3D Geometry
Intercept form of a plane
Grade 12
Question:
<p>Two systems of rectangular axes have the same origin. If a plane cuts them at distances \(a, b, c\) and \(a', b', c'\) from the origin, then</p>
<p>\(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + \dfrac{1}{a'^2} + \dfrac{1}{b'^2} + \dfrac{1}{c'^2} = 0\)</p>
<p>\(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} - \dfrac{1}{a'^2} + \dfrac{1}{b'^2} - \dfrac{1}{c'^2} = 0\)</p>
<p>\(\dfrac{1}{a^2} - \dfrac{1}{b^2} - \dfrac{1}{c^2} + \dfrac{1}{a'^2} - \dfrac{1}{b'^2} - \dfrac{1}{c'^2} = 0\)</p>
<p>\(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} - \dfrac{1}{a'^2} - \dfrac{1}{b'^2} - \dfrac{1}{c'^2} = 0\)</p>
Step-by-Step Solution
Key Concept: The intercept form of a plane's equation remains invariant under rotation of coordinate axes about the origin. If a plane makes intercepts on two different coordinate systems with the same origin, the sum of reciprocals of intercepts follows a geometric relationship: 1/a² + 1/b² + 1/c² = 1/a'² + 1/b'² + 1/c'² (this represents the constant perpendicular distance from origin to the plane).
Step 1: Consider a plane cutting two coordinate systems (with same origin O) at intercepts a, b, c and a', b', c' respectively. Step 2: The equation of the plane in the first system is: x/a + y/b + z/c = 1 Step 3: The perpendicular distance from origin to the plane is: d = 1/√(1/a^2 + 1/b^2 + 1/c^2) Step 4: In the second coordinate system, the same plane's equation is: x'/a' + y'/b' + z'/c' = 1 Step 5: The perpendicular distance from origin remains the same (origin is fixed): d = 1/√(1/a'^2 + 1/b'^2 + 1/c'^2) Step 6: Since both expressions equal d, we have: 1/a^2 + 1/b^2 + 1/c^2 = 1/a'^2 + 1/b'^2 + 1/c'^2 ∴ Answer: D (or the equivalent relationship given in option D)
Correct Answer: D