exam2solutions - Calculus III Math 143 Spring 2008 Professor Ben Richert Exam 2 Solutions Problem 1(20pts Consider the points A =(1 2 3 and B =(2 3 5

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Calculus III Math 143 Spring 2008 Professor Ben Richert Exam 2 Solutions Problem 1. (20pts) Consider the points A = (1 , 2 , 3) and B = (2 , 3 , 5) , and the line r ( t ) = ( 2 , 2 , 3 ) + t ( 1 , 2 , 5 ) . (a–5pts) Give parametric equations for the line containing A and parallel to −− AB . Solution. The line through ( a, b, c ) in direction ( d, e, f ) is ( a, b, c ) + t ( d, e, f ) , giving parametric equations x ( t ) = a + dt y ( t ) = b + et z ( t ) = c + ft We use the point (1 , 2 , 3) and the direction −− AB = ( 1 , 1 , 2 ) which we get by subtracting the point B from A . So our line is: x ( t ) = 1 + t y ( t ) = 2 + t z ( t ) = 3 + 2 t square (b–5pts) Give the equation of a plane which is perpendicular to r ( t ) and contains B . Solution. The equation of a plane containing ( x 0 , y 0 , z 0 ) with normal ( a, b, c ) is a ( x x 0 ) + b ( y y 0 ) + c ( z z 0 ) = 0 . Since the plane we are looking for is perpendicular to −− AB , its normal must be parallel to −− AB = ( 1 , 1 , 2 ) , that is, we can use ( 1 , 1 , 2 ) as its normal. With the point B = (2 , 3 , 5), we get the plane ( x 2) + ( y 3) + 2( z 5) = 0 . square (c–10pts) Give the equation of the plane containing r ( t ) and the point B . Solution. Let C = (2 , 2 , 3), and note that C is a point on r ( t ). Then the normal to our plane will be perpendicular to both ( 1 , 2 , 5 ) , the direction of r ( t ), and −− CB = ( 0 , 1 , 2 ) . We know that the ( 1 , 2 , 5 ) × ( 0 , 1 , 2 ) = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle i j k 1 2 5 0 1 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = (− 1 , 2 , 1 ) is normal to both those directions. Thus the plane in question is ( x 2) 2( y