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21c_02f_1as

# 21c_02f_1as - y = t Then x =-2 t and z = 0 In other words h...

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Math 21C First Midterm Version A Oct. 21, 2002 1. dx/dt =6 t 2 +6 t and dy/dt =2 t - 3. (a) length = R 2 - 4 p (6 t 2 +6 t ) 2 +(2 t - 3) 2 dt (b) To be vertical, dx/dt = 0, which means t =0o r t = - 1. This gives the two points ( - 1 , 2) and (0 , 6). 2. (a) Let c = --→ BA = h 1 , 1 , - 1 i and b = -→ AC = h x - 2 , 2 , 4 i . Since b must be perpendic- ular to c , b · c =0. Thus x = 4 and so C is (4 , 3 , 4) and b = h 2 , 2 , 4 i . (b) Let a = --→ BC = h 3 , 3 , 3 i . The cosine is a · b | a || b | = 6+6+12 4+4+16 9+9+9 = 24 24 27 = r 24 27 = 2 2 3 . 3. (a) First plane : The equation is h 1 , 2 , 0 i·h x - 0 ,y - 0 ,z - 0 i = 0; that is, x +2 y =0. Second plane : Since the line is in the plane, h 1 , 1 , 0 i is parallel to the plane. With t = 0, we see that (0 , 2 , 0) is in the plane. (Any other value of t would work.) Since the origin is in the plane, h 0 , 2 , 0 i is parallel to the plane. Taking the cross product of the two vectors, we get the normal h 0 , 0 , 2 i and the equation of the plane is 2 z = 0 or, equivalently, z =0. (b) We have the equations x +2 y = 0 and z = 0 for the planes. Since both must
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Unformatted text preview: y = t . Then x =-2 t and z = 0. In other words, h x, y, z i = t h-2 , 1 , i . Of course, other answers are also valid, for example, h x, y, z i = t h 2 ,-1 , i + h-4 , 2 , i . Alternatively, we could take the cross product of the two normals to the planes, h 1 , 2 , i and h , , 1 i to get a vector in the direction of the line. We also need a point on the line. The origin works since the plane passes through the origin. Alternatively, we could ﬁnd two points on the line, P and Q and then the line would be t--→ P Q + Q . An obvious point is the origin. We can ﬁnd another by choosing any nonzero value for x or y ; for example, with x = 12, x + 2 y = 0 gives us y =-6. Since we have z = 0 from the second plane, our point is (12 ,-6 , 0)....
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