MATH 307 Vectors in Rn Dr. Neal, WKU Matrices of dimension 1 n can be thoght of as coordinates, or ectors, in n- dimensional space R n. We can perform special calclations on these ectors. In particlar, we can find special geometric interpretations for ectors in R 2 (the x y - plane) and for ectors in R 3. In fact, most of the definitions and theorems for R n are generalizations of reslts that can be seen geometrically in R 2 and R 3. Throghot, let = ( 1,..., n ) and = ( 1,..., n ) be ectors in R n. 0. Scalar Mltiplcation/Sm/Difference: (i) For any scalar c, c = (c 1,..., c n ). (ii) + = ( 1 + 1,..., n + n ) (iii) = ( 1 1,..., n n ), which gies the directional ector from to. I. Norm: The norm (or length or magnitde) of the ector is defined by = 1 2 +... + n 2. This definition of norm is an extension of the Pythagorean Theorem. Consider the ector = (5, 8) in R 2. Then the length of the line segment from (0, 0) to (5, 8) is = 5 2 + 8 2 = 89. In R 3, the norm of the ector (x, y, z) is simply the length of the segment from the origin to the point (x, y, z), and is gien by x 2 + y 2 + z 2. Howeer for n 4, we lose the geometric interpretation of length. So for = (4, 3, 5, 6) in R 4, the norm is simply = 4 2 +( 3) 2 + 5 2 + 6 2 = 86 withot the concept of length attached to it. Some Properties of Norm: (i) = (0, 0,, 0), (iii) 0, (ii) 2 = 2 1 +... + n 2 and (i) = 0 if and only if is the zero ector 0 c = c for any scalar c. II. Distance: The distance d(, ) between ectors and is gien by the length of the ector from to : d(, ) = = ( 1 1 ) 2 +... +( n n ) 2. This definition is an extension of the reslt in R 2. If = (5, 8) and = (10, 6), then the distance between these points is fond by the sal distance formla from algebra: d(, ) = (x 2 x 1 ) 2 +(y 2 y 1 ) 2 = 5 2 + ( 14) 2 = 221. Althogh we do not really hae a distance in R n for n 4, we still se the name distance for this measrement. Some Properties of Distance: (i) d(, ) 0, (ii) d(, ) = 0 and d(, ) > 0 if, and (iii) d(, ) = d(, ).
III. Dot Prodct: We define the dot prodct (also called inner prodct) between ectors and by = 1 1 +... + n n. In terms of matrix mltiplication, is really the matrix prodct T which is a 1 1 matrix. If = (4, 3, 5, 6) and = (8, 4, 2, 10), then = 32 12 10 + 60 = 70. Properties of Dot Prodct (a) = 2 1 +... + n 2 = 2 0. (b) = (c) = (d) For any scalar c, (c ) = c ( ) and (c ) = c ( ). (e) For another ector w, ( + ) w = ( w ) + ( w ). Parallelogram Diagonals: In R 2 and R 3, two non-collinear ectors and form a parallelogram. θ (3, 5) (7, 4) + Vectors = (3, 5) and = (7, 4) The long diagonal of the parallelogram aboe is gien by the ector + = (10, 9). The short diagonal gies the ector from to, which is gien by the difference = (4, 1). In R 2 and R 3, we can find the lengths of these diagonals sing the Law of Cosines. First, we let θ be the measre of the angle between and. Then 2 = 2 + 2 2 and cosθ, + 2 = 2 + 2 2 = 2 + 2 + 2 cos(π θ) cosθ.
These formlas are neer sed directly to find the lengths of the diagonals. For instance with = (3, 5) and = (7, 4), then + = (10, 9). Ths, + = 10 2 + 9 2 = 181. Howeer, the first formla can be sed to find the measre of the angle θ. Angle Between Vectors in R 2 and R 3 : Becase 2 = we can sole for cosθ and simplify the expression: 2 + 2 2 cosθ, cosθ = 2 + 2 ( 2 2 + ( ) ( )) = 2 + + = 2 = 2( ) 2 =. ( ) Ths, cosθ = and θ = cos 1. Note: It is always the case that 0º θ 180º. The angle θ between = (3, 5) and = (7, 4) is gien by θ = cos 1 41 34 65 29.29. We can compte cos 1 for any two ectors and in R n ; howeer, only in R 2 and R 3 does the ale gie the geometric interpretation of the measre of the angle between the ectors. Orthogonal Vectors: In R 2 and R 3, two ectors and are perpendiclar if and only if the angle θ between them is 90 ; that is, if and only if cosθ = 0. cosθ =, then cosθ = 0 if and only if the dot prodct eqals 0. Becase Ths in R 2 and R 3. we can say that and are perpendiclar if and only if = 0. For R n in general, we say that ectors and are orthogonal if and only if = 0. For example, if = ( 2, 0, 3, 0, 6) and = (3, 5, 2, 10, 0), then = 6 + 6 = 0; so and are orthogonal in R 5.
Identities in R n : Only in R 2 and R 3 do we hae the geometric interpretation of angle between ectors. Howeer the parallelogram identities can be generalized to R n as follows: 2 = 2 + 2 2( ) and + 2 = 2 + 2 + 2( ). We qickly obtain these reslts sing the properties of dot prodct: 2 = ( ) ( ) = + = + 2 = ( + ) ( + ) = + + + = 2 + 2 2( ), 2 + 2 + 2( ). Alternately in R 2 and R 3, 2 = 2 + 2 2 cosθ = 2 + 2 2 = 2 + 2 + 2( ) Cachy-Schwarz Ineqality: Let and be ectors in R n. Then Proof. If = assme 0 and 0, then = 0 and. = 0, so the ineqality holds. Ths we may > 0. Now let k be a scalar and consider the ector k +. Then 0 k + 2 = (k + ) (k + ) = k 2 ( ) + k( ) + k( ) + ( ) = 2 k 2 + 2( ) k + 2. This expression defines a qadratic in k which is always non-negatie; so the qadratic has zero roots or jst one root. Ths the discriminant b 2 4ac mst be less than or eqal to 0, where a = 2, b = 2( ), and c = 2. (If the discriminant were positie, then the qadratic wold hae two roots and therefore wold hae to be negatie oer some interal.) Ths, b 2 4ac = 4( ) 2 4 2 2 0, which implies that ( ) 2 2 2. By taking sqare roots we obtain. QED Triangle Ineqality: Let and be ectors in R n. Then + + Proof: We simply expand + 2. The first ineqality below arises from the fact that the nmber is less than or eqal to its absolte ale. The second ineqality is from Cachy-Schwarz..
+ 2 = ( + ) ( + ) = 2 + 2 + 2( ) 2 + 2 + 2 2 + 2 + 2 = + ( ) 2. Becase the sqare root fnction is increasing and + + by taking sqare roots. + 2 + ( ) 2, we obtain QED Note: For ectors and in R 2, the triangle ineqality states that length of the diagonal + can be no more than the sm of the lengths of the two sides and (hence, the name triangle ineqality). The reslt can be generalized to the distance between ectors as shown next. + Shortest Distance Between Points: Let and be ectors in R n. For any other ector w, we hae d(, ) d(, w ) + d( w, ). Proof. We se the definition of distance and apply the triangle ineqality: d(, ) = = ( w ) + ( w w + ) w = d( w, ) + d(, w ) = d(, w ) + d( w, ). QED The reslt states that the planar distance directly from and is less than or eqal to the sm of the distances from to w then from w to. We note that if,, and w are collinear with w between and, then d(, ) will eqal d(, w ) + d( w, ). Some Applications of Determinants to Vectors R 2 and R 3 Area of a Triangle: Three non-collinear points ( x 1, y 1 ), ( x 2, y 2 ), and ( x 3, y 3 ) form a triangle in the x y -plane. The area of the triangle is gien by Area = 1 x 1 y 1 1 2 det x 2 y 2 1. x 3 y 3 1 The aboe determinant eqals 0 if and only if the three points are on the same line.
With the points (10, 9), (7, 4), and (3, 5), 10 9 1 then det 7 4 1 = 23; ths the area of 3 5 1 the triangle is 23/2 = 11.5 sq. nits. Volme of a Tetrahedron: For non-planar points ( x 1, y 1, z 1 ), ( x 2, y 2, z 2 ), ( x 3, y 3, z 3 ), and ( x 4, y 4, z 4 ) form a tetrahedron in R 3. The olme is gien by x 1 y 1 z 1 1 Volme = 1 6 det x 2 y 2 z 2 1 x 3 y 3 z 3 1 x 4 y 4 z 4 1. The aboe determinant eqals 0 if and only if the for points are on the same plane. For the points (3, 0, 0), (0, 4, 0), (0, 0, 6), and (0, 0, 0), 3 0 0 1 0 4 0 1 then det = 72; ths the tetrahedral 0 0 6 1 0 0 0 1 olme is 72/6 = 12 nits cbed. (0, 0, 6) (0, 4, 0) (3, 0, 0) Example. Let = (5, 8, 5) and = ( 4, 6, 12). (a) Plot the points in R 3 and find their norms (i.e., lengths). (b) Find the ector from to, the direct distance from to, and the angle in between and. (c) Verify the Cachy-Schwarz Ineqality and the Triangle Ineqality sing and. (d) Let w = (1, 2, 3). Verify that d(, ) d(, w ) + d( w, ).
Soltion. (a) Below are the points = (5, 8, 5) and = ( 4, 6, 12) and some other points in R 3. z (5, 8, 5) (0, 8, 5) 5 4 ( 4, 6, 0) 8 6 y (5, 8, 0) 5 x ( 4, 6, 12) 12 (0, 6, 12) Below are the ectors = (5, 8, 5) and = ( 4, 6, 12). z (5, 8, 5) 4 8 6 y 5 x ( 4, 6, 12) The length of is = 5 2 + 8 2 + 5 2 = 114 10.677 and the length of is = 4 2 + 6 2 +12 2 = 196 =14.
= (5, 8, 5) and = ( 4, 6, 12) and w = (1, 2, 3). (b) The ector from to is = ( 9, 14, 17). Dr. Neal, WKU The direct distance from to is = 9 2 +14 2 +17 2 = 566 23.79. As before, = 114 and =14. The dot prodct is = 20 48 60 = 128. So the angle in between ectors and is θ = cos 1 z 128 148.9º. 114 14 = (5, 8, 5) 114 θ 14 x 566 = ( 4, 6, 12) (c) Cachy-Schwarz: =128 and ths, =14 114 149.479;. Triangle Ineqality: + = (1, 2, 7) and + = 1+ 4 + 49 = 54 7.35. Howeer, + = + 114 +14 24.677; ths, +. (d) From (b), d(, ) = = 566 23.79. Also, d(, w ) = w = 4 2 +10 2 + 2 2 = 120 d( w, ) = w = 5 2 + 4 2 +15 2 = 266 So d(, w ) + d( w, ) = 120 + 266 27.264 > d(, ). That is, d(, ) d(, w ) + d( w, ).