A, B, C and D are four points in space. Using vector methods, prove that `AC^(2)+BD^(2)+AC^(2)+BC^(2)geAB^(2)+CD^(2)` what is the implication of the sign of equality.

Options

A. `AB^2+CD^2`
B. `(1)/(AB^2)-(1)/(CD^2)`
C. `(1)/(CD^2)-(1)/(AB^2)`
D. None of these

Answer

To prove the inequality \( AC^2 + BD^2 + AC^2 + BC^2 \geq AB^2 + CD^2 \) using vector methods, we can follow these steps:

### Step 1: Define Position Vectors
Let the position vectors of points A, B, C, and D be represented as:
- \( \vec{a} \) for point A
- \( \vec{b} \) for point B
- \( \vec{c} \) for point C
- \( \vec{d} \) for point D

### Step 2: Express Squared Distances
We can express the squared distances in terms of the position vectors:
- \( AC^2 = |\vec{c} - \vec{a}|^2 = (\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = |\vec{c}|^2 - 2\vec{a} \cdot \vec{c} + |\vec{a}|^2 \)
- \( BD^2 = |\vec{d} - \vec{b}|^2 = (\vec{d} - \vec{b}) \cdot (\vec{d} - \vec{b}) = |\vec{d}|^2 - 2\vec{b} \cdot \vec{d} + |\vec{b}|^2 \)
- \( BC^2 = |\vec{c} - \vec{b}|^2 = |\vec{c}|^2 - 2\vec{b} \cdot \vec{c} + |\vec{b}|^2 \)
- \( AB^2 = |\vec{b} - \vec{a}|^2 = |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{a}|^2 \)
- \( CD^2 = |\vec{d} - \vec{c}|^2 = |\vec{d}|^2 - 2\vec{c} \cdot \vec{d} + |\vec{c}|^2 \)

### Step 3: Substitute and Rearrange
Now, substitute these expressions into the inequality:
\[ AC^2 + BD^2 + AC^2 + BC^2 \geq AB^2 + CD^2 \]
This becomes:
\[ \left( |\vec{c}|^2 - 2\vec{a} \cdot \vec{c} + |\vec{a}|^2 \right) + \left( |\vec{d}|^2 - 2\vec{b} \cdot \vec{d} + |\vec{b}|^2 \right) + \left( |\vec{c}|^2 - 2\vec{b} \cdot \vec{c} + |\vec{b}|^2 \right) \geq \left( |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{a}|^2 \right) + \left( |\vec{d}|^2 - 2\vec{c} \cdot \vec{d} + |\vec{c}|^2 \right) \]

### Step 4: Combine Like Terms
Combine all the terms on both sides:
- Left-hand side (LHS):
\[ 2|\vec{c}|^2 + 2|\vec{b}|^2 + |\vec{d}|^2 - 2\vec{a} \cdot \vec{c} - 2\vec{b} \cdot \vec{d} - 2\vec{b} \cdot \vec{c} \]
- Right-hand side (RHS):
\[ |\vec{b}|^2 + |\vec{a}|^2 + |\vec{d}|^2 + |\vec{c}|^2 - 2\vec{a} \cdot \vec{b} - 2\vec{c} \cdot \vec{d} \]

### Step 5: Rearranging the Inequality
Now rearranging gives:
\[ |\vec{c}|^2 + |\vec{b}|^2 - 2\vec{b} \cdot \vec{c} - 2\vec{a} \cdot \vec{c} + 2\vec{a} \cdot \vec{b} + 2\vec{c} \cdot \vec{d} \geq 0 \]

### Step 6: Recognizing the Square
This expression can be recognized as a squared term:
\[ |\vec{a} + \vec{b} - \vec{c} - \vec{d}|^2 \geq 0 \]
This is always true since the square of a vector's magnitude is always non-negative.

### Step 7: Implication of Equality
The equality holds when:
\[ \vec{a} + \vec{b} = \vec{c} + \vec{d} \]
This implies that the points A, B, C, and D are coplanar and lie on a straight line.