The unit vector parallel to the resultant of the vectors `vecA=4hati+3hatj+6hatk` and `vecB=-hati+3hatj-8hatk` is

Options

A. `(1)/(7)(3hati+6hatj-2hatk)`
B. `(1)/(7)(3hati+6hatj+2hatk)`
C. `(1)/(49)(3hati+6hatj-2hatk)`
D. `(1)/(49)(3hati-6hatj+2hatk)`

Explanation

To find the unit vector parallel to the resultant of the vectors \(\vec{A} = 4\hat{i} + 3\hat{j} + 6\hat{k}\) and \(\vec{B} = -\hat{i} + 3\hat{j} - 8\hat{k}\), we can follow these steps:

### Step 1: Find the resultant vector \(\vec{C}\)
The resultant vector \(\vec{C}\) is given by the sum of vectors \(\vec{A}\) and \(\vec{B}\):
\[ \vec{C} = \vec{A} + \vec{B} \]
Substituting the values of \(\vec{A}\) and \(\vec{B}\):
\[ \vec{C} = (4\hat{i} + 3\hat{j} + 6\hat{k}) + (-\hat{i} + 3\hat{j} - 8\hat{k}) \]
Now, combine the components:
\[ \vec{C} = (4 - 1)\hat{i} + (3 + 3)\hat{j} + (6 - 8)\hat{k} \]
\[ \vec{C} = 3\hat{i} + 6\hat{j} - 2\hat{k} \]

### Step 2: Calculate the magnitude of the resultant vector \(\vec{C}\)
The magnitude of \(\vec{C}\) is calculated using the formula:
\[ |\vec{C}| = \sqrt{C_x^2 + C_y^2 + C_z^2} \]
Substituting the components of \(\vec{C}\):
\[ |\vec{C}| = \sqrt{3^2 + 6^2 + (-2)^2} \]
Calculating each term:
\[ |\vec{C}| = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]

### Step 3: Find the unit vector in the direction of \(\vec{C}\)
The unit vector \(\hat{c}\) in the direction of \(\vec{C}\) is given by:
\[ \hat{c} = \frac{\vec{C}}{|\vec{C}|} \]
Substituting \(\vec{C}\) and its magnitude:
\[ \hat{c} = \frac{3\hat{i} + 6\hat{j} - 2\hat{k}}{7} \]
This simplifies to:
\[ \hat{c} = \frac{3}{7}\hat{i} + \frac{6}{7}\hat{j} - \frac{2}{7}\hat{k} \]

### Final Answer
The unit vector parallel to the resultant of the vectors \(\vec{A}\) and \(\vec{B}\) is:
\[ \hat{c} = \frac{1}{7}(3\hat{i} + 6\hat{j} - 2\hat{k}) \]

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