Math Problem Statement

There are 3 vectors in a 2D coordinate with unit vector π‘₯Μ‚ and 𝑦̂ along x and y axes respectively. The magnitude of π‘Žβƒ— is 5 and it forms an angle of 60π‘œ with the positive x-axis. Vector 𝑏⃗⃗ = 5π‘₯Μ‚ + 8𝑦̂. The coordinates of the starting and ending points of 𝑐⃗ are (4, 5) and (12, 7). Calculate the resultant of π‘Žβƒ— + 𝑏⃗⃗ + 𝑐⃗

Solution

We are asked to calculate the resultant of three vectors a⃗+b⃗+c⃗\vec{a} + \vec{b} + \vec{c}, where:

  1. Vector aβƒ—\vec{a} has a magnitude of 5 and forms an angle of 60∘60^\circ with the positive x-axis.
  2. Vector b⃗=5i^+8j^\vec{b} = 5\hat{i} + 8\hat{j}.
  3. Vector c⃗\vec{c} has starting and ending points at (4, 5) and (12, 7), respectively.

Step 1: Express each vector in component form

Vector a⃗\vec{a}

The components of a⃗\vec{a} can be derived using trigonometry:

  • The x-component is ax=5cos⁑(60∘)a_x = 5 \cos(60^\circ).
  • The y-component is ay=5sin⁑(60∘)a_y = 5 \sin(60^\circ).

Since cos⁑(60∘)=12\cos(60^\circ) = \frac{1}{2} and sin⁑(60∘)=32\sin(60^\circ) = \frac{\sqrt{3}}{2}, we get: ax=5Γ—12=2.5a_x = 5 \times \frac{1}{2} = 2.5 ay=5Γ—32β‰ˆ4.33a_y = 5 \times \frac{\sqrt{3}}{2} \approx 4.33 Thus, aβƒ—=2.5i^+4.33j^\vec{a} = 2.5\hat{i} + 4.33\hat{j}.

Vector b⃗\vec{b}

This vector is already given in component form: b⃗=5i^+8j^\vec{b} = 5\hat{i} + 8\hat{j}

Vector c⃗\vec{c}

To find the components of cβƒ—\vec{c}, we calculate the differences between the ending and starting points: cx=12βˆ’4=8c_x = 12 - 4 = 8 cy=7βˆ’5=2c_y = 7 - 5 = 2 Thus, cβƒ—=8i^+2j^\vec{c} = 8\hat{i} + 2\hat{j}.

Step 2: Add the vectors

Now, sum the components of a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c} in the x- and y-directions separately.

x-component:

Rx=ax+bx+cx=2.5+5+8=15.5R_x = a_x + b_x + c_x = 2.5 + 5 + 8 = 15.5

y-component:

Ry=ay+by+cy=4.33+8+2=14.33R_y = a_y + b_y + c_y = 4.33 + 8 + 2 = 14.33

Step 3: Calculate the magnitude of the resultant vector

The magnitude ∣Rβƒ—βˆ£|\vec{R}| of the resultant vector is given by the Pythagorean theorem: ∣Rβƒ—βˆ£=Rx2+Ry2=(15.5)2+(14.33)2|\vec{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(15.5)^2 + (14.33)^2}

Let's compute that.

The magnitude of the resultant vector R⃗=a⃗+b⃗+c⃗\vec{R} = \vec{a} + \vec{b} + \vec{c} is approximately 21.11.

Would you like more details on any part of the calculation? Here are some questions to consider:

  1. How can the direction of the resultant vector be determined?
  2. What are unit vectors and how do they affect vector addition?
  3. How would the result change if the angle for a⃗\vec{a} were different?
  4. How can you represent the vectors graphically?
  5. What would the calculation look like in 3D?

Tip: Always ensure the angles used in vector calculations are in the correct unit (degrees or radians).

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Math Problem Analysis

Mathematical Concepts

Vector addition
Trigonometry
Pythagorean theorem
Coordinate geometry

Formulas

Vector components: a_x = magnitude * cos(angle), a_y = magnitude * sin(angle)
Resultant vector magnitude: |R| = √(R_x^2 + R_y^2)

Theorems

Pythagorean theorem

Suitable Grade Level

Grades 10-12