Homework 3 Solution

Then the answer is true see review of lectures xiv

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Unformatted text preview: 2 ′ X + b2 ′ Y + c2 ′ Z + d2 ′ W = 0 defines a line ℓ′ ⊆ P3 (X :Y :Z :W ) . Suppose P l ℓ = P l ℓ′ . Then ℓ = ℓ′ . ” • The answer is true . See “Review of Lectures – XIV”, page 2, Proposition B. True or false : (5) “For an arbitrary point ∈ P5 (P0 :P1 :P2 :P3 :P4 :P5 ) , P0 : P1 : P2 : P3 : P4 : P5 there exists a line ℓ ⊆ P3 (X :Y :Z :W ) such that P0 : P1 : P2 : P3 : P4 : P5 = Pl ℓ . ” • The answer is false . See “Review of Lectures – XIV”, page 7, Example 1 for counterexample. (6) If your answer in (5) is false, then provide the exact ‘if and only if’ condition = necessary and sufficient condition for a point in P 5 (P0 :P1 :P2 :P3 :P4 :P5 ) to be equal to P l ℓ in terms of P0 , P1 , P2 , P3 , P4 and P5 . P0 : P1 : P2 : P3 : P4 : P5 for some line ℓ ⊆ P3 (X :Y :Z :W ) , The answer is a one-line equation. Also, give the name of this equation. • The answer is ( ∗) P0 P5 − P1 P4 + P2 P3 = 0 . The name of this equation is Pl¨cker relation . See “Review of Lectures – XIV”, u page 28, Corollary 2. 3...
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This note was uploaded on 11/06/2013 for the course MATH 660 taught by Professor Yasuyukikachi during the Fall '13 term at Kansas.

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