Engineering Calculus Notes 80

# Engineering Calculus Notes 80 - − 2 y 4 z = 0 so any...

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

68 CHAPTER 1. COORDINATES AND VECTORS Given a linear equation, we can parametrize its locus by fnding three noncollinear points on the locus and using the procedure above. For example, to parametrize the plane given by 3 x 2 y + 4 z = 12 we need to fnd three noncollinear points in this plane. I± we set y = z = 0 , we have x = 4 , and so we can take our basepoint P to be (4 , 0 , 0), or −→ p 0 = 4 −→ ı . To fnd two other points, we could note that i± x = 4 then
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: − 2 y + 4 z = 0 , so any choice with y = 2 z will work, ±or example Q (4 , 2 , 1), or −→ v = −−→ PQ = 2 −→ + −→ k gives one such point. Un±ortunately, any third point given by this scheme will produce −→ w a scalar multiple o± −→ v , so won’t work. However, i± we set x = 0 we have − 2 y + 4 z = 12 , and one solution o± this is y = − 4 , z = 1 ,...
View Full Document

## This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

Ask a homework question - tutors are online