100hw1_merged_merged

# 100hw1_merged_merged - Math 100 Homework 1 fall 2010 due...

This preview shows pages 1–3. 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: Math 100 Homework 1 fall 2010 due 15/9 1. (11.1, 24) Describe the surface whose equation is given by x 2 + y 2 + z 2- y = 0 . 2. (11.3, 22) Find the acute angle formed by two diagonals of a cube. 3. (11.3, 28) Determine if it is true that for any vectors a , b , c such that a 6 = 0 and a · b = a · c , then b = c . 4. (11.4, 14) Determine if it is true that for any three vectors a , b , c , a × ( b × c ) = ( a × b ) × c. 5. (11.4, 30) Show that in the 3-space the distance d from a point P to the line through points A and B can be expressed as d = ||-→ AP ×--→ AB || ||--→ AB || . Where--→ AB stands for the vector representing the point B minus the vector representing the point A . 6. (11.5, 22) Find the parametric equation of the line through the origin that is parallel to the line given by x = 2 t , y =- 1 + t , z = 2. 7. (11.5, 52) Let L be the line that passes through the point ( x , y , z ) and is parallel to v = ( a, b, c ), where a , b , c are nonzero. Show that a point ( x, y, z ) lies on the line L if and only if x- x a = y- y b = z- z c . These equations, which are called the symmetric equations of L , provides a nonparametric representation of L . 1 Math 100 Homework 1 fall 2010 1. If ( x, y, z ) belongs to the given surface, x 2 + y 2 + z 2- y = 0 x 2 + y 2- y + 1 4 + z 2 = 1 4 x 2 + ( y- 1 2 ) 2 + z 2 = 1 4 the distance from ( x, y, z ) to (0 , 1 2 ) is 1 2 . Therefore, the given surface is the sphere centered at (0 , 1 2 , 0) with radius 1 2 . 2. Choose coordinate axes so that the vertices of the given cube are at (0 , , 0) , (1 , , 0) , (0 , 1 , 0) , (0 , , 1) , (1 , 1 , 0) , (1 , , 1) , (0 , 1 , 1) , (1 , 1 , 1) . Since (0 , , 0), (1 , 1 , 1) are a pair of opposite vertices, a diagonal of the cube is along (1 , 1 , 1). Since (1 , , 0), (0 , 1 , 1) are a pair of opposite vertices, a diagonal of the cube is along (- 1 , 1 , 1). Now, (1 , 1 , 1) · (- 1 , 1 , 1) || (1 , 1 , 1) || || (- 1 , 1 , 1) || = 1 3 . The angle between these two diagonals is cos- 1 1 3 . 3. No! For instance (1 , 0) · (0 , 1) = (1 , 0) · (0 ,- 1). 4. No! For instance (1 , , 0) × ((0 , 1 , 0) × (0 , 1 , 0)) = (0 , , 0) ((1 , , 0) × (0 , 1 , 0)) × (0 , 1 , 0) = (- 1 , , 0) . 5. Let Δ be the triangle with vertices A , B and P . Then, the area of Δ is half of the area of the parallelogram spanned by-→ AP and--→ AB which is 1 2 ||-→ AP ×--→ AB || . On the other hand, if ||--→ AB || is regarded as the base of Δ, the corresponding altitude would be d . Therefore 1 2 ||-→ AP ×--→ AB || = 1 2 d ||--→ AB || or d = ||-→ AP ×--→ AB || ||--→ AB || . 6. The line we want passes through (0 , , 0) and it is along the same direction as the given line, that is, along (2 , 1 , 0). Therefore, a point ( x, y, z ) lies on the line we want if x = 2 t , y = t , z = 0 7. Suppose that x- x a = y- y b = z- z c ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 12

100hw1_merged_merged - Math 100 Homework 1 fall 2010 due...

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

View Full Document
Ask a homework question - tutors are online