Homework 3 Solution

# Then the answer is true see review of lectures xiv

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 11/06/2013 for the course MATH 660 taught by Professor Yasuyukikachi during the Fall '13 term at Kansas.

Ask a homework question - tutors are online