To find the value of \( \alpha \) such that the plane \( 2x - 3y + 6z - 11 = 0 \) 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:
\[
2x - 3y + 6z - 11 = 0
\]
From this equation, we can identify the normal vector \( \mathbf{n} \) of the plane, which is given by the coefficients of \( x \), \( y \), and \( z \):
\[
\mathbf{n} = \langle 2, -3, 6 \rangle
\]
### Step 2: Identify the direction vector of the x-axis
The x-axis can be represented by the direction vector:
\[
\mathbf{b} = \langle 1, 0, 0 \rangle
\]
### Step 3: Use the formula for the sine of the angle between two vectors
The sine of the angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be calculated using the formula:
\[
\sin(\theta) = \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}| |\mathbf{b}|}
\]
However, since we are dealing with the angle between the normal vector and the x-axis, we can use the dot product to find the cosine of the angle and then relate it to sine.
### Step 4: Calculate the magnitudes of the vectors
First, we find the magnitude of the normal vector \( \mathbf{n} \):
\[
|\mathbf{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7
\]
Next, we find the magnitude of the direction vector \( \mathbf{b} \):
\[
|\mathbf{b}| = \sqrt{1^2 + 0^2 + 0^2} = 1
\]
### Step 5: Calculate the dot product of the vectors
Now, we calculate the dot product \( \mathbf{b} \cdot \mathbf{n} \):
\[
\mathbf{b} \cdot \mathbf{n} = \langle 1, 0, 0 \rangle \cdot \langle 2, -3, 6 \rangle = 1 \cdot 2 + 0 \cdot (-3) + 0 \cdot 6 = 2
\]
### Step 6: Relate sine and cosine
Using the relationship between sine and cosine:
\[
\sin(\theta) = \sqrt{1 - \cos^2(\theta)}
\]
We can find \( \cos(\theta) \) using the dot product:
\[
\cos(\theta) = \frac{\mathbf{b} \cdot \mathbf{n}}{|\mathbf{b}| |\mathbf{n}|} = \frac{2}{1 \cdot 7} = \frac{2}{7}
\]
Now, we find \( \sin(\theta) \):
\[
\sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \left(\frac{2}{7}\right)^2 = 1 - \frac{4}{49} = \frac{49 - 4}{49} = \frac{45}{49}
\]
Thus,
\[
\sin(\theta) = \sqrt{\frac{45}{49}} = \frac{\sqrt{45}}{7} = \frac{3\sqrt{5}}{7}
\]
### Step 7: Find the value of \( \alpha \)
Since \( \sin(\theta) = \alpha \), we have:
\[
\alpha = \frac{3\sqrt{5}}{7}
\]
### Final Answer
The value of \( \alpha \) is:
\[
\alpha = \frac{3\sqrt{5}}{7}
\]
