So the point on the plane closest to 1 1 is P 1 3 1 3 1 3 Problem 13b Spring

So the point on the plane closest to 1 1 is p 1 3 1 3

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So the point on the plane closest to (1 , 0 , - 1) is P = ( 1 3 , - 1 3 , - 1 3 ).
Problem 13(b) - Spring 2007 Find the center and radius of the sphere x 2 + y 2 + 2 y + z 2 + 4 z = 20 .
Problem 13(b) - Spring 2007 Find the center and radius of the sphere x 2 + y 2 + 2 y + z 2 + 4 z = 20 . Solution: Completing the square in the y and z variables, we get x 2 + ( y 2 + 2 y + 1) + ( z 2 + 4 z + 4) = 20 + 1 + 4 . Rewriting, we have x 2 + ( y + 1) 2 + ( z + 2) 2 = 25 = 5 2 . Hence, the center is C = (0 , - 1 , - 2) and the radius is r = 5.
Problem 15(a) - Spring 2008 Consider the points A (2 , 1 , 0), B (1 , 0 , 2) and C (0 , 2 , 1). Find the area A of the triangle ABC . (Hint: If you know how to find the area of a parallelogram spanned by 2 vectors, then you should be able to solve this problem.)
Problem 15(a) - Spring 2008 Consider the points A (2 , 1 , 0), B (1 , 0 , 2) and C (0 , 2 , 1). Find the area A of the triangle ABC . (Hint: If you know how to find the area of a parallelogram spanned by 2 vectors, then you should be able to solve this problem.) Solution: The area of the parallelogram is | -→ AB × -→ AC | = i j k - 1 - 1 2 - 2 1 1 = - 1 2 1 1 i - - 1 2 - 2 1 j + - 1 - 1 - 2 1 k = |h- 3 , - 3 , - 3 i| = 27 . So the area of the triangle ABC is A = 27 2 .
Problem 22(a) - Fall 2006 Find parametric equations for the line r which contains A (2 , 0 , 1) and B ( - 1 , 1 , - 1).
Problem 22(a) - Fall 2006 Find parametric equations for the line r which contains A (2 , 0 , 1) and B ( - 1 , 1 , - 1). Solution: Note that -→ AB = h- 3 , 1 , - 2 i and the vector equation is: r ( t ) = ~ A + t -→ AB = h 2 , 0 , 1 i + t h- 3 , 1 , - 2 i = h 2 - 3 t , t , 1 - 2 t i . The parametric equations are: x = 2 - 3 t y = t z = 1 - 2 t .
Problem 23(a) - Fall 2006 Find an equation of the plane which contains the points P ( - 1 , 2 , 1), Q (1 , - 2 , 1) and R (1 , 1 , - 1).
Problem 23(a) - Fall 2006 Find an equation of the plane which contains the points P ( - 1 , 2 , 1), Q (1 , - 2 , 1) and R (1 , 1 , - 1). Solution: Consider the vectors -→ PQ = h 2 , - 4 , 0 i and -→ PR = h 2 , - 1 , - 2 i which are parallel to the plane. The normal vector to the plane is: n = -→ PQ × -→ PR = i j k 2 - 4 0 2 - 1 - 2 = 8 i + 4 j + 6 k . Since P ( - 1 , 2 , 1) lies on the plane, the equation of the plane is: h 8 , 4 , 6 i · h x + 1 , y - 2 , z - 1 i = 8( x + 1) + 4( y - 2) + 6( z - 1) = 0 .
Problem 23(b) - Fall 2006 Find the distance D from the point (1 , 2 , - 1) to the plane 2 x + y - 2 z = 1.
Problem 23(b) - Fall 2006 Find the distance D from the point (1 , 2 , - 1) to the plane 2 x + y - 2 z = 1. Solution: The normal to the plane is n = h 2 , 1 , - 2 i and the point P 0 = (0 , 1 , 0) lies on this plane. Consider the vector from P 0 to P 1 = (1 , 2 , - 1) which is b = h 1 , 1 , - 1 i . The distance D from (1 , 2 , - 1) to the plane is equal to: | comp n b | = b · n | n | = |h 1 , 1 , - 1 i · 1 3 h 2 , 1 , - 2 i| = 5 3 .
Problem 26(b) - Fall 2006 Find the center and radius of the sphere x 2 + y 2 + z 2 + 6 z = 16 .
Problem 26(b) - Fall 2006 Find the center and radius of the sphere x 2 + y 2 + z 2 + 6 z = 16 . Solution: Complete the square in order to put the equation in the form: ( x - x 0 ) 2 + ( y - y 0 ) + ( z - z 0 ) 2 = r 2 . We get: x 2 + y 2 + ( z 2 + 6 z ) = x 2 + y 2 + ( z 2 + 6 z + 9) - 9 = 16 . This gives the equation ( x - 0) 2 + ( y - 0) 2 + ( z + 3) 2 = 25 = 5 2 . Hence, the center is C = (0 , 0 , - 3) and the radius is r = 5.