lab2_sol_f_08revised2

lab2_sol_f_08revised2 - Math 115 - Lab 2 - Fall 2008....

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

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

Unformatted text preview: Math 115 - Lab 2 - Fall 2008. Topics: Projections and planes Distance of a point to a line or a plane in R 3 Solutions and elementary operations to linear systems, Gaussian elimination, Homogeneous equations You are to provide full solutions to the following problems. You are allowed, and encouraged to collaborate with your classmates, use your notes and textbook and ask the TA for guidance. Direct copying of solutions is not encouraged, nor is it allowed or ethical. Last name: First name: Student number: Math 115 - Lab 2 - Fall 2008. Student number: 1. Let y = (- 1 ,- 2 , 2) and x = (1 ,- 2 , 1). a) Find the projection proj x y of the vector y onto the vector x . Solution: proj x y = < x , y > x bardbl x bardbl 2 = < (- 1 ,- 2 , 2) , (1 ,- 2 , 1) > (1 ,- 2 , 1) 6 = 5 6 (1 ,- 2 , 1) b) Find the point on the line L : x = (2 , 2 ,- 1) + t (1 ,- 2 , 1) which is closest to y = (1 , , 1). Solution: Let a = (2 , 2 ,- 1) and d = (1 ,- 2 , 1) (the direction vector of the line L ). We want the point proj d ( y- a ) + a . proj d ( y- a ) + a = < (1 , , 1)- (2 , 2 ,- 1) , (1 ,- 2 , 1) > (1 ,- 2 , 1) bardbl (1 ,- 2 , 1) bardbl 2 + (2 , 2 ,- 1) = < (- 1 ,- 2 , 2) , (1 ,- 2 , 1) > (1 ,- 2 , 1) 6 + (2 , 2 ,- 1) = 5 6 (1 ,- 2 , 1) + (2 , 2 ,- 1) = parenleftbigg 17 6 , 1 3 ,- 1 6 parenrightbigg c) Find the minimum distance of the point y = (1 , , 1) to the line L : x = (2 , 2 ,- 1)+ t (1 ,- 2 , 1)....
View Full Document

Page1 / 8

lab2_sol_f_08revised2 - Math 115 - Lab 2 - Fall 2008....

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online