The resultant of the two vectors `overset(rarr)a` and `overset(rarr)b` of magnitudes 6 and 10 respectively is at right angles to the vector `overset(rarr)a` Find their resultant and also the angle between the given vectors.

Options

A. `127^(@)`
B. `143^(@)`
C. `120^(@)`
D. `150^(@)`

Explanation

To solve the problem, we need to find the resultant vector \( \vec{R} \) of two vectors \( \vec{a} \) and \( \vec{b} \) with magnitudes 6 and 10 respectively, given that \( \vec{R} \) is at right angles to \( \vec{a} \). We also need to determine the angle \( \theta \) between the two vectors.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two vectors \( \vec{a} \) and \( \vec{b} \) with magnitudes \( |\vec{a}| = 6 \) and \( |\vec{b}| = 10 \).
- The resultant vector \( \vec{R} \) is perpendicular to \( \vec{a} \).

2. **Setting Up the Vectors**:
- Let \( \vec{a} = 6 \hat{i} \) (along the x-axis).
- Let \( \vec{b} \) make an angle \( \theta \) with \( \vec{a} \). Thus, we can express \( \vec{b} \) in terms of its components:
\[ \vec{b} = |\vec{b}| \cos(\theta) \hat{i} + |\vec{b}| \sin(\theta) \hat{j} = 10 \cos(\theta) \hat{i} + 10 \sin(\theta) \hat{j} \]

3. **Finding the Resultant Vector**:
- The resultant vector \( \vec{R} \) is given by:
\[ \vec{R} = \vec{a} + \vec{b} = 6 \hat{i} + (10 \cos(\theta) \hat{i} + 10 \sin(\theta) \hat{j}) \]
- Simplifying this, we get:
\[ \vec{R} = (6 + 10 \cos(\theta)) \hat{i} + 10 \sin(\theta) \hat{j} \]

4. **Condition for Perpendicularity**:
- Since \( \vec{R} \) is perpendicular to \( \vec{a} \), the dot product \( \vec{R} \cdot \vec{a} \) must equal zero:
\[ \vec{R} \cdot \vec{a} = (6 + 10 \cos(\theta)) \cdot 6 + 10 \sin(\theta) \cdot 0 = 0 \]
- This simplifies to:
\[ 6 + 10 \cos(\theta) = 0 \]
- Solving for \( \cos(\theta) \):
\[ 10 \cos(\theta) = -6 \implies \cos(\theta) = -\frac{3}{5} \]

5. **Finding \( \theta \)**:
- To find \( \theta \), we can use the cosine inverse function:
\[ \theta = \cos^{-1}\left(-\frac{3}{5}\right) \]

6. **Finding the Magnitude of the Resultant**:
- The magnitude of the resultant \( |\vec{R}| \) can be calculated using Pythagoras' theorem:
\[ |\vec{R}| = \sqrt{(6 + 10 \cos(\theta))^2 + (10 \sin(\theta))^2} \]
- Substituting \( \cos(\theta) = -\frac{3}{5} \):
\[ 6 + 10 \cos(\theta) = 6 - 6 = 0 \]
- Therefore, the x-component of \( \vec{R} \) is zero, and we only have the y-component:
\[ |\vec{R}| = |10 \sin(\theta)| \]
- Using \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \sin^2(\theta) = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \implies \sin(\theta) = \frac{4}{5} \]
- Thus, the magnitude of \( \vec{R} \) is:
\[ |\vec{R}| = 10 \cdot \frac{4}{5} = 8 \]

### Final Results:
- The magnitude of the resultant vector \( |\vec{R}| = 8 \).
- The angle \( \theta = \cos^{-1}\left(-\frac{3}{5}\right) \).