Let \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are two unit vectors such that \(\vec{\text{a}}+2 \vec{\text{b}}\) and \(5\vec{\text{a}}−4\vec{\text{b}}\) are perpendicular. What is the angle between \(\vec{\text{a}}\) and \(\vec{\text{b}}\) ?
Let's find the angle between two unit vectors given a condition about their linear combinations being perpendicular. The problem involves vectors, unit vectors, perpendicularity, and the dot product.
Understanding Unit Vectors and Perpendicularity
A unit vector is a vector with a magnitude of 1. We are given that \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are unit vectors, which means \(|\vec{\text{a}}| = 1\) and \(|\vec{\text{b}}| = 1\).
Two vectors are perpendicular if their dot product (or scalar product) is zero. We are told that the vector \(\vec{\text{a}} + 2\vec{\text{b}}\) and the vector \(5\vec{\text{a}} - 4\vec{\text{b}}\) are perpendicular. Therefore, their dot product must be zero:
\((\vec{\text{a}} + 2\vec{\text{b}}) \cdot (5\vec{\text{a}} - 4\vec{\text{b}}) = 0\)
Using the Dot Product Property
We can expand the dot product using the distributive property, similar to multiplying binomials:
\(\vec{\text{a}} \cdot (5\vec{\text{a}} - 4\vec{\text{b}}) + 2\vec{\text{b}} \cdot (5\vec{\text{a}} - 4\vec{\text{b}}) = 0\)
\((\vec{\text{a}} \cdot 5\vec{\text{a}}) - (\vec{\text{a}} \cdot 4\vec{\text{b}}) + (2\vec{\text{b}} \cdot 5\vec{\text{a}}) - (2\vec{\text{b}} \cdot 4\vec{\text{b}}) = 0\)
Using the properties of the dot product, such as \(c(\vec{\text{x}} \cdot \vec{\text{y}}) = \vec{\text{x}} \cdot (c\vec{\text{y}})\) and \(\vec{\text{x}} \cdot \vec{\text{y}} = \vec{\text{y}} \cdot \vec{\text{x}}\):
\(5(\vec{\text{a}} \cdot \vec{\text{a}}) - 4(\vec{\text{a}} \cdot \vec{\text{b}}) + 10(\vec{\text{b}} \cdot \vec{\text{a}}) - 8(\vec{\text{b}} \cdot \vec{\text{b}}) = 0\)
Since \(\vec{\text{a}} \cdot \vec{\text{a}} = |\vec{\text{a}}|^2\), \(\vec{\text{b}} \cdot \vec{\text{b}} = |\vec{\text{b}}|^2\), and \(\vec{\text{b}} \cdot \vec{\text{a}} = \vec{\text{a}} \cdot \vec{\text{b}}\), we get:
\(5|\vec{\text{a}}|^2 - 4(\vec{\text{a}} \cdot \vec{\text{b}}) + 10(\vec{\text{a}} \cdot \vec{\text{b}}) - 8|\vec{\text{b}}|^2 = 0\)
Combine the terms involving the dot product \(\vec{\text{a}} \cdot \vec{\text{b}}\):
\(5|\vec{\text{a}}|^2 + 6(\vec{\text{a}} \cdot \vec{\text{b}}) - 8|\vec{\text{b}}|^2 = 0\)
Substituting Unit Vector Magnitudes
Since \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are unit vectors, we substitute \(|\vec{\text{a}}| = 1\) and \(|\vec{\text{b}}| = 1\) into the equation:
\(5(1)^2 + 6(\vec{\text{a}} \cdot \vec{\text{b}}) - 8(1)^2 = 0\)
\(5 + 6(\vec{\text{a}} \cdot \vec{\text{b}}) - 8 = 0\)
Simplify the equation:
\(6(\vec{\text{a}} \cdot \vec{\text{b}}) - 3 = 0\)
Solve for the dot product \(\vec{\text{a}} \cdot \vec{\text{b}}\):
\(6(\vec{\text{a}} \cdot \vec{\text{b}}) = 3\)
\(\vec{\text{a}} \cdot \vec{\text{b}} = \frac{3}{6} = \frac{1}{2}\)
Finding the Angle Between Vectors
The dot product of two vectors is also defined as \(\vec{\text{a}} \cdot \vec{\text{b}} = |\vec{\text{a}}| |\vec{\text{b}}| \cos \theta\), where \(\theta\) is the angle between \(\vec{\text{a}}\) and \(\vec{\text{b}}\). We know \(|\vec{\text{a}}| = 1\), \(|\vec{\text{b}}| = 1\), and \(\vec{\text{a}} \cdot \vec{\text{b}} = \frac{1}{2}\).
Substitute these values into the formula:
\(\frac{1}{2} = (1)(1) \cos \theta\)
\(\cos \theta = \frac{1}{2}\)
We need to find the angle \(\theta\) whose cosine is \(\frac{1}{2}\). In the range \(0 \le \theta \le \pi\) (which is typically considered for the angle between two vectors), the angle is \(\theta = \frac{\pi}{3}\).
Therefore, the angle between the unit vectors \(\vec{\text{a}}\) and \(\vec{\text{b}}\) is \(\frac{\pi}{3}\).
Revision Table: Key Concepts in Vector Problems
| Concept | Description | Formula/Property |
|---|---|---|
| Unit Vector | A vector with magnitude 1. | \(|\vec{\text{v}}| = 1\) |
| Perpendicular Vectors | Two vectors whose dot product is zero. | \(\vec{\text{u}} \cdot \vec{\text{v}} = 0\) |
| Dot Product (\(\vec{\text{a}} \cdot \vec{\text{b}}\)) | Scalar value representing the projection of one vector onto another scaled by the magnitude of the other. | \(\vec{\text{a}} \cdot \vec{\text{b}} = |\vec{\text{a}}| |\vec{\text{b}}| \cos \theta\) |
| Dot Product Property | Distributes over vector addition; commutative. | \(\vec{\text{a}} \cdot (\vec{\text{b}} + \vec{\text{c}}) = \vec{\text{a}} \cdot \vec{\text{b}} + \vec{\text{a}} \cdot \vec{\text{c}}\), \(\vec{\text{a}} \cdot \vec{\text{b}} = \vec{\text{b}} \cdot \vec{\text{a}}\) |
| Magnitude Squared | The dot product of a vector with itself. | \(\vec{\text{a}} \cdot \vec{\text{a}} = |\vec{\text{a}}|^2\) |
Additional Information on Vector Angle Calculation
When finding the angle between two vectors \(\vec{\text{a}}\) and \(\vec{\text{b}}\), the formula \(\cos \theta = \frac{\vec{\text{a}} \cdot \vec{\text{b}}}{|\vec{\text{a}}| |\vec{\text{b}}|}\) is fundamental. This formula is derived directly from the definition of the dot product. The result \(\theta\) is usually taken to be in the range \([0, \pi]\) radians or \([0, 180]\) degrees.
In this specific problem, the condition that linear combinations of the vectors are perpendicular provides the necessary equation to solve for the dot product \(\vec{\text{a}} \cdot \vec{\text{b}}\). Once the dot product is known, and given that the vectors are unit vectors (so their magnitudes are 1), we can easily find the cosine of the angle and thus the angle itself.
This method is a common approach for solving vector problems involving angles and perpendicularity conditions. It reinforces the importance of the dot product as a tool to relate vector algebra to geometric properties like angles and orthogonality.
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