Linear Algebra Solutions 91

Linear Algebra Solutions 91 - or x−2 y−1 z−4 −1...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: or x−2 y−1 z−4 −1 −2 −2 2 −2 −3 = 2x − 7y + 6z − 21 = 0. 16. Non–parallel lines L and M in three dimensional space are given by equations P = A + sX, Q = B + tY. E (i) Suppose P Q is orthogonal to both X and Y . Now E E P Q= Q − P = (B + tY ) − (A + sX ) =AB +tY − sX. Hence E (AB +tY + sX ) · X = 0 E (AB +tY + sX ) · Y = 0. More explicitly E t(Y · X ) − s(X · X ) = − AB ·X E t(Y · Y ) − s(X · Y ) = − AB ·Y. However the coefficient determinant of this system of linear equations in t and s is equal to Y · X −X · X Y · Y −X · Y = −(X · Y )2 + (X · X )(Y · Y ) = ||X × Y ||2 = 0, as X = 0, Y = 0 and X and Y are not proportional (L and M are not parallel). (ii) P and Q can be viewed as the projections of C and D onto the line P Q, where C and D are arbitrary points on the lines L and M, respectively. Hence by equation (8.14) of Theorem 8.5.3, we have P Q ≤ CD. Finally we derive a useful formula for P Q. Again by Theorem 8.5.3 E E E | AB · P Q | ˆ = | AB ·n|, PQ = PQ 94 ...
View Full Document

This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

Ask a homework question - tutors are online