# Through the point 111 and perpendicular to the plane

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through the point (1,1,1) and perpendicular to the plane with vector equation The parameter value t for the point of intersection satisfies the equation Substituting this into the equation for the line reveals that the point has coordinates ( x , y , z ) = (5/3,7/3,7/3)

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TEST1/MAC2313 Page 2 of 5 ______________________________________________________________________ 5. (5 pts.) What point (x 0 ,y 0 ) is four-fifths of the way from P = (-2,-3) to Q = (48,7) ?? The vector v with initial point P and terminal point Q has coordinates <50,10> in standard position. Consequently, the terminal point of <x 0 ,y 0 > = <-2,-3> + (4/5)<50,10> = <38,8> in standard position provides us with the coordinates for (x 0 ,y 0 ). Vector magic!! ______________________________________________________________________ 6. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,1,1>. Then v w = (-3)(-1) + (-2)(1) + (1)(1) = 3 - 2 + 1 = 2 ______________________________________________________________________ 7. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,1,1>. Then v × w = < -3, -(2), -5 > = < -3, 2, -5 > ______________________________________________________________________ 8. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,1,1>. Then proj w ( v ) = < -2/3, 2/3, 2/3 > , and the component of v perpendicular to w is w 2 = < -7/3, -8/3, 1/3 > . ______________________________________________________________________ The yz-plane part of Problem 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
TEST1/MAC2313 Page 3 of 5 ______________________________________________________________________ 9. (5 pts.) Suppose v = <-3,-4, 5> and w = <-1,1,1>. If α , β , and γ are the direction angles of v , then cos( α ) = , cos( β ) = ,and cos( γ ) = . ______________________________________________________________________ 10. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,1,1>. What is the exact value of the angle θ between v and w ?? ______________________________________________________________________ 11. (5 pts.) Write a point-normal equation for the plane perpendicular to v = <-3,-2,1> and containing the point (-1,2,-3). -3(x - (-1)) - 2(y - 2) + (z - (-3)) = 0 ______________________________________________________________________ 12. (5 pts.) Which point on the line defined the vector equation <x,y,z> = <1,1,1> + t<2,-1,-1> is nearest the point, (0,-1,0)? Build the vector with initial point (0,-1,0) to an arbitrary point on the line with parameter t , The norm of this vector is the distance from the point (0,-1,0) to the point on the line with parameter value t . The norm will be smallest when the vector is perpendicular to the vector <2,-1,-1>. Consequently, provides us with the value of t needed for the closest point.

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