If the plane `2x-3y+6z=11` makes an angle `sin^(-1) (alpha)` with the x-axis, then the value of `alpha` is

Options

A. `(sqrt3)/(2)`
B. `(2)/( sqrt3)`
C. `(1)/(7)`
D. `(2)/(7)`

Answer

To find the value of \( \alpha \) such that the plane \( 2x - 3y + 6z = 11 \) makes an angle \( \sin^{-1}(\alpha) \) with the x-axis, we can follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane is given as \( 2x - 3y + 6z = 11 \). The normal vector \( \mathbf{n} \) to the plane can be derived from the coefficients of \( x, y, z \) in the equation. Thus, we have:
\[ \mathbf{n} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k} \]

### Step 2: Identify the direction vector along the x-axis
The direction vector along the x-axis can be represented as:
\[ \mathbf{b} = \mathbf{i} + 0\mathbf{j} + 0\mathbf{k} \]

### Step 3: Calculate the magnitudes of the vectors
Now, we need to find the magnitudes of the vectors \( \mathbf{n} \) and \( \mathbf{b} \).

For \( \mathbf{n} \):
\[ |\mathbf{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

For \( \mathbf{b} \):
\[ |\mathbf{b}| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1 \]

### Step 4: Find the dot product of the vectors
Next, we calculate the dot product \( \mathbf{n} \cdot \mathbf{b} \):
\[ \mathbf{n} \cdot \mathbf{b} = (2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) \cdot (\mathbf{i}) = 2 \cdot 1 + (-3) \cdot 0 + 6 \cdot 0 = 2 \]

### Step 5: Use the formula for the sine of the angle
The sine of the angle \( \theta \) between the normal vector and the direction vector is given by:
\[ \sin(\theta) = \frac{\mathbf{n} \cdot \mathbf{b}}{|\mathbf{n}| \cdot |\mathbf{b}|} \]
Substituting the values we calculated:
\[ \sin(\theta) = \frac{2}{7 \cdot 1} = \frac{2}{7} \]

### Step 6: Relate sine of the angle to alpha
Since the problem states that \( \sin(\theta) = \sin^{-1}(\alpha) \), we can equate:
\[ \sin^{-1}(\alpha) = \frac{2}{7} \]
Thus, we find:
\[ \alpha = \frac{2}{7} \]

### Final Answer
The value of \( \alpha \) is \( \frac{2}{7} \).

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