Calculate the x and y components of the vectors that are orthogonal to and have

the same length as the vectors in previous exercise.

Given vector Orthogonal vector

Vector Length l Angle α Angle x component y component

𝑎⃗ 1 0° 90°

𝑏$⃗ 2 30° 120°

𝑐⃗ 3 45° 135°

𝑑⃗ 4 90° 180°

To solve this, we need to find vectors that are orthogonal (perpendicular) to each of the given vectors $ \vec{a}, \vec{b}, \vec{c}, \vec{d} $, and that have the same length as those vectors.

### Steps to calculate orthogonal vectors:
1. The length of the orthogonal vector will be the same as the original vector.
2. If a vector has an angle $ \alpha $, then a vector orthogonal to it will have an angle $ \alpha + 90^\circ $ (or $ \alpha - 90^\circ $, since both directions are valid for orthogonality).
3. To find the components of the orthogonal vector, use the following formulas:
   - The $ x $-component is: $ x = l \cdot \cos(\alpha) $
   - The $ y $-component is: $ y = l \cdot \sin(\alpha) $
   where $ l $ is the length and $ \alpha $ is the angle of the orthogonal vector.

We will apply this logic for each vector.

### Vector $ \vec{a} $ (Length = 1, Angle = $ 0^\circ $)
- Orthogonal angle: $ 0^\circ + 90^\circ = 90^\circ $
- Length: $ l = 1 $
- Components of orthogonal vector:
  - $ x = 1 \cdot \cos(90^\circ) = 0 $
  - $ y = 1 \cdot \sin(90^\circ) = 1 $
- Orthogonal vector components: $ (0, 1) $

### Vector $ \vec{b} $ (Length = 2, Angle = $ 30^\circ $)
- Orthogonal angle: $ 30^\circ + 90^\circ = 120^\circ $
- Length: $ l = 2 $
- Components of orthogonal vector:
  - $ x = 2 \cdot \cos(120^\circ) = 2 \cdot (-\frac{1}{2}) = -1 $
  - $ y = 2 \cdot \sin(120^\circ) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} $
- Orthogonal vector components: $ (-1, \sqrt{3}) $

### Vector $ \vec{c} $ (Length = 3, Angle = $ 45^\circ $)
- Orthogonal angle: $ 45^\circ + 90^\circ = 135^\circ $
- Length: $ l = 3 $
- Components of orthogonal vector:
  - $ x = 3 \cdot \cos(135^\circ) = 3 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\frac{3\sqrt{2}}{2} $
  - $ y = 3 \cdot \sin(135^\circ) = 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} $
- Orthogonal vector components: $ \left(-\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2}\right) $

### Vector $ \vec{d} $ (Length = 4, Angle = $ 90^\circ $)
- Orthogonal angle: $ 90^\circ + 90^\circ = 180^\circ $
- Length: $ l = 4 $
- Components of orthogonal vector:
  - $ x = 4 \cdot \cos(180^\circ) = 4 \cdot (-1) = -4 $
  - $ y = 4 \cdot \sin(180^\circ) = 4 \cdot 0 = 0 $
- Orthogonal vector components: $ (-4, 0) $

### Final Answer:
1. $ \vec{a} $ orthogonal vector: $ (0, 1) $
2. $ \vec{b} $ orthogonal vector: $ (-1, \sqrt{3}) $
3. $ \vec{c} $ orthogonal vector: $ \left(-\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2}\right) $
4. $ \vec{d} $ orthogonal vector: $ (-4, 0) $

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Do you need further clarification on these steps? Here are some related questions to explore:

1. How can we verify that two vectors are orthogonal?
2. What is the significance of the dot product in vector calculations?
3. How can the magnitude of a vector be calculated from its components?
4. How would the result change if we used the negative orthogonal direction?
5. Can you apply these techniques to vectors in three dimensions?

**Tip:** When checking orthogonality, the dot product of two vectors should equal zero.

Calculate the x and y components of the vectors that are orthogonal to and have the same length as the vectors in the previous exercise.

This page explains how to calculate the x and y components of vectors that are orthogonal and have the same length as given vectors. Using vector component formulas and trigonometric principles, the solution covers orthogonality and vector decomposition. Suitable for Grades 10-12 students.

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