Quiz 1 Solutions

1 k 0 1 0 0 4 0 0 0 1 5 0 0 1 0 0 5 2 let

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: 0, 4, 0 + 0, 0, 1 . +5 0, 0, 1 0, 0, 5 (2) Let a be the answer for (1). Thus, a= 3, 4, 5 . Then 3, 4, 5 = a = 32 + 42 + 52 = (3) √ √ 50 =5 0, 1, 0 · j ·a = √ 2. 3, 4, 5 = 0 ·3 + 1 ·4 + 0 ·5 = 4. [III] (6pts) For a= proja b = 1, 2, 1 a ·b a 2 and b= 5, −1, 0 , a 1, 2, 1 · = 5, −1, 0 1, 2, 1 2 1, 2, 1 = = 1 · 5 + 2 · (−1) + 1 · 0 12 + 22 + 12 1 2 1, 2, 1 = 2 1, 2, 1 1 , 1, 2 1 2 . [IV] (8pts) (1) We set x−1 2 y−2 4 = = z−4 6 = t. We may solve this equation for x, y and z : = x, y, z (2) 2t + 1, 4t + 2, 6t + 4 . The line in (1) is contained i...
View Full Document

This note was uploaded on 11/06/2013 for the course MATH 223 taught by Professor Staff during the Fall '08 term at Kansas.

Ask a homework question - tutors are online