If a and b are two non-parallel vectors having equal magnitude, then the vector `(a-b) times (a times b)` is parallel to

Options

A. b
B. a - b
C. a + b
D. a

Answer

To solve the problem, we need to find the vector \((\mathbf{a} - \mathbf{b}) \times (\mathbf{a} \times \mathbf{b})\) and determine what it is parallel to. Let's break this down step by step.

### Step 1: Understand the Cross Product
We know that the cross product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by:
\[ \mathbf{u} \times \mathbf{v} = |\mathbf{u}||\mathbf{v}|\sin(\theta) \, \mathbf{n} \]
where \(\theta\) is the angle between the vectors and \(\mathbf{n}\) is the unit vector perpendicular to the plane formed by \(\mathbf{u}\) and \(\mathbf{v}\).

### Step 2: Calculate \(\mathbf{a} \times \mathbf{b}\)
Since \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel vectors with equal magnitude, we can denote their magnitudes as \( |\mathbf{a}| = |\mathbf{b}| = m \). The cross product \(\mathbf{a} \times \mathbf{b}\) will yield a vector perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).

### Step 3: Use the Vector Triple Product Identity
We can use the vector triple product identity:
\[ \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{u} \cdot \mathbf{v})\mathbf{w} \]
In our case, let \(\mathbf{u} = \mathbf{a} - \mathbf{b}\), \(\mathbf{v} = \mathbf{a}\), and \(\mathbf{w} = \mathbf{b}\).

### Step 4: Apply the Identity
Applying the identity, we have:
\[ (\mathbf{a} - \mathbf{b}) \times (\mathbf{a} \times \mathbf{b}) = ((\mathbf{a} - \mathbf{b}) \cdot \mathbf{b})\mathbf{a} - ((\mathbf{a} - \mathbf{b}) \cdot \mathbf{a})\mathbf{b} \]

### Step 5: Calculate the Dot Products
Now we calculate the dot products:
1. \((\mathbf{a} - \mathbf{b}) \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{b} - \mathbf{b} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{b} - |\mathbf{b}|^2 = \mathbf{a} \cdot \mathbf{b} - m^2\)
2. \((\mathbf{a} - \mathbf{b}) \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{a} = |\mathbf{a}|^2 - \mathbf{a} \cdot \mathbf{b} = m^2 - \mathbf{a} \cdot \mathbf{b}\)

### Step 6: Substitute Back
Substituting these back into our expression gives:
\[ (\mathbf{a} - \mathbf{b}) \times (\mathbf{a} \times \mathbf{b}) = (\mathbf{a} \cdot \mathbf{b} - m^2)\mathbf{a} - (m^2 - \mathbf{a} \cdot \mathbf{b})\mathbf{b} \]
This simplifies to:
\[ = (\mathbf{a} \cdot \mathbf{b} - m^2)\mathbf{a} - (m^2 - \mathbf{a} \cdot \mathbf{b})\mathbf{b} \]

### Step 7: Analyze the Result
The resulting vector is a linear combination of \(\mathbf{a}\) and \(\mathbf{b}\). Therefore, it is parallel to the vector \(\mathbf{a} + \mathbf{b}\).

### Final Conclusion
Thus, the vector \((\mathbf{a} - \mathbf{b}) \times (\mathbf{a} \times \mathbf{b})\) is parallel to \(\mathbf{a} + \mathbf{b}\).