Let `veca,vecb,vecc` be three non coplanar and `vecd`be a vector which is perpendicular to `veca + vecb + vecc`. If `vecd = xvecb xx vecc + yvecc xx veca + zveca xx vecb` the-

Options

A. xy + yz +zx=0
B. x=y=z
C. `x^(3) =y^(3) =z^(3) = 3xyz`
D. `x +y +z=0`

Answer

To solve the problem step by step, we will analyze the given vectors and their relationships.

### Step 1: Understand the Problem
We have three non-coplanar vectors \( \vec{a}, \vec{b}, \vec{c} \) and a vector \( \vec{d} \) which is defined as:
\[ \vec{d} = x (\vec{b} \times \vec{c}) + y (\vec{c} \times \vec{a}) + z (\vec{a} \times \vec{b}) \]
We know that \( \vec{d} \) is perpendicular to \( \vec{a} + \vec{b} + \vec{c} \).

### Step 2: Use the Perpendicular Condition
Since \( \vec{d} \) is perpendicular to \( \vec{a} + \vec{b} + \vec{c} \), we can express this condition using the dot product:
\[ \vec{d} \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]
This can be expanded as:
\[ \vec{d} \cdot \vec{a} + \vec{d} \cdot \vec{b} + \vec{d} \cdot \vec{c} = 0 \]
Let’s denote this as Equation (1).

### Step 3: Calculate \( \vec{d} \cdot \vec{a} \)
Now, we compute \( \vec{d} \cdot \vec{a} \):
\[ \vec{d} \cdot \vec{a} = \left( x (\vec{b} \times \vec{c}) + y (\vec{c} \times \vec{a}) + z (\vec{a} \times \vec{b}) \right) \cdot \vec{a} \]
Using the properties of the dot product:
- \( \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot \vec{b} \times \vec{c} = \text{scalar triple product} = \vec{a} \cdot (\vec{b} \times \vec{c}) \)
- \( \vec{a} \cdot (\vec{c} \times \vec{a}) = 0 \) (since \( \vec{a} \) is perpendicular to \( \vec{c} \times \vec{a} \))
- \( \vec{a} \cdot (\vec{a} \times \vec{b}) = 0 \) (since \( \vec{a} \) is perpendicular to \( \vec{a} \times \vec{b} \))

Thus, we have:
\[ \vec{d} \cdot \vec{a} = x (\vec{a} \cdot (\vec{b} \times \vec{c})) + 0 + 0 = x (\vec{a} \cdot (\vec{b} \times \vec{c})) \]
Let’s denote this as Equation (2).

### Step 4: Calculate \( \vec{d} \cdot \vec{b} \)
Next, we compute \( \vec{d} \cdot \vec{b} \):
\[ \vec{d} \cdot \vec{b} = \left( x (\vec{b} \times \vec{c}) + y (\vec{c} \times \vec{a}) + z (\vec{a} \times \vec{b}) \right) \cdot \vec{b} \]
Using similar properties:
- \( \vec{b} \cdot (\vec{b} \times \vec{c}) = 0 \)
- \( \vec{b} \cdot (\vec{c} \times \vec{a}) = y (\vec{b} \cdot (\vec{c} \times \vec{a})) \)
- \( \vec{b} \cdot (\vec{a} \times \vec{b}) = 0 \)

Thus, we have:
\[ \vec{d} \cdot \vec{b} = y (\vec{b} \cdot (\vec{c} \times \vec{a})) \]
Let’s denote this as Equation (3).

### Step 5: Calculate \( \vec{d} \cdot \vec{c} \)
Now, we compute \( \vec{d} \cdot \vec{c} \):
\[ \vec{d} \cdot \vec{c} = \left( x (\vec{b} \times \vec{c}) + y (\vec{c} \times \vec{a}) + z (\vec{a} \times \vec{b}) \right) \cdot \vec{c} \]
Using similar properties:
- \( \vec{c} \cdot (\vec{b} \times \vec{c}) = 0 \)
- \( \vec{c} \cdot (\vec{c} \times \vec{a}) = 0 \)
- \( \vec{c} \cdot (\vec{a} \times \vec{b}) = z (\vec{c} \cdot (\vec{a} \times \vec{b})) \)

Thus, we have:
\[ \vec{d} \cdot \vec{c} = z (\vec{c} \cdot (\vec{a} \times \vec{b})) \]
Let’s denote this as Equation (4).

### Step 6: Combine the Results
Now, substituting Equations (2), (3), and (4) into Equation (1):
\[ x (\vec{a} \cdot (\vec{b} \times \vec{c})) + y (\vec{b} \cdot (\vec{c} \times \vec{a})) + z (\vec{c} \cdot (\vec{a} \times \vec{b})) = 0 \]
This can be rewritten as:
\[ x + y + z = 0 \]
This implies that:
\[ X + Y + Z = 0 \]

### Step 7: Conclusion
Since \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar, the scalar triple product \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) is non-zero. Therefore, we conclude that:
\[ X + Y + Z = 0 \]