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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 3space 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 ....
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 Spring '11
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 Calculus, Vectors

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