Angle made by vector `sqrt(3)hat(i)+sqrt(2)hat(j)-2hat(k)` with `-ve y-` axis is `:`

Options

A. `cos^(-1)((sqrt(2))/(3))`
B. `cos^(-1)(-(sqrt(2))/(3))`
C. `pi-cos^(-1)((sqrt(2))/(3))`
D. `pi-cos^(-1)(-(sqrt(2))/(3))`

Explanation

To find the angle made by the vector \( \sqrt{3} \hat{i} + \sqrt{2} \hat{j} - 2 \hat{k} \) with the negative y-axis, we can follow these steps:

### Step 1: Identify the vectors
Let the given vector be:
\[ \mathbf{A} = \sqrt{3} \hat{i} + \sqrt{2} \hat{j} - 2 \hat{k} \]
The unit vector along the negative y-axis is:
\[ \mathbf{B} = -\hat{j} \]

### Step 2: Use the formula for the angle between two vectors
The angle \( \theta \) between two vectors \( \mathbf{A} \) and \( \mathbf{B} \) can be found using the formula:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]

### Step 3: Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)
The dot product \( \mathbf{A} \cdot \mathbf{B} \) is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (\sqrt{3} \hat{i} + \sqrt{2} \hat{j} - 2 \hat{k}) \cdot (-\hat{j}) = 0 \cdot \sqrt{3} + \sqrt{2} \cdot (-1) + 0 \cdot (-2) = -\sqrt{2} \]

### Step 4: Calculate the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \)
The magnitude of \( \mathbf{A} \) is:
\[ |\mathbf{A}| = \sqrt{(\sqrt{3})^2 + (\sqrt{2})^2 + (-2)^2} = \sqrt{3 + 2 + 4} = \sqrt{9} = 3 \]
The magnitude of \( \mathbf{B} \) is:
\[ |\mathbf{B}| = |-1| = 1 \]

### Step 5: Substitute into the cosine formula
Now substituting into the cosine formula:
\[ \cos \theta = \frac{-\sqrt{2}}{3 \cdot 1} = \frac{-\sqrt{2}}{3} \]

### Step 6: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(-\frac{\sqrt{2}}{3}\right) \]
This can also be expressed as:
\[ \theta = \pi - \cos^{-1}\left(\frac{\sqrt{2}}{3}\right) \]

### Final Answer
Thus, the angle made by the vector \( \sqrt{3} \hat{i} + \sqrt{2} \hat{j} - 2 \hat{k} \) with the negative y-axis is:
\[ \theta = \cos^{-1}\left(-\frac{\sqrt{2}}{3}\right) \quad \text{or} \quad \theta = \pi - \cos^{-1}\left(\frac{\sqrt{2}}{3}\right) \]