3D Geometry
Equidistant Points
Grade 12
Question:
<p>The vertices of △ABC are A(2, 0, 0), B(0, 1, 0), C(0, 0, 2). Its orthocentre is H and circumcentre is S. P is a point equidistant from A, B, C and the origin O. PA is equal to:</p>
<p>(a) 1</p>
<p>(b) \(\sqrt{2}\)</p>
<p>(c) \(\frac{3}{2}\)</p>
<p>(d) \(\frac{\sqrt{3}}{2}\)</p>
Step-by-Step Solution
Key Concept: P is equidistant from four points (A, B, C, O), so it must satisfy PA = PB = PC = PO simultaneously.
P is equidistant from A, B, C and O. This means P lies on the perpendicular bisector planes. Let P = (x, y, z). Then PA = PB = PC = PO. Solving these equations gives PA = 3/2.
Correct Answer: c