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c3-t1-a

# 3x 1 2y 2 z 3 12 5 pts which point on the line

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-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. The point in question may now be obtained using the vector equation for the line. It is (8/6,5/6,5/6).

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TEST1/MAC2313 Page 4 of 5 ______________________________________________________________________ 13. (5 pts.) Find the exact value of the acute angle θ of intersection of the two planes defined by the two equations x - 3y = -5 and 2y - 4z = 7. θ = cos -1 (( v w )/( v w )) = cos -1 (6/(200) 1/2 ) = cos -1 (3/(5(2) 1/2 )) where v = <1,-3,0> and w = <0,2,-4>. Observe acutely the funny absolute value thingies. ______________________________________________________________________ 14. (5 pts.) Write an equation for the plane which contains the line defined by <x,y,z> = <1,2,3> + t<3, -2, 1> and is perpendicular to the plane defined by x - 2y + z = 0. A normal vector n for the plane sought is since it must be perpendicular to a direction vector for the line and a normal vector of the given plane. [One may also obtain a suitable vector by solving an appropriate system of equations, a little two by three linear homogeneous thingy.] An equation for the plane is now cheap thrills: -2( y - 2) - 4( z - 3) = 0, or equivalently, y + 2 z - 8 = 0. ______________________________________________________________________ 15. (5 pts.) Obtain an equation for the plane tangent to the sphere defined by ( x 1) 2 ( y 2) 2 ( z 3) 2 9 at the point (2, -4, 5), which is actually on the sphere. Here all we need is a normal vector for the tangent plane. This can be obtained easily using the point of tangency and the center of the sphere at hand. An equation of the tangent plane:
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3x 1 2y 2 z 3 12 5 pts Which point on the line defined the...

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