Let `bar a,bar b,bar c` be three non-coplanar vectors and `bar d` be a non-zero vector, which is perpendicularto `bar a+bar b+bar c`. Now, if `bar d= (sin x)(bar a xx bar b) + (cos y)(bar b xx bar c) + 2(bar c xx bar a)` then minimum value of `x^2+y^2` is equal to

Options

A. `pi^(2)`
B. `(pi^(2))/(2)`
C. `(pi^(2))/(4)`
D. `(5pi^(2))/(4)`

Answer

To solve the problem, we need to find the minimum value of \( x^2 + y^2 \) given the conditions of the vectors. Let's break it down step by step.

### Step 1: Understanding the Problem
We have three non-coplanar vectors \( \vec{a}, \vec{b}, \vec{c} \) and a vector \( \vec{d} \) that is perpendicular to \( \vec{a} + \vec{b} + \vec{c} \). The vector \( \vec{d} \) is expressed as:
\[ \vec{d} = \sin x (\vec{a} \times \vec{b}) + \cos y (\vec{b} \times \vec{c}) + 2 (\vec{c} \times \vec{a}) \]

### Step 2: Setting Up the Perpendicular Condition
Since \( \vec{d} \) is perpendicular to \( \vec{a} + \vec{b} + \vec{c} \), we can write:
\[ \vec{d} \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

### Step 3: Substituting \( \vec{d} \)
Substituting the expression for \( \vec{d} \):
\[ \left( \sin x (\vec{a} \times \vec{b}) + \cos y (\vec{b} \times \vec{c}) + 2 (\vec{c} \times \vec{a}) \right) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

### Step 4: Expanding the Dot Product
Using the properties of the dot product and cross product, we can evaluate each term:
1. \( \sin x (\vec{a} \times \vec{b}) \cdot \vec{a} = 0 \) (since \( \vec{a} \times \vec{b} \) is perpendicular to \( \vec{a} \))
2. \( \sin x (\vec{a} \times \vec{b}) \cdot \vec{b} = 0 \) (since \( \vec{a} \times \vec{b} \) is perpendicular to \( \vec{b} \))
3. \( \sin x (\vec{a} \times \vec{b}) \cdot \vec{c} = \sin x (\vec{a} \vec{b} \vec{c}) \)

Similarly for the other terms:
- \( \cos y (\vec{b} \times \vec{c}) \cdot \vec{a} = \cos y (\vec{b} \vec{c} \vec{a}) \)
- \( 2 (\vec{c} \times \vec{a}) \cdot \vec{b} = 2 (\vec{c} \vec{a} \vec{b}) \)

### Step 5: Combining the Results
Combining these results gives us:
\[ \sin x (\vec{a} \vec{b} \vec{c}) + \cos y (\vec{b} \vec{c} \vec{a}) + 2 (\vec{c} \vec{a} \vec{b}) = 0 \]

### Step 6: Solving for \( \sin x + \cos y \)
Rearranging gives:
\[ \sin x + \cos y + 2 = 0 \]
Thus,
\[ \sin x + \cos y = -2 \]

### Step 7: Finding Minimum Values
The maximum value of \( \sin x \) is 1 and the maximum value of \( \cos y \) is also 1. Therefore, the only way for their sum to be -2 is if both are at their minimum:
- \( \sin x = -1 \) when \( x = -\frac{\pi}{2} \)
- \( \cos y = -1 \) when \( y = \pi \)

### Step 8: Calculating \( x^2 + y^2 \)
Now we calculate:
\[ x^2 + y^2 = \left(-\frac{\pi}{2}\right)^2 + (\pi)^2 = \frac{\pi^2}{4} + \pi^2 = \frac{\pi^2}{4} + \frac{4\pi^2}{4} = \frac{5\pi^2}{4} \]

### Final Answer
Thus, the minimum value of \( x^2 + y^2 \) is:
\[ \frac{5\pi^2}{4} \]