If `vecA=4hati-3hatj` and `vecB=6hati+8hatj` then magnitude and direction of `vecA+vecB` will be

Options

A. `5, tan^(-1)(3//4)`
B. `5sqrt(5), tan ^(-1)(1//2)`
C. `10, tan ^(-1)(5)`
D. `25 ,tan ^(-1)(3//4)`

Explanation

To find the magnitude and direction of the vector \(\vec{A} + \vec{B}\), we will follow these steps:

### Step 1: Add the vectors \(\vec{A}\) and \(\vec{B}\)

Given:
\[ \vec{A} = 4\hat{i} - 3\hat{j} \]
\[ \vec{B} = 6\hat{i} + 8\hat{j} \]

Now, we add the vectors:
\[ \vec{A} + \vec{B} = (4\hat{i} - 3\hat{j}) + (6\hat{i} + 8\hat{j}) \]

Combine the components:
\[ \vec{A} + \vec{B} = (4 + 6)\hat{i} + (-3 + 8)\hat{j} = 10\hat{i} + 5\hat{j} \]

### Step 2: Calculate the magnitude of \(\vec{A} + \vec{B}\)

The magnitude of a vector \(\vec{V} = x\hat{i} + y\hat{j}\) is given by:
\[ |\vec{V}| = \sqrt{x^2 + y^2} \]

For \(\vec{A} + \vec{B} = 10\hat{i} + 5\hat{j}\):
\[ |\vec{A} + \vec{B}| = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} \]

Simplifying \(\sqrt{125}\):
\[ \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \]

### Step 3: Calculate the direction of \(\vec{A} + \vec{B}\)

The direction (angle \(\theta\)) of the vector can be found using the tangent function:
\[ \tan \theta = \frac{y}{x} \]

For our vector:
\[ \tan \theta = \frac{5}{10} = \frac{1}{2} \]

To find \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]

### Final Result

Thus, the magnitude and direction of \(\vec{A} + \vec{B}\) are:
- Magnitude: \(5\sqrt{5}\)
- Direction: \(\theta = \tan^{-1}\left(\frac{1}{2}\right)\)