The angle subtended by vector `vecA = 4 hati + 3hatj + 12hatk` with the x-axis is :

Options

A. `sin^(-1) (3/13)`
B. `sin^(-1) (4/13)`
C. `cos^(-1)(4/13)`
D. `cos^(-1)(3/13)`

Answer

To find the angle subtended by the vector \(\vec{A} = 4\hat{i} + 3\hat{j} + 12\hat{k}\) with the x-axis, we can follow these steps:

### Step 1: Identify the vectors
We have:
- Vector \(\vec{A} = 4\hat{i} + 3\hat{j} + 12\hat{k}\)
- Vector along the x-axis, \(\vec{B} = \hat{i}\)

### Step 2: Calculate the dot product \(\vec{A} \cdot \vec{B}\)
The dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:
\[ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z \]
For our vectors:
- \(A_x = 4\), \(A_y = 3\), \(A_z = 12\)
- \(B_x = 1\), \(B_y = 0\), \(B_z = 0\)

Calculating the dot product:
\[ \vec{A} \cdot \vec{B} = (4)(1) + (3)(0) + (12)(0) = 4 + 0 + 0 = 4 \]

### Step 3: Calculate the magnitudes of \(\vec{A}\) and \(\vec{B}\)
The magnitude of vector \(\vec{A}\) is calculated as:
\[ |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} = \sqrt{4^2 + 3^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 \]

The magnitude of vector \(\vec{B}\) is:
\[ |\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2} = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1 \]

### Step 4: Use the dot product to find the cosine of the angle
The cosine of the angle \(\theta\) between the two vectors is given by:
\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]
Substituting the values we found:
\[ \cos \theta = \frac{4}{13 \cdot 1} = \frac{4}{13} \]

### Step 5: Find the angle \(\theta\)
To find the angle \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{4}{13}\right) \]

Thus, the angle subtended by the vector \(\vec{A}\) with the x-axis is \(\theta = \cos^{-1}\left(\frac{4}{13}\right)\).