x- component of a vector `vecA` is twice of its y-component and `sqrt(2)` times of its z-component. Find out the angle made by the vector from y-axis.

Options

A. `cos^(-1)""((2)/(sqrt(7)))`
B. `cos^(-1)""((1)/(sqrt(7)))`
C. `cos^(-1)""((1)/(sqrt(6)))`
D. `cos^(-1)""((2)/(sqrt(6)))`

Answer

To solve the problem step by step, we will follow the logical sequence as presented in the video transcript.

### Step 1: Define the components of the vector
Let the components of vector \(\vec{A}\) be:
- \(A_x\) = x-component
- \(A_y\) = y-component
- \(A_z\) = z-component

### Step 2: Establish relationships between components
According to the problem:
- The x-component is twice the y-component:
\[ A_x = 2A_y \]
- The x-component is \(\sqrt{2}\) times the z-component:
\[ A_x = \sqrt{2} A_z \]

### Step 3: Express \(A_y\) and \(A_z\) in terms of \(A_y\)
From the first equation, we can express \(A_x\) in terms of \(A_y\):
\[ A_x = 2A_y \]
From the second equation, substituting \(A_x\) gives:
\[ 2A_y = \sqrt{2} A_z \implies A_z = \frac{2A_y}{\sqrt{2}} = \sqrt{2} A_y \]

### Step 4: Substitute the components into the magnitude formula
The magnitude of vector \(\vec{A}\) is given by:
\[ |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \]
Substituting the expressions we found:
\[ |\vec{A}| = \sqrt{(2A_y)^2 + A_y^2 + (\sqrt{2} A_y)^2} \]
Calculating this:
\[ |\vec{A}| = \sqrt{4A_y^2 + A_y^2 + 2A_y^2} = \sqrt{7A_y^2} = A_y \sqrt{7} \]

### Step 5: Find the cosine of the angle with the y-axis
The cosine of the angle \(\theta\) between vector \(\vec{A}\) and the y-axis is given by:
\[ \cos \theta = \frac{A_y}{|\vec{A}|} \]
Substituting the magnitude we found:
\[ \cos \theta = \frac{A_y}{A_y \sqrt{7}} = \frac{1}{\sqrt{7}} \]

### Step 6: Determine the angle \(\theta\)
To find the angle \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{7}}\right) \]

### Final Answer
The angle made by the vector \(\vec{A}\) from the y-axis is:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{7}}\right) \]

---