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 it makes with vector \(\vec{A}\), we can follow these steps:

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

### Step 2: Find the magnitude of \(\vec{A} - \vec{B}\)
To find the magnitude of the vector difference \(\vec{A} - \vec{B}\), we can use the formula for the magnitude of the difference of two vectors:
\[ |\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos(\theta)} \]
where \(\theta\) is the angle between the two vectors.

Substituting the known values:
- \(|\vec{A}| = 10\)
- \(|\vec{B}| = 6\)
- \(\theta = 60^\circ\)

We have:
\[ |\vec{A} - \vec{B}| = \sqrt{10^2 + 6^2 - 2 \cdot 10 \cdot 6 \cdot \cos(60^\circ)} \]

### Step 3: Calculate the cosine of the angle
Since \(\cos(60^\circ) = \frac{1}{2}\), we can substitute this into the equation:
\[ |\vec{A} - \vec{B}| = \sqrt{100 + 36 - 2 \cdot 10 \cdot 6 \cdot \frac{1}{2}} \]
\[ = \sqrt{100 + 36 - 60} \]
\[ = \sqrt{76} \]

### Step 4: Simplify the magnitude
The magnitude can be simplified:
\[ |\vec{A} - \vec{B}| = \sqrt{76} = 2\sqrt{19} \]

### Step 5: Find the angle with vector \(\vec{A}\)
To find the angle \(\phi\) that the vector \(\vec{A} - \vec{B}\) makes with vector \(\vec{A}\), we can use the tangent function:
\[ \tan(\phi) = \frac{|\vec{B}| \sin(60^\circ)}{|\vec{A}| - |\vec{B}| \cos(60^\circ)} \]

Substituting the known values:
- \(|\vec{B}| = 6\)
- \(|\vec{A}| = 10\)
- \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\)
- \(\cos(60^\circ) = \frac{1}{2}\)

We have:
\[ \tan(\phi) = \frac{6 \cdot \frac{\sqrt{3}}{2}}{10 - 6 \cdot \frac{1}{2}} \]
\[ = \frac{3\sqrt{3}}{10 - 3} = \frac{3\sqrt{3}}{7} \]

### Step 6: Final results
Thus, the magnitude of the vector difference \(\vec{A} - \vec{B}\) is \(2\sqrt{19}\) units, and the angle it makes with vector \(\vec{A}\) is given by:
\[ \phi = \tan^{-1}\left(\frac{3\sqrt{3}}{7}\right) \]

### Summary of the solution:
- Magnitude of \(\vec{A} - \vec{B} = 2\sqrt{19}\) units
- Angle with vector \(\vec{A} = \tan^{-1}\left(\frac{3\sqrt{3}}{7}\right)\)