X- component of `vec(a)` is twice of its Y- component. If the magnitude of the vector is `5sqrt(2)` and it makes an angle of `135^(@)` with z-axis then the components of vector is:

Options

A. `2sqrt(3), sqrt(3), -3`
B. `2sqrt(6), sqrt(6), -6`
C. `2sqrt(5), sqrt(5), -5`
D. None of these

Explanation

To solve the problem step by step, we need to find the components of the vector \( \vec{A} \) given the conditions provided.

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

According to the problem, we know that:
\[ A_x = 2A_y \]

### Step 2: Use the magnitude of the vector
The magnitude of the vector \( \vec{A} \) is given as \( 5\sqrt{2} \). The magnitude can be expressed in terms of its components:
\[ |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \]
Substituting the magnitude:
\[ 5\sqrt{2} = \sqrt{A_x^2 + A_y^2 + A_z^2} \]

### Step 3: Find the z-component using the angle with the z-axis
The angle \( \gamma \) with the z-axis is given as \( 135^\circ \). The z-component can be calculated using:
\[ A_z = |\vec{A}| \cos(\gamma) \]
Substituting the values:
\[ A_z = 5\sqrt{2} \cos(135^\circ) \]
Since \( \cos(135^\circ) = -\frac{1}{\sqrt{2}} \):
\[ A_z = 5\sqrt{2} \left(-\frac{1}{\sqrt{2}}\right) = -5 \]

### Step 4: Substitute \( A_x \) in terms of \( A_y \)
Now we substitute \( A_x = 2A_y \) into the magnitude equation:
\[ 5\sqrt{2} = \sqrt{(2A_y)^2 + A_y^2 + (-5)^2} \]
This simplifies to:
\[ 5\sqrt{2} = \sqrt{4A_y^2 + A_y^2 + 25} \]
\[ 5\sqrt{2} = \sqrt{5A_y^2 + 25} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (5\sqrt{2})^2 = 5A_y^2 + 25 \]
\[ 50 = 5A_y^2 + 25 \]

### Step 6: Solve for \( A_y \)
Rearranging the equation:
\[ 5A_y^2 = 50 - 25 \]
\[ 5A_y^2 = 25 \]
\[ A_y^2 = 5 \]
Taking the square root:
\[ A_y = \sqrt{5} \]

### Step 7: Find \( A_x \)
Now substitute \( A_y \) back to find \( A_x \):
\[ A_x = 2A_y = 2\sqrt{5} \]

### Step 8: Summarize the components
We have found:
- \( A_x = 2\sqrt{5} \)
- \( A_y = \sqrt{5} \)
- \( A_z = -5 \)

Thus, the components of the vector \( \vec{A} \) are:
\[ \vec{A} = (2\sqrt{5}, \sqrt{5}, -5) \]