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Tutorial 5 solution

# Tutorial 5 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA 1505 Mathematics I Tutorial 5 (Solution Notes) 1. Write the vector 3 i +2 j + k as a sum of two vectors u + v such that u is parallel to w = i +3 j +4 k and v is perpendicular to w . (Hint: Use projection) Preliminaries Dot Product DEFINITION Dot Product The dot product of vectors and is u # v = u 1 v 1 + u 2 v 2 + u 3 v 3 . v = 8 v 1 , v 2 , v 3 9 u = 8 u 1 , u 2 , u 3 9 u # v s u dot v d Properties of the Dot Product If u , v , and w are any vectors and c is a scalar, then 1. 2. 3. 4. 5. 0 # u = 0. u # u = ƒ u ƒ 2 u # s v + w d = u # v + u # w s c u d # v = u # s c v d = c s u # v d u # v = v # u DEFINITION Orthogonal Vectors Vectors u and v are orthogonal (or perpendicular ) if and only if u # v = Important Result The angle θ between two nonzero vectors u and v is given by cos θ = u · v ± u ± v ± . It follows many results, for example: u v ⇐⇒ u · v =0 . ( u v θ = π 2 cos θ u · v =0) 104

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Projection The vector projection of u onto v is given by: u v ± proj v u Length ± ² u ² cos ± Proj v u = ± ± u ± cos θ ² v ± v ± = ± ± u ± u · v ± u ± v ± ² v ± v ± = ± u · v ± v ± 2 ² v Solution We are given the vector a =3 i +2 j + k , and we want to Fnd two vectors u and v in some directions, such that a = u + v . Since the two directions are perpendicular, then we consider the projection: u =p ro j w a , which is parallel to w ; and deFne v = a u , which is perpendicular to w . Therefore, a = u + v . u = a · w ± w ± 2 w = 3+6+4 1+9+16 ( i +3 j +4 k )= 1 2 ( i j k ) v = a u =(3 i j + k ) 1 2 ( i j k 1 2 (5 i + j 2 k ) . ± Remark: If you know Parallelogram Law , then the physical meanings of vector addition and subtraction are quite natural (two diagonal lines of the parallelogram): 105
2. Consider the two lines: ± 1 : x =2+2 t, y =2+ t, z =3+3 t ± 2 : x = 12 + 4 t, y = 5+2 t, z = 3+ t Show that ± 1 and ± 2 intersect. Find the point of intersection and an equation of the plane containing ± 1 and ± 2 . Preliminaries Intersection of Two Lines An Easy Case Recall what we learnt before: two lines in a plane, y = ax + b and y = cx + d , intersect each other, then the intersection point is given by solving ax + b = cx + d. A Puzzle But if the two lines are given by parameter equations, for example: l 1 : x = t +3 ,y =2 t +7 l 2 : x =3 t +1 t We use the same scheme to solve it: ( x =) t +3=3 t +1 = t =1 ( y =) 2 t +7=3 t t =6 We get two answers! So that means the two di±erent lines have two intersection points? (IMPOSSIBLE!!) !" !" If you substitute t = 1 back to the parameter equations, we ²nd two di±erent points, and the same case for t =6. Observation Actually, the equation of the two lines are y x +1 and y = x ,andthey intersect at the point ( 1 , 1) when t = 4for l 1 ,and t = 2 3 for l 2 . Notice that the values of the parameters are di±erent, so we should change one of the parameters!

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Tutorial 5 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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