A vector `vec(A)` of length `10` units makes an angle of `60^(@)` with a vector `vec(B)` of length `6` units. Find the magnitude of the vector difference `vec(A)-vec(B)` & the angles with vector `vec(A)`.

Explanation

To solve the problem of finding the magnitude of the vector difference \(\vec{A} - \vec{B}\) and the angle with vector \(\vec{A}\), we can follow these steps:

### Step 1: Identify the given values
- Length of vector \(\vec{A} = 10\) units
- Length of vector \(\vec{B} = 6\) units
- Angle between \(\vec{A}\) and \(\vec{B} = 60^\circ\)

### Step 2: Use the formula for the magnitude of the vector difference
The magnitude of the vector difference \(\vec{A} - \vec{B}\) can be calculated using the formula:
\[ |\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos(\theta)} \]

### Step 3: Substitute the values into the formula
Substituting the values we have:
- \(|\vec{A}| = 10\)
- \(|\vec{B}| = 6\)
- \(\theta = 60^\circ\) (where \(\cos(60^\circ) = \frac{1}{2}\))

Now substituting these values into the formula:
\[ |\vec{A} - \vec{B}| = \sqrt{10^2 + 6^2 - 2 \cdot 10 \cdot 6 \cdot \cos(60^\circ)} \]
\[ = \sqrt{100 + 36 - 2 \cdot 10 \cdot 6 \cdot \frac{1}{2}} \]
\[ = \sqrt{100 + 36 - 60} \]
\[ = \sqrt{76} \]

### Step 4: Simplify the result
The magnitude of the vector difference is:
\[ |\vec{A} - \vec{B}| = \sqrt{76} = 2\sqrt{19} \]

### Step 5: Find the angle with vector \(\vec{A}\)
To find the angle \(\alpha\) between \(\vec{A}\) and \(\vec{A} - \vec{B}\), we can use the formula:
\[ \tan(\alpha) = \frac{|\vec{B}| \sin(\theta)}{|\vec{A} - \vec{B}|} \]
Where \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\).

Substituting the values:
\[ \tan(\alpha) = \frac{6 \cdot \frac{\sqrt{3}}{2}}{2\sqrt{19}} \]
\[ = \frac{3\sqrt{3}}{2\sqrt{19}} \]

### Step 6: Calculate \(\alpha\)
Now, we can find \(\alpha\) using the arctangent function:
\[ \alpha = \tan^{-1}\left(\frac{3\sqrt{3}}{2\sqrt{19}}\right) \]

### Final Result
1. The magnitude of the vector difference \(|\vec{A} - \vec{B}| = 2\sqrt{19}\).
2. The angle \(\alpha\) with vector \(\vec{A}\) is given by \(\alpha = \tan^{-1}\left(\frac{3\sqrt{3}}{2\sqrt{19}}\right)\).