How can I find the component form of a vector v that lies in the first quadrant and forms an angle of π/3 with the positive x-axis, given that the magnitude of v is 8?

To find the component form of the vector v that lies in the first quadrant, we start by using the information provided: the angle θ is π/3 (or 60 degrees) with respect to the positive x-axis and the magnitude of the vector is 8.

The component form of a vector can be expressed as:

v = 
  < vx, vy >

Where:

• v_x is the horizontal component of the vector, and
• v_y is the vertical component of the vector.

To find these components, we use trigonometric functions:

• v_x = |v| * cos(θ)
• v_y = |v| * sin(θ)

Substituting the given values:

• |v| = 8
• θ = π/3

Calculating the components:

• v_x = 8 * cos(π/3)
• v_y = 8 * sin(π/3)

Using the known values of the cosine and sine for π/3:

• cos(π/3) = 1/2
• sin(π/3) = √3/2

So:

• v_x = 8 * (1/2) = 4
• v_y = 8 * (√3/2) = 4√3

Therefore, the component form of the vector v is:

v = < 4, 4√3 >

This means that the vector v has a horizontal component of 4 and a vertical component of 4√3, and it lies in the first quadrant as desired.