20Chw1

# 20Chw1 - . Solution: Since u · v = h a,a,b i · h 1 ,-1 ,...

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

MATH 20C graded homework 1. (18+2 points) 12.2 # 50 (6 points): Determine if the lines r 1 ( t ) = h 2 , 1 , 1 i + t h- 4 , 0 , 1 i and r 2 ( s ) = h- 4 , 1 , 5 i + s h 2 , 1 , - 2 i intersect and, if so, ﬁnd the point of intersection. Solution: If the lines intersect, then there are parameters t and s which satisfy the system of equations 2 - 4 t = - 4 + 2 s (1) 1 = 1 + s (2) 1 + t = 5 - 2 s. (3) Equation (2) implies s = 0. Equation (3) then becomes 1 + t = 5, so t = 4. However, the values s = 0, t = 4 do not satisfy equation (1): 2 - 16 6 = - 4 + 0 . Therefore, this system of equations has no solution, and the lines do not intersect. 12.3 # 48 (6 points): Find the projection proj v ( u ) where u = h a,a,b i and v = i - j
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . Solution: Since u · v = h a,a,b i · h 1 ,-1 , i = a-a = 0 , then u and v are perpendicular to each other, and the projection of u onto v is . 12.4 # 34 (6 points): Find the two unit vectors orthogonal to both a = h 3 , 1 , 1 i and b = h-1 , 2 , 1 i . Solution : The cross product is a vector perpendicular to both a and b : a × b = h 1-2 ,-1-3 , 6-(-1) i = h-1 ,-4 , 7 i . So the two unit vectors perpendicular to both a and b are ± e a × b = a × b || a × b || = ± 1 √ 66 h-1 ,-4 , 7 i ....
View Full Document

## This note was uploaded on 03/28/2011 for the course MATH 20C taught by Professor Helton during the Winter '08 term at UCSD.

Ask a homework question - tutors are online