Then n a a 1 7 a unit vector parallel to a So proj a b b n n b a a 2 a 1 49h 1

Then n a a 1 7 a unit vector parallel to a so proj a

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Then n = a | a | = 1 7 a = unit vector parallel to a . So, proj a b = ( b · n ) n = b · a | a | 2 a = 1 49 h 1 , 2 , 3 i · h 3 , 6 , - 2 ih 3 , 6 , - 2 i = 9 49 h 3 , 6 , - 2 i .
Problem 45(b) - Spring 2009 Given a = h 3 , 6 , - 2 i , b = h 1 , 2 , 3 i . Write b as a sum of a vector parallel to a and a vector orthogonal to a . (Hint: Use projections.) Solution: We have b = h 1 , 2 , 3 i = h 1 , 2 , 3 i - 9 49 h 3 , 6 , - 2 i + 9 49 h 3 , 6 , - 2 i = 1 49 h 22 , 44 , 165 i + 9 49 h 3 , 6 , - 2 i . Here 9 49 h 3 , 6 , - 2 i parallel to a = h 3 , 6 , - 2 i and 1 49 h 22 , 44 , 165 i orthogonal to a = h 3 , 6 , - 2 i .
Problem 45(b) Continuation - Spring 2009 Given a = h 3 , 6 , - 2 i , b = h 1 , 2 , 3 i . Write b as a sum of a vector parallel to a and a vector orthogonal to a . (Hint: Use projections.) Solution: Why so? All we did was to write b = b - ( b · n ) n + ( b · n ) n where n = a 7 , n 2 = 1 . Of course this is the same as b = ( b - proj a b ) + proj a b . That is, we write b as proj a b plus “the rest”. But “the rest” is orthogonal to n (and to a ), since ( b - ( b · n ) n ) · n = b · n - ( b · n )( n · n ) = 0 , as n · n = 1 .
Problem 45(c) - Spring 2009 Given a = h 3 , 6 , - 2 i , b = h 1 , 2 , 3 i . Let θ be the angle between a and b . Find cos θ . Solution: cos( θ ) = a · b | a || b | = h 3 , 6 , - 2 i · h 1 , 2 , 3 i |h 3 , 6 , - 2 i||h 1 , 2 , 3 i| = 9 49 14 = 9 7 14 .
Problem 46(a) - Spring 2009 Given A = ( - 1 , 7 , 5), B = (3 , 2 , 2) and C = (1 , 2 , 3). Let L be the line which passes through the points A = ( - 1 , 7 , 5) and B = (3 , 2 , 2). Find the parametric equations for L . Solution: To get parametric equations for L you need a point through which the line passes and a vector parallel to the line. For example, take the point to be A and the vector to be -→ AB . The vector equation of L is r ( t ) = -→ OA + t -→ AB = h- 1 , 7 , 5 i + t h 4 , - 5 , - 3 i = h- 1 + 4 t , 7 - 5 t , 5 - 3 t i , where O is the origin. The parametric equations are: x = - 1 + 4 t y = 7 - 5 t , t R . z = 5 - 3 t
Problem 46(b) - Spring 2009 Given A = ( - 1 , 7 , 5), B = (3 , 2 , 2) and C = (1 , 2 , 3). A , B and C are three of the four vertices of a parallelogram , while CA and CB are two of the four edges. Find the fourth vertex. Solution: Denote the fourth vertex by D . Then -→ OD = -→ OA + -→ CB = h- 1 , 7 , 5 i + h 2 , 0 , - 1 i = h 1 , 7 , 4 i , where O is the origin. That is, D = (1 , 7 , 4) .
Problem 47(a) - Spring 2009 Consider the points P (1 , 3 , 5), Q ( - 2 , 1 , 2), R (1 , 1 , 1) in R 3 . Find an equation of the plane containing P , Q and R . Solution: Since a plane is determined by its normal vector n and a point on it, say the point P , it suffices to find n . Note that: n = -→ PQ × -→ PR = i j k - 3 - 2 - 3 0 - 2 - 4 = h 2 , - 12 , 6 i = 2 h 1 , - 6 , 3 i . So the equation of the plane is: ( x - 1) - 6( y - 3) + 3( z - 5) = 0 .
Problem 47(b) - Spring 2009 Consider the points P (1 , 3 , 5), Q ( - 2 , 1 , 2), R (1 , 1 , 1) in R 3 . Find the area of the triangle with vertices P , Q , R . Solution: The area of the triangle Δ with vertices P , Q , R can be found by taking the area of the parallelogram spanned by -→ PQ and -→ PR and dividing it by 2. Thus, using a) , we have: Area ( Δ ) = | -→ PQ × -→ PR | 2 = 1 2 | 2 h 1 , - 6 , 3 i| = 1 + 36 + 9 = 46 .
Problem 48 - Spring 2009 Find parametric equations for the line of intersection of the planes x + y + 3 z = 1 and x - y + 2 z = 0 . Solution: A vector v parallel to the line is the cross product of the normal vectors of the planes: v = h 1 , 1 , 3 i × h 1 , - 1 , 2 i = i j k 1 1 3 1 - 1 2 = h 5 , 1 , - 2 i .