<p>If OABC is a tetrahedron such that $OA^2 + BC^2 = OB^2 + CA^2 = OC^2 + AB^2$, then which of the following is not true?</p>
Step-by-Step Solution
Key Concept: Use the condition $|\mathbf{a}|^2 + |\mathbf{b} - \mathbf{c}|^2 = |\mathbf{b}|^2 + |\mathbf{c} - \mathbf{a}|^2$ to derive dot product relationships showing perpendicularity.
Step 1: Let $OA = \mathbf{a}$, $OB = \mathbf{b}$, $OC = \mathbf{c}$ Step 2: From the given conditions: $|\mathbf{a}|^2 + (\mathbf{b} - \mathbf{c}) \cdot (\mathbf{b} - \mathbf{c}) = |\mathbf{b}|^2 + (\mathbf{c} - \mathbf{a}) \cdot (\mathbf{c} - \mathbf{a})$ Step 3: Simplifying: $\mathbf{a} \cdot \mathbf{a} - 2\mathbf{b} \cdot \mathbf{c} = \mathbf{b} \cdot \mathbf{b} - 2\mathbf{c} \cdot \mathbf{a}$ $\Rightarrow \mathbf{c} \cdot (\mathbf{b} - \mathbf{a}) = 0$ $\Rightarrow \mathbf{BA} \cdot \mathbf{OC} = 0$ This shows $OC \perp AB$, and similarly we can prove $OA \perp BC$ and $OB \perp AC$. ∴ Answer is (d), as $AB \perp AC$ is not necessarily true from the given conditions.
Correct Answer: D