Finding the Angle Between Unit Vectors

The problem asks us to find the angle between two unit vectors, â and b̂, given a condition related to two other vectors formed by them. The condition is that the vectors â + 2b̂ and 5â - 4b̂ are perpendicular to each other. Understanding the properties of unit vectors and perpendicular vectors is key to solving this.

Recall that a unit vector has a magnitude of 1. So, |â| = 1 and |b̂| = 1. Also, two vectors are perpendicular if their dot product is zero.

Using the Perpendicularity Condition

Since the vectors â + 2b̂ and 5â - 4b̂ are perpendicular, their dot product is zero. We can write this as:

\( (\hat{a} + 2\hat{b}) \cdot (5\hat{a} - 4\hat{b}) = 0 \)

Now, let's expand the dot product using the distributive property:

\( \hat{a} \cdot (5\hat{a}) + \hat{a} \cdot (-4\hat{b}) + (2\hat{b}) \cdot (5\hat{a}) + (2\hat{b}) \cdot (-4\hat{b}) = 0 \)

This simplifies to:

\( 5(\hat{a} \cdot \hat{a}) - 4(\hat{a} \cdot \hat{b}) + 10(\hat{b} \cdot \hat{a}) - 8(\hat{b} \cdot \hat{b}) = 0 \)

Properties of Dot Product and Unit Vectors

We use the following properties of the dot product:

• \( \mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2 \)
• \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \) (Commutative property)
• For unit vectors â and b̂, \( \hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1^2 = 1 \) and \( \hat{b} \cdot \hat{b} = |\hat{b}|^2 = 1^2 = 1 \).

Substituting these properties into the expanded dot product equation:

\( 5(1) - 4(\hat{a} \cdot \hat{b}) + 10(\hat{a} \cdot \hat{b}) - 8(1) = 0 \)

\( 5 - 4(\hat{a} \cdot \hat{b}) + 10(\hat{a} \cdot \hat{b}) - 8 = 0 \)

Combine like terms:

\( (10 - 4)(\hat{a} \cdot \hat{b}) + (5 - 8) = 0 \)

\( 6(\hat{a} \cdot \hat{b}) - 3 = 0 \)

Solve for the dot product \( \hat{a} \cdot \hat{b} \):

\( 6(\hat{a} \cdot \hat{b}) = 3 \)

\( \hat{a} \cdot \hat{b} = \frac{3}{6} = \frac{1}{2} \)

Calculating the Angle Between the Unit Vectors

The dot product of two vectors is also defined as \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta \), where \( \theta \) is the angle between the vectors. For our unit vectors â and b̂, the angle between unit vectors is \( \theta \).

So, \( \hat{a} \cdot \hat{b} = |\hat{a}| |\hat{b}| \cos \theta \).

Since |â| = 1 and |b̂| = 1, this becomes:

\( \hat{a} \cdot \hat{b} = (1)(1) \cos \theta = \cos \theta \)

We found that \( \hat{a} \cdot \hat{b} = \frac{1}{2} \). Therefore:

\( \cos \theta = \frac{1}{2} \)

To find the angle \( \theta \), we need to find the value whose cosine is 1/2. This is a standard trigonometric value.

\( \theta = \cos^{-1} \left( \frac{1}{2} \right) \)

The angle whose cosine is 1/2 is \( \pi/3 \) radians (or 60 degrees).

Final Angle Between Unit Vectors

Thus, the angle between the unit vectors â and b̂ is \( \pi/3 \). This demonstrates how the properties of perpendicularity and unit vectors can be used to find the angle between unit vectors.