If the lines are to intersect there must be numbers t

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If the lines are to intersect, there must be numbers t 1 and t 2 so 4t 1 = 1 + t 2 , 1 - 2t 1 = 1 - t 2 , and 2 + 2t 1 = -1 + 4t 2 . [Here t 1 is the parameter for the putative point in terms of the first equation and t 2 is the parameter value for the second equation.] Solving this system yields t 1 = 1/2 and t 2 = 1. Using either t 1 or t 2 in the appropriate vector equation yields the point of intersection, (2, 0, 3). Note, however, you really do not have to obtain the point itself to answer this question!! ______________________________________________________________________ 19. (5 pts.) What is the area of the triangle in three space with vertices at P = (-3, 0, 0), Q = (0, 4, 0), and R = (0 , 0, 4). Let v be the vector with initial point P and terminal point Q, and let w be the vector with initial point P and terminal point R. Then the area A of the triangle is given by A = v × w /2 = <3,4,0> × <3,0,4> /2 = <16,-12,-12> /2 = 4 <4,-3,-3> /2 = 2 <4,-3,-3> = 2(34) 1/2 = (544) 1/2 /2 ???? regressing...??? ______________________________________________________________________ 20. (5 pts.) Do the three 2-space sketches of the traces in each of the coordinate planes of the surface defined by . z 1 x 2 9 y 2 4 Do not attempt to do a 3 - space sketch. If you don’t have enough space below, say where any additional work is. The xy-plane and the xz-plane appear below. The yz-plane appears on Page 2 of 5.
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