The plane 2x – 3y + 6z – 11 = 0 makes an angle sin-1 (α) with .e-axis. The value of a is equal to

The correct option is 2/7.

Let us understand how to find the angle between a given plane and the x-axis step-by-step.

First, let the equation of the plane be:
$$2x-3y+6z-11=0$$

The normal vector n to this plane can be determined from the coefficients of $x$, $y$, and $z$:
$$\overrightarrow{n}=2\hat{i}-3\hat{j}+6\hat{k}$$

Next, we consider the x-axis. The unit vector along the x-axis is:
$$\overrightarrow{u}=1\hat{i}+0\hat{j}+0\hat{k}=\hat{i}$$

If $\theta$ is the angle between the plane and the x-axis, then the angle between the normal vector of the plane and the x-axis is ${90}^{\circ }-\theta$. Therefore, the relationship is given by:
$$sin\left(\theta \right)=\frac{|\overrightarrow{n}\cdot \overrightarrow{u}|}{\left|\overrightarrow{n}\right|\left|\overrightarrow{u}\right|}$$

Let us compute the dot product of $\overrightarrow{n}$ and $\overrightarrow{u}$:
$$\overrightarrow{n}\cdot \overrightarrow{u}=\left(2\right)\left(1\right)+(-3)\left(0\right)+\left(6\right)\left(0\right)=2$$

Now, let us calculate the magnitude of the normal vector $\overrightarrow{n}$:
$$\left|\overrightarrow{n}\right|=\sqrt{2^2+{(-3)}^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$$

Since the magnitude of the unit vector $\overrightarrow{u}$ is $1$, we substitute these values into the formula:
$$sin\left(\theta \right)=\frac{2}{7\cdot 1}=\frac{2}{7}$$

Thus, the angle $\theta$ is:
$$\theta ={sin}^{-1}\left(\frac{2}{7}\right)$$

Comparing this with the given format ${sin}^{-1}\left(\alpha \right)$, we find that:
$$\alpha =\frac{2}{7}$$