Let x, y, z be the vector, such that `|x|=|y|=|z|=sqrt(2)` and x, y, z make angles of `60^(@)` with each other also, `xtimes(ytimesz)=a and ytimes(ztimesx)=b xtimesy=c,` . The value of z is

Options

A. `(1)/(2)[(b-a)timesc+(a+b)]`
B. `(1)/(2)[(b-a)+ctimes(a+b)]`
C. `[(b-a)timesc+(a+b)]`
D. None of these

Answer

To solve the given problem step by step, we will use the properties of vectors, particularly focusing on the dot product and cross product.

### Given:
- \( |x| = |y| = |z| = \sqrt{2} \)
- The angles between the vectors \( x, y, z \) are \( 60^\circ \).

### Step 1: Calculate the dot products
The dot product of two vectors can be expressed as:
\[ x \cdot y = |x| |y| \cos(60^\circ) \]
Since \( |x| = |y| = \sqrt{2} \) and \( \cos(60^\circ) = \frac{1}{2} \):
\[ x \cdot y = \sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2} = 1 \]
Thus, we have:
\[ x \cdot y = y \cdot z = z \cdot x = 1 \]

### Step 2: Express \( a \) in terms of \( x, y, z \)
Given:
\[ a = x \times (y \times z) \]
Using the vector triple product identity:
\[ x \times (y \times z) = (x \cdot z) y - (x \cdot y) z \]
Substituting the known dot products:
\[ a = (1) y - (1) z = y - z \tag{1} \]

### Step 3: Express \( b \) in terms of \( x, y, z \)
Given:
\[ b = y \times (z \times x) \]
Using the vector triple product identity:
\[ y \times (z \times x) = (y \cdot x) z - (y \cdot z) x \]
Substituting the known dot products:
\[ b = (1) z - (1) x = z - x \tag{2} \]

### Step 4: Add equations (1) and (2)
From equations (1) and (2):
\[ a + b = (y - z) + (z - x) = y - x \]
Thus:
\[ y - x = a + b \tag{3} \]

### Step 5: Express \( c \)
Given:
\[ c = x \times y \]
We will use this later.

### Step 6: Calculate \( x \times c \)
Using the property of the cross product:
\[ x \times (x \times y) = (x \cdot y) x - (x \cdot x) y \]
Substituting the known values:
\[ x \times c = (1) x - (2) y = x - 2y \tag{4} \]

### Step 7: Calculate \( y \times c \)
Similarly:
\[ y \times (y \times x) = (y \cdot x) y - (y \cdot y) x \]
Substituting the known values:
\[ y \times c = (1) y - (2) x = y - 2x \tag{5} \]

### Step 8: Subtract equations (4) and (5)
Subtracting equation (4) from equation (5):
\[ (y - 2x) - (x - 2y) = y - 2x - x + 2y = 3y - 3x \]
Thus:
\[ y - x = \frac{1}{3}(y \times c - x \times c) \tag{6} \]

### Step 9: Substitute back to find \( z \)
From equation (3):
\[ y - x = a + b \]
Substituting this into equation (6):
\[ a + b = \frac{1}{3}(y \times c - x \times c) \]
This implies:
\[ z = \frac{1}{2}(a + b \times c) + b - a \]

### Final Expression for \( z \)
Thus, we find:
\[ z = \frac{1}{2}(a + b \times c) + b - a \]