Which of the following is not a unit vector for all values of `theta`

Options

A. `(cos theta)i- (sin theta) j`
B. `(sin theta)i+ (cos theta)j`
C. `(sin 2 theta)i- (cos theta)j`
D. `(cos 2theta)i- (sin 2 theta)j`

Answer

To determine which of the given options is not a unit vector for all values of \(\theta\), we need to analyze each option and find their magnitudes. A unit vector has a magnitude of 1.

### Step-by-Step Solution:

1. **Understanding Unit Vectors**:
A unit vector is defined as a vector whose magnitude is equal to 1. For a vector \(\mathbf{A} = x \hat{i} + y \hat{j}\), the magnitude is given by:
\[ |\mathbf{A}| = \sqrt{x^2 + y^2} \]
We need to check if the magnitude equals 1 for each option.

2. **Option 1: \(\cos \theta \hat{i} - \sin \theta \hat{j}\)**:
- Calculate the magnitude:
\[ |\mathbf{A}_1| = \sqrt{(\cos \theta)^2 + (-\sin \theta)^2} = \sqrt{\cos^2 \theta + \sin^2 \theta} \]
- Using the Pythagorean identity:
\[ |\mathbf{A}_1| = \sqrt{1} = 1 \]
- **Conclusion**: This is a unit vector.

3. **Option 2: \(\sin \theta \hat{i} + \cos \theta \hat{j}\)**:
- Calculate the magnitude:
\[ |\mathbf{A}_2| = \sqrt{(\sin \theta)^2 + (\cos \theta)^2} = \sqrt{\sin^2 \theta + \cos^2 \theta} \]
- Again, using the Pythagorean identity:
\[ |\mathbf{A}_2| = \sqrt{1} = 1 \]
- **Conclusion**: This is also a unit vector.

4. **Option 3: \(\sin^2(2\theta) \hat{i} + (-\cos \theta) \hat{j}\)**:
- Calculate the magnitude:
\[ |\mathbf{A}_3| = \sqrt{(\sin^2(2\theta))^2 + (-\cos \theta)^2} = \sqrt{\sin^4(2\theta) + \cos^2 \theta} \]
- Substitute \(\sin(2\theta) = 2 \sin \theta \cos \theta\):
\[ |\mathbf{A}_3| = \sqrt{(2 \sin \theta \cos \theta)^4 + \cos^2 \theta} = \sqrt{16 \sin^4 \theta \cos^4 \theta + \cos^2 \theta} \]
- This expression does not simplify to 1 for all \(\theta\).
- **Conclusion**: This is not a unit vector.

5. **Option 4: \(\cos^2(2\theta) \hat{i} + \sin^2(2\theta) \hat{j}\)**:
- Calculate the magnitude:
\[ |\mathbf{A}_4| = \sqrt{(\cos^2(2\theta))^2 + (\sin^2(2\theta))^2} = \sqrt{\cos^4(2\theta) + \sin^4(2\theta)} \]
- Using the identity \(a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2\):
\[ |\mathbf{A}_4| = \sqrt{1 - 2\cos^2(2\theta)\sin^2(2\theta)} \text{ (not always equal to 1)} \]
- This expression also does not simplify to 1 for all \(\theta\).
- **Conclusion**: This is a unit vector.

### Final Conclusion:
The option that is **not a unit vector for all values of \(\theta\)** is **Option 3**.