{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution1

# solution1 - M341 Homework Solution 1 Section 1.1 7(f 2 x 3...

This preview shows pages 1–2. Sign up to view the full content.

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

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

Unformatted text preview: M341 Homework Solution 1 Section 1.1 - 7(f) : 2 x + 3 y − 4 z = [ − 4 , 8 , 10] + [ − 3 , , 9] − [16 , − 4 , 8] = [ − 23 , 12 , 11] Section 1.1 - 9 : Let A , B , C denote the points (7 , − 3 , 6) , (11 , − 5 , 3) and (10 , − 7 , 8) respectively. Then by distance formula we have AB = √ 16 + 4 + 9 = √ 29 , BC = √ 1 + 4 + 25 = √ 30 , AC = √ 9 + 16 + 4 = √ 29 Thus AB = AC ̸ = BC and triangle ABC is isosceles but not equilateral. Section 1.1 - 19 : First notice that f 1 = 22 · [9 , 6 , − 2] √ 81 + 36 + 4 = [18 , 12 , − 4] . f 2 = 27 · [7 , − 4 , 4] √ 49 + 16 + 16 = [21 , − 12 , 12] . Hence the total force f is f = f 1 + f 2 = [39 , , 8] and the acceleration vector a is a = f m = [39 , , 8] 6 = [ 13 2 , , 4 3 ] ( m/sec 2 ) Section 1.1 - 22(a) : Let x = [ x 1 , x 2 , . . . , x n ] be any vector in R n , then ∥ x ∥ = √ x 2 1 + x 2 2 + . . . + x 2 n ≥ since √ a ≥ for any nonnegative number a ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

solution1 - M341 Homework Solution 1 Section 1.1 7(f 2 x 3...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online