00ˆj,

• Draw the vector sum C⃗ = A⃗ + B⃗ on the x − y plane.

• Draw the vector difference D⃗ = A⃗ + B⃗ on the x − y plane.

• Calculate C⃗ and D⃗ with the components.

• Calculate A⃗ · B⃗ .

Solution

Let's work through each part of the problem step by step.

1. Draw the vector sum C=A+B\vec{C} = \vec{A} + \vec{B} on the x-y plane.

  • The vectors are given as:

    • A=2.00i^+3.00j^\vec{A} = 2.00 \hat{i} + 3.00 \hat{j}
    • B=3.00i^5.00j^\vec{B} = 3.00 \hat{i} - 5.00 \hat{j}
  • The vector sum C=A+B\vec{C} = \vec{A} + \vec{B}: C=(2.00i^+3.00j^)+(3.00i^5.00j^)=(2.00+3.00)i^+(3.005.00)j^\vec{C} = (2.00 \hat{i} + 3.00 \hat{j}) + (3.00 \hat{i} - 5.00 \hat{j}) = (2.00 + 3.00) \hat{i} + (3.00 - 5.00) \hat{j} C=5.00i^2.00j^\vec{C} = 5.00 \hat{i} - 2.00 \hat{j}

    The sum vector C\vec{C} is 5.00i^2.00j^5.00 \hat{i} - 2.00 \hat{j}, which is a vector pointing 5 units along the positive x-axis and 2 units down along the y-axis.

2. Draw the vector difference D=AB\vec{D} = \vec{A} - \vec{B} on the x-y plane.

  • The vector difference D=AB\vec{D} = \vec{A} - \vec{B}: D=(2.00i^+3.00j^)(3.00i^5.00j^)=(2.003.00)i^+(3.00(5.00))j^\vec{D} = (2.00 \hat{i} + 3.00 \hat{j}) - (3.00 \hat{i} - 5.00 \hat{j}) = (2.00 - 3.00) \hat{i} + (3.00 - (-5.00)) \hat{j} D=1.00i^+8.00j^\vec{D} = -1.00 \hat{i} + 8.00 \hat{j}

    The difference vector D\vec{D} is 1.00i^+8.00j^-1.00 \hat{i} + 8.00 \hat{j}, which is a vector pointing 1 unit left along the x-axis and 8 units up along the y-axis.

3. Calculate C\vec{C} and D\vec{D} with the components.

We already calculated the components:

  • C=5.00i^2.00j^\vec{C} = 5.00 \hat{i} - 2.00 \hat{j}
  • D=1.00i^+8.00j^\vec{D} = -1.00 \hat{i} + 8.00 \hat{j}

4. Calculate AB\vec{A} \cdot \vec{B} (dot product).

The dot product of two vectors A=Axi^+Ayj^\vec{A} = A_x \hat{i} + A_y \hat{j} and B=Bxi^+Byj^\vec{B} = B_x \hat{i} + B_y \hat{j} is calculated as:

AB=AxBx+AyBy\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y

For the given vectors:

AB=(2.00)(3.00)+(3.00)(5.00)\vec{A} \cdot \vec{B} = (2.00)(3.00) + (3.00)(-5.00) AB=6.0015.00=9.00\vec{A} \cdot \vec{B} = 6.00 - 15.00 = -9.00

So, the dot product AB=9.00\vec{A} \cdot \vec{B} = -9.00.


Now that we've computed the necessary values, let me know if you'd like further details or a visual representation!

Related Questions:

  1. How do you graphically represent vector addition and subtraction?
  2. What are the real-world applications of dot products in physics?
  3. How do vector components help in calculating the magnitude of a vector?
  4. What is the significance of the dot product result being negative?
  5. Can we use vectors in higher dimensions? How would this change the operations?

Tip:

For vector addition, always align vectors tail-to-head graphically to make the process easier to visualize.

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Math Problem Analysis

Mathematical Concepts

Vector Addition
Vector Subtraction
Dot Product
Vector Components

Formulas

Vector addition: C⃗ = A⃗ + B⃗
Vector subtraction: D⃗ = A⃗ − B⃗
Dot product: A⃗ · B⃗ = AₓBₓ + AᵧBᵧ

Theorems

Properties of vectors
Dot product properties

Suitable Grade Level

Grades 11-12