`veca, vecb, vecc` are unit vectors such that if the angles between the vectors `veca` and `vecb`, `vecb` and `vecc`, `vecc` and `veca` are respectively `pi/6, pi/4` and `pi/3`, then find out the angle the vector `veca` makes with the plane containing `vecb` and `vecc` is

Options

A. (a) `cos^(-1)sqrt(1-sqrt(2//3))`
B. (b) `cos^(-1)sqrt(2-sqrt(3//2))`
C. (c) `cos^(-1)sqrt(sqrt(3//2)-1)`
D. (d) `cos^(-1)sqrt(sqrt(2//3))`

Explanation

To find the angle that the vector \(\vec{a}\) makes with the plane containing the vectors \(\vec{b}\) and \(\vec{c}\), we can follow these steps:

### Step 1: Understand the relationship between the angle and the cross product
Let \(\theta\) be the angle between \(\vec{a}\) and the plane containing \(\vec{b}\) and \(\vec{c}\). The vector \(\vec{b} \times \vec{c}\) is perpendicular to the plane formed by \(\vec{b}\) and \(\vec{c}\). Therefore, the angle between \(\vec{a}\) and \(\vec{b} \times \vec{c}\) can be expressed as:
\[ \text{Angle between } \vec{a} \text{ and } \vec{b} \times \vec{c} = 90^\circ - \theta \]

### Step 2: Use the dot product to express the angles
The dot product of \(\vec{a}\) with the cross product \(\vec{b} \times \vec{c}\) can be expressed as:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = |\vec{a}| |\vec{b} \times \vec{c}| \cos(90^\circ - \theta) = |\vec{a}| |\vec{b} \times \vec{c}| \sin(\theta) \]
Since \(\vec{a}\) is a unit vector, we have \(|\vec{a}| = 1\).

### Step 3: Calculate the dot products of the vectors
We need to find \(\vec{a} \cdot \vec{b}\) and \(\vec{a} \cdot \vec{c}\):
1. The angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\):
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\left(\frac{\pi}{6}\right) = 1 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

2. The angle between \(\vec{a}\) and \(\vec{c}\) is \(\frac{\pi}{3}\):
\[ \vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos\left(\frac{\pi}{3}\right) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2} \]

### Step 4: Calculate \(\vec{b} \cdot \vec{c}\)
The angle between \(\vec{b}\) and \(\vec{c}\) is \(\frac{\pi}{4}\):
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos\left(\frac{\pi}{4}\right) = 1 \cdot 1 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 5: Calculate the magnitude of \(\vec{b} \times \vec{c}\)
The magnitude of the cross product \(\vec{b} \times \vec{c}\) is given by:
\[ |\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin\left(\frac{\pi}{4}\right) = 1 \cdot 1 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 6: Substitute into the expression for \(\vec{a} \cdot (\vec{b} \times \vec{c})\)
Now, we can find \(\vec{a} \cdot (\vec{b} \times \vec{c})\):
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot \left(\vec{b} \times \vec{c}\right) = \left(\vec{a} \cdot \vec{b}\right) \cdot \vec{c} - \left(\vec{a} \cdot \vec{c}\right) \cdot \vec{b} \]
Substituting the values we calculated:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2} \cdot \vec{c} - \frac{1}{2} \cdot \vec{b} \]

### Step 7: Set up the equation
We can now set up the equation:
\[ 1 \cdot \frac{1}{\sqrt{2}} \sin(\theta) = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} - \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \]
Simplifying gives:
\[ \sin(\theta) = \frac{\sqrt{3} - 1}{2} \]

### Step 8: Find \(\theta\)
Finally, we can find \(\theta\) using the inverse sine function:
\[ \theta = \sin^{-1}\left(\frac{\sqrt{3} - 1}{2}\right) \]

### Summary of the Steps
1. Relate the angle \(\theta\) to the cross product.
2. Calculate the dot products \(\vec{a} \cdot \vec{b}\) and \(\vec{a} \cdot \vec{c}\).
3. Find the dot product \(\vec{b} \cdot \vec{c}\).
4. Calculate the magnitude of \(\vec{b} \times \vec{c}\).
5. Substitute into the expression for \(\vec{a} \cdot (\vec{b} \times \vec{c})\).
6. Set up the equation and solve for \(\theta\).