3D Geometry
Locus in 3D
Grade 12
Question:
<p>Let <em>P</em>(<em>x</em>, <em>y</em>, <em>z</em>) be any point on the locus, then the distances from the six faces are |<em>x</em> + 1|, |<em>x</em> − 1|, |<em>y</em> + 1|, |<em>y</em> − 1|, |<em>z</em> + 1| and |<em>z</em> − 1|. According to the given condition, find the locus of <em>P</em>.<br>\(|x+1|^2 + |x-1|^2 + |y+1|^2 + |y-1|^2 + |z+1|^2 + |z-1|^2 = 10\)</p>
<p>\(x^2 + y^2 + z^2 = 1\)</p>
<p>\(x^2 + y^2 + z^2 = 2\)</p>
<p>\(x^2 + y^2 + z^2 = 3\)</p>
<p>\(x^2 + y^2 + z^2 = 4\)</p>
Step-by-Step Solution
Key Concept: Expand each squared distance term |a±b|² = a² ± 2ab + b² and simplify the sum by collecting like terms. The cross terms cancel while squared terms combine to form a clean equation of a sphere.
Step 1: Expand each squared absolute value term. |x+1|^2 + |x−1|^2 = (x+1)^2 + (x−1)^2 = x^2 + 2x + 1 + x^2 − 2x + 1 = 2x^2 + 2 |y+1|^2 + |y−1|^2 = (y+1)^2 + (y−1)^2 = y^2 + 2y + 1 + y^2 − 2y + 1 = 2y^2 + 2 |z+1|^2 + |z−1|^2 = (z+1)^2 + (z−1)^2 = z^2 + 2z + 1 + z^2 − 2z + 1 = 2z^2 + 2 Step 2: Add all six terms together. (2x^2 + 2) + (2y^2 + 2) + (2z^2 + 2) = 10 2x^2 + 2y^2 + 2z^2 + 6 = 10 Step 3: Simplify to standard form. 2x^2 + 2y^2 + 2z^2 = 4 x^2 + y^2 + z^2 = 2 ∴ The locus is a sphere with center at origin (0,0,0) and radius √2 . Answer: B
Correct Answer: B