If `vec(A) = 4hat(i) - 2hat(j) + 6hat(k)` and `B = hat(i) - 2hat(j) - 3hat(k)` the angle which the `vec(A) vec(B)` makes with x-axis is :

Options

A. `cos^(-1)((1)/(10))`
B. `cos^(-1)((2)/(sqrt(10)))`
C. `cos^(-1)((1)/(sqrt(10)))`
D. `cos^(-1)((1)/(5))`

Answer

To find the angle that the vector \( \vec{A} \) makes with the x-axis, we will follow these steps:

### Step 1: Identify the vectors
Given:
\[ \vec{A} = 4\hat{i} - 2\hat{j} + 6\hat{k} \]
\[ \vec{B} = \hat{i} - 2\hat{j} - 3\hat{k} \]

### Step 2: Calculate the resultant vector \( \vec{AB} \)
We need to find the vector \( \vec{AB} \) which is given by:
\[ \vec{AB} = \vec{A} - \vec{B} \]
Calculating \( \vec{AB} \):
\[ \vec{AB} = (4\hat{i} - 2\hat{j} + 6\hat{k}) - (\hat{i} - 2\hat{j} - 3\hat{k}) \]
\[ = (4 - 1)\hat{i} + (-2 + 2)\hat{j} + (6 + 3)\hat{k} \]
\[ = 3\hat{i} + 0\hat{j} + 9\hat{k} \]
Thus,
\[ \vec{AB} = 3\hat{i} + 9\hat{k} \]

### Step 3: Find the magnitude of \( \vec{AB} \)
The magnitude of \( \vec{AB} \) is given by:
\[ |\vec{AB}| = \sqrt{(3)^2 + (0)^2 + (9)^2} = \sqrt{9 + 0 + 81} = \sqrt{90} = 3\sqrt{10} \]

### Step 4: Find the dot product with the x-axis vector
The unit vector along the x-axis is \( \hat{i} \). The dot product of \( \vec{AB} \) and \( \hat{i} \) is:
\[ \vec{AB} \cdot \hat{i} = (3\hat{i} + 0\hat{j} + 9\hat{k}) \cdot \hat{i} = 3 \]

### Step 5: Use the dot product to find the angle
The formula for the dot product is:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) \]
Here, since we are taking the dot product with \( \hat{i} \):
\[ 3 = |\vec{AB}| \cdot 1 \cdot \cos(\theta) \]
Substituting the magnitude:
\[ 3 = (3\sqrt{10}) \cos(\theta) \]
Thus,
\[ \cos(\theta) = \frac{3}{3\sqrt{10}} = \frac{1}{\sqrt{10}} \]

### Step 6: Find the angle \( \theta \)
To find \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{10}}\right) \]

### Final Result
The angle that the vector \( \vec{A} \) makes with the x-axis is:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{10}}\right) \]