The Relationship Between Vectors A and B in Geometry and Trigonometry

The problem presented involves the relationship between two vectors, A and B, in terms of their magnitudes and angles. We are given that the resultant vector R has half the magnitude of A and is perpendicular to B. This article will explore the detailed steps and calculations necessary to determine the magnitude of B and the angle between A and B.

Understanding the Problem

Given that the magnitude of vector A is 5 units, and the resultant vector R has half the magnitude of A and is perpendicular to B, we need to find the magnitude of vector B and the angle between vectors A and B. This problem involves both vector algebra and trigonometry, specifically the application of sine and cosine rules.

Step-by-Step Solution

To solve this, we need to use both vector properties and trigonometric identities. Here, we assume that the magnitude of A is given in a specific unit (e.g., inches or feet).

Using Vector Properties to Find the Magnitude of B

Given that R is perpendicular to B, the dot product of R and B is 0:

R · B 0

Using the vector properties, we can express the dot product as:

(A B) · B 0

Expanding this, we get:

A · B B · B 0

Since A and R are perpendicular, we can use the properties of the dot product to simplify:

A · B -B · B

Given that R A/2, we can write:

R · B 0

Substituting R with A/2:

(A/2) · B 0

This implies:

A · B 0

However, from the problem, we know that R is perpendicular to B and we can use the cosine rule for vectors:

R · B |R| |B| cos(90 x)

Since R is perpendicular to B, the angle between them is 90 degrees, and thus:

R · B |R| |B| cos(90) 0

This confirms that the dot product is zero, which is correct.

Determining the Magnitude of B

To find the magnitude of B, we use the cosine rule:

R A/2 B sin(90 - x)

Since R is perpendicular to B, the angle x can be found using the sine function:

R A/2 B sin(60)

Solving for B:

B (A/2) / sin(60)

Given that A 5 units:

B (5/2) / (sqrt(3)/2) 5 / sqrt(3)

Multiplying the numerator and denominator by sqrt(3):

B 5sqrt(3) / 3

This is the magnitude of vector B.

Finding the Angle Between Vectors A and B

The angle between A and B can be found using the sine rule and the given conditions:

sin(90 - x) 1/2

Solving for the angle 90 - x:

90 - x 30

This implies:

x 60

The angle between A and B is:

150 degrees (since 90 60 150)

Conclusion

In summary, the magnitude of vector B is 5sqrt(3)/3 units, and the angle between vectors A and B is 150 degrees. This solution provides a comprehensive approach to understanding and calculating vector relationships using basic trigonometric principles and vector properties.

Keywords

Vector magnitude, angle between vectors, vector perpendicular