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  1. Convert the following vectors to Cartesian form.
    1. 13 m on a bearing of 165°
    2. 120 km/h 20° West of North
  2. Convert the following vectors to direction/magnitude form, giving the direction as a bearing.
    1. [12, 5]
    2. [–31, –11]

Mohit Dhankhar , 3 Years ago
Grade 12th pass
anser 1 Answers
Askiitians Tutor Team

Last Activity: 13 Days ago

To convert vectors into Cartesian form and then back into direction/magnitude form, we need to understand a few concepts about vector representation. Let's tackle each part of your question step by step.

Converting Vectors to Cartesian Form

Cartesian form represents a vector in terms of its horizontal (x) and vertical (y) components. The first vector is given as 13 m on a bearing of 165°. A bearing is measured clockwise from North.

Vector 1: 13 m on a bearing of 165°

To convert this to Cartesian coordinates, we can use trigonometric functions. The angle from the positive x-axis (East) can be found by subtracting the bearing from 90°:

  • Angle from East = 90° - 165° = -75° (or 285° if you prefer positive angles)

Now, we can find the x and y components:

  • x = 13 m * cos(285°)
  • y = 13 m * sin(285°)

Calculating these:

  • x ≈ 13 * 0.2588 ≈ 3.37 m
  • y ≈ 13 * (-0.9659) ≈ -12.55 m

Thus, the Cartesian form of the first vector is approximately:

(3.37 m, -12.55 m)

Vector 2: 120 km/h, 20° West of North

For this vector, we first need to determine the angle from the North. Since it is 20° West of North, we can directly use this angle:

  • Angle from North = 20°

Now, we can find the components:

  • x = 120 km/h * sin(20°)
  • y = 120 km/h * cos(20°)

Calculating these:

  • x ≈ 120 * 0.3420 ≈ 41.04 km/h
  • y ≈ 120 * 0.9397 ≈ 112.76 km/h

Since the x-component is directed West, we represent it as negative:

(-41.04 km/h, 112.76 km/h)

Converting Vectors to Direction/Magnitude Form

Now, let’s convert the given Cartesian vectors into direction/magnitude form. This involves calculating the magnitude and the angle (bearing) of each vector.

Vector 1: [12, 5]

First, we calculate the magnitude:

  • Magnitude = √(12² + 5²) = √(144 + 25) = √169 = 13

Next, we find the angle:

  • Angle = tan⁻¹(5/12) ≈ 22.62°

Since this vector is in the first quadrant, the bearing is:

  • Bearing = 90° - 22.62° ≈ 67.38°

Thus, the direction/magnitude form is:

13 at a bearing of 67.38°

Vector 2: [-31, -11]

For this vector, we again start with the magnitude:

  • Magnitude = √((-31)² + (-11)²) = √(961 + 121) = √1082 ≈ 32.88

Next, we find the angle:

  • Angle = tan⁻¹(11/31) ≈ 19.74°

This vector is in the third quadrant, so we adjust the angle to find the bearing:

  • Bearing = 180° + 19.74° ≈ 199.74°

Therefore, the direction/magnitude form is:

32.88 at a bearing of 199.74°

Summary

In summary, we converted the vectors to Cartesian form as follows:

  • 13 m on a bearing of 165° → (3.37 m, -12.55 m)
  • 120 km/h, 20° West of North → (-41.04 km/h, 112.76 km/h)

And we converted the Cartesian vectors to direction/magnitude form:

  • [12, 5] → 13 at a bearing of 67.38°
  • [-31, -11] → 32.88 at a bearing of 199.74°

Understanding these conversions is crucial in physics and engineering, as it allows for the analysis of forces, velocities, and other vector quantities in a clear and systematic way.

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