{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

prelim1_fall05solutions

# prelim1_fall05solutions - Math 192 Prelim 1 Solution 1 a In...

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

Math 192, Prelim 1. Solution September 29, 2005 1) a) In order for the lines to intersect, there must be a point of intersection at which x 1 = x 2 , so 3 + 2 t = 15 7 s , t = 6 7 s/ 2. The point of intersection also must have y 1 = y 2 , so 3 + 3 t = 2 + s . Substituting t = 6 7 t/ 2 and solving yields s = 2, t = 1. With these values of t and s find z 1 = 1 = z 2 . So there is a point of intersection, and its coordinates are (1 , 0 , 1). b) Let the plane be given by A ( x x o )+ B ( y y o )+ C ( z z 0 ) = 0. The point of intersection is a point in the plane, so can take ( x o , y o , z o ) = (1 , 0 , 1). The direction of L 1 is given by v 1 = 2 i + 3 j + 2 k , that of L 2 by v 2 = 7 i + j + 3 k . The normal to the plane is v 1 × v 2 = i j k 2 3 2 7 1 3 = 7 i 20 j + 23 k . So A = 7, B = 20 and C = 23. The equation for the plane is 7( x 1) 20( y 0) + 23( z + 1) = 0, or 7 x 20 y + 23 z = 16 . 2) The velocity is v = a ( t ) dt = 2 tj + tk + C . At time t = 0, v (0) = 2 i = C , so v ( t ) = 2 i +2 tj + tk . The speed is v ( t ) = 4 + 4 t + t 2 = t +2. The length of the trajectory between t = 0 and t = 3 is d = 3 0 v

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.

{[ snackBarMessage ]}