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 solve the problem, we need to find the magnitude and direction of the vector sum \(\vec{A} + \vec{B}\), where \(\vec{A} = 4\hat{i} - 3\hat{j}\) and \(\vec{B} = 6\hat{i} + 8\hat{j}\).

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

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

Combine the \(\hat{i}\) components and the \(\hat{j}\) components:

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

\[ \vec{A} + \vec{B} = 10\hat{i} + 5\hat{j} \]

### Step 2: Find 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 \(\vec{A} + \vec{B} = 10\hat{i} + 5\hat{j}\):

\[ |\vec{A} + \vec{B}| = \sqrt{(10)^2 + (5)^2} \]

Calculating this:

\[ |\vec{A} + \vec{B}| = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \]

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

The direction (angle \(\theta\)) of a 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 of \(\vec{A} + \vec{B}\) is \(5\sqrt{5}\) and the direction is \(\tan^{-1}\left(\frac{1}{2}\right)\).

### Summary

- Magnitude: \(5\sqrt{5}\)
- Direction: \(\tan^{-1}\left(\frac{1}{2}\right)\)

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