If `veca, vecb` and `vecc` are three units vectors equally inclined to each other at an angle `alpha`. Then the angle between `veca` and plane of `vecb` and `vecc` is

Options

A. `theta=cos^(-1)(cosalpha)/(cosalpha/2)`
B. `theta=sin^(-1)(cosalpha)/(cosalpha/2)`
C. `theta=cos^(-1)(sinalpha/2)/(sinalpha)`
D. `theta=sin^(-1)(sinalpha/2)/(sinalpha)`

Explanation

To find the angle between the unit vector \(\vec{a}\) and the plane formed by the unit vectors \(\vec{b}\) and \(\vec{c}\), which are equally inclined to each other at an angle \(\alpha\), we can follow these steps:

### Step 1: Understand the Geometry
We have three unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) that are equally inclined to each other at an angle \(\alpha\). This means that the angle between any two of these vectors is \(\alpha\).

### Step 2: Define the Angle
Let \(\theta\) be the angle between the vector \(\vec{a}\) and the plane formed by the vectors \(\vec{b}\) and \(\vec{c}\).

### Step 3: Use the Dot Product
The cosine of the angle \(\theta\) can be expressed using the dot product:
\[ \cos \theta = \frac{\vec{a} \cdot (\vec{b} + \vec{c})}{|\vec{a}| |\vec{b} + \vec{c}|} \]
Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are unit vectors, we have \(|\vec{a}| = 1\).

### Step 4: Calculate \(|\vec{b} + \vec{c}|\)
To find \(|\vec{b} + \vec{c}|\), we use the formula:
\[ |\vec{b} + \vec{c}| = \sqrt{|\vec{b}|^2 + |\vec{c}|^2 + 2 \vec{b} \cdot \vec{c}} \]
Since \(\vec{b}\) and \(\vec{c}\) are unit vectors:
\[ |\vec{b}|^2 = 1, \quad |\vec{c}|^2 = 1, \quad \text{and} \quad \vec{b} \cdot \vec{c} = \cos \alpha \]
Thus,
\[ |\vec{b} + \vec{c}| = \sqrt{1 + 1 + 2 \cos \alpha} = \sqrt{2(1 + \cos \alpha)} = \sqrt{2(2 \cos^2(\alpha/2))} = 2 \cos(\alpha/2) \]

### Step 5: Calculate \(\vec{a} \cdot (\vec{b} + \vec{c})\)
Now we compute \(\vec{a} \cdot (\vec{b} + \vec{c})\):
\[ \vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = \cos \alpha + \cos \alpha = 2 \cos \alpha \]

### Step 6: Substitute into the Cosine Formula
Now substituting back into the cosine formula:
\[ \cos \theta = \frac{2 \cos \alpha}{1 \cdot 2 \cos(\alpha/2)} = \frac{\cos \alpha}{\cos(\alpha/2)} \]

### Step 7: Find \(\theta\)
Finally, we can express \(\theta\) as:
\[ \theta = \cos^{-1}\left(\frac{\cos \alpha}{\cos(\alpha/2)}\right) \]

### Final Result
Thus, the angle between the vector \(\vec{a}\) and the plane formed by the vectors \(\vec{b}\) and \(\vec{c}\) is:
\[ \theta = \cos^{-1}\left(\frac{\cos \alpha}{\cos(\alpha/2)}\right) \]

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