Section 10.6 Vectors in Space

Transcription

1 36 Section 10.6 Vectors in Space Up to now, we have discussed vectors in two-dimensional plane. We now want to epand our ideas into three-dimensional space. Thus, each point in three-dimensional space will have three coordinates, an -coordinate, a -coordinate, and a -coordinate. Our coordinate sstem will consist of three aes, the -ais, -ais and the -ais. If we want to plot the point (,, 3), we would move units along the positive -ais, units in the direction of the positive -ais, and 3 units in the direction of the positive -ais. (,, 3)

2 The opposite vertices of a rectangular bo whose edges are parallel to the coordinate aes are given. Find the coordinates of the other vertices: E. 1 (0, 0, 0) and (,, 3) If we plot the points (0, 0, 0) and (,, 3) and draw the sides of the bo, we get: 37 On the aes, we have vertices of (, 0, 0), (0,, 0) and (0, 0, 3). Moving three units in the positive -direction from the points (, 0, 0) and (0,, 0), we get two more vertices (, 0, 3), (0,, 3). Moving two units in the positive -direction from (, 0, 0), we get (,, 0). Objective #1: Find the Distance between Two Points in Space. To find the distance between two points in space, we use an etension of the Pthagorean Theorem which include the distance between the two -coordinates. Distance Formula in Space Theorem Let ( 1, 1, 1 ) and (,, ) be two points in space. Then the distance d between the two points is given b: d = ( 1 ) +( 1 ) +( 1 )

3 38 Find the distance between the following points: E. (3,, 1) and (1, 5, 3) d = (1 3) +( 5 ) +(3 ( 1)) = = 101 Objective #: Find Position Vectors in Space. Recall that i represented the unit vector along the positive -ais and j represented the unit vector along the positive -ais. Now, we will add the unit vector k along the positive -ais. Thus, if v has an initial point at the origin (0, 0, 0) and a terminal point at (a, b, c), then v can be written as: v = ai + bj + ck where a, b, and c are called the components of v. Theorem If v is a vector with an initial point of ( 1, 1, 1 ) and a terminal point of (,, ), then v is equal to the position vector: v = ( 1 )i + ( 1 )j + ( 1 )k Find the position vector of the vector v: E. 3 v = P 1 P if P 1 = (3,, 5) and P = (1, 5, ) v = (1 3)i + (5 ( ))j + ( 5)k = i + 7j 9k Objective #3: Perform Operations on Vectors. We will now take man of the definitions and properties established in section 9. and epand them to include three dimensions. Definition Let v = a 1 i + b 1 j + c 1 k and w = a i + b j + c k be two vectors and let α be a scalar. Then 1) v = w if and onl if a 1 = a, b 1 = b, and c 1 = c. ) v + w = (a 1 + a )i + (b 1 + b )j + (c 1 + c )k 3) v w = (a 1 a )i + (b 1 b )j + (c 1 c )k ) αv = (αa 1 )i + (αb 1 )j + (αc 1 )k 5) v = a 1 +b1 + c1

4 Given v = 3i + j + k and w = i 7j 3k, find E. a v + w E. b v w E. c v E. d 3v w E. e w a) v + w = ( 3i + j + k) + (i 7j 3k) = i 5j + k b) v w = ( 3i + j + k) (i 7j 3k) = 5i + 9j + 7k c) v = ( 3i + j + k) = 1i + 8j + 16k d) 3v w = 3( 3i + j + k) (i 7j 3k) = 9i + 6j + 1k 8i + 8j + 1k = 17i + 3j + k e) w = () +( 7) +( 3) = = 6 Theorem Unit Vector in the Direction of v Given a nonero vector v, the unit vector in the same direction as v is u = v v This also implies that v = v u. Find a unit vector in the same direction as v: E. 5 v = 6i j + 3k v = (6) +( ) +(3) = = 9 = 7 u = v v = 6i j+3k 7 = 6 7 i 7 j k 39 Objective #: Find the Dot Product We will now epand the dot product and its properties to include vectors in space: Dot Product Definition Let v = a 1 i + b 1 j + c 1 k and w = a i + b j + c k. Then, the dot product v w is: v w = a 1 a + b 1 b + c 1 c Find the following if v = i + 3j k and w = 3i + j + k: E. 6a v w E. 6b w v E. 6c v v E. 6d w w E. 6e v E. 6f w

5 0 a) v w = ( 3) + 3() (1) = = b) w v = 3() + (3) + 1() = = c) v v = () + 3(3) () = = 9 d) w w = 3( 3) + () + 1(1) = = 6 e) v = () +(3) +( ) = = 9 f) w = ( 3) +() +(1) = = 6 Properties of the Dot Product If u, v, and w are vectors, then a) u v = v u Commutative Propert b) u (v + w) = u v + u w Distributive Propert c) v v = v d) 0 v = 0 Objective #5: Find the Angle between Two Vectors. Angle between Two Vectors Theorem Let u and v be two non-ero vectors and let θ be the angle between the vectors, where 0 θ π. Then cos(θ) = u v u v Find the angle between the two given vectors: E. 7 v = 3i j + k and w = 7i + 6j 3k v w = 3(7) (6) + ( 3) = 1 6 = 9 v = (3) +( ) +() = 9 w = (7) +(6) +( 3) = 9 Thus, u v cos(θ) = = 9 = u v 9 9 θ = cos 1 ( ) =

6 1 Objective #6: Find the Direction Angles of a Vector. In two dimensions, a position vector can be described b its magnitude and the angle between the positive -ais and its terminal side. In space, to describe a vector v, we need the magnitude and its three direction angles. The direction angles are defined below: α is the angle between v and i (the positive -ais) where 0 α π. β is the angle between v and j (the positive -ais) where 0 β π. γ is the angle between v and k (the positive -ais) where 0 γ π. γ α β cos(α) = v i v i cos(β) = v j v j cos(γ) = v k v k But, v i = a and But, v j = b and But, v k = c and i = 1, so j = 1, so k = 1, so cos(α) = a v cos(β) = b v cos(γ) = c v Since v = a +b + c, then we get the following theorem: Direction Angles Theorem Let v = ai + bj + ck be a non-ero vector with direction angles α, β, and γ. Then the direction cosines of the vector v are: cos(α) = a = a cos(β) = b = b v a +b +c v a +b +c cos(γ) = c = c v a +b +c As a consequence, a = v cos(α), b = v cos(β), and c = v cos(γ).

7 Find the direction angles of the following vector: E. 8 v = 1i 15j 16k v = (1) +( 15) +( 16) = = 65 = 5 So, cos(α) = a v = 1 5 cos(β) = b v = 15 5 cos(γ) = c v = 16 5 α = cos 1 (0.8) β = cos 1 ( 0.6) γ = cos 1 ( 0.6) α 61.3 β 16.9 γ 19.8 Propert of the Direction Cosines Let α, β, and γ be the direction angles of a non-ero vector v in space. Then, cos (α) + cos (β) + cos (γ) = 1 Proof: cos (α) + cos (β) + cos (γ) = ( a a +b +c ) + ( = a a +b +c + b a +b +c + b ) + ( a +b +c c ) a +b +c c a +b +c = a +b +c a +b +c = 1 Find the missing direction angle: E. 9 For vector v, the direction angles α and β are both 10. We need to find γ: cos (10 ) + cos (10 ) + cos (γ) = 1 cos(γ) = ( 1 ) + ( 1 ) + cos (γ) = cos (γ) = 1 cos (γ) = 1 cos(γ) = ± 1 = ± or cos(γ) = γ = 5 or γ = 135 Thus, the third angle is either 5 or 135. Notice that knowing two of the direction angles is not enough information to uniquel determine the direction of the vector.

8 3 Magnitude and Direction Angles Theorem Let v be a non-ero vector and let α, β, and γ be the direction angles. Then v = v [cos(α)i + cos(β)j + cos(γ)k] Proof: Let v = ai + bj + ck be a non-ero vector with direction angles α, β, and γ. As a consequence of the Direction Angles Theorem, we saw that: a = v cos(α), b = v cos(β), and c = v cos(γ). Plugging into v, we get: v = v cos(α)i + v cos(β)j + v cos(γ)k = v [cos(α)i + cos(β)j + cos(γ)k] Find the direction angles of each vector and write the vector in the magnitude and direction angle form: E. 10 v = i + 6j + 3k v = ( ) +(6) +(3) = = 9 = 7 cos(α) = a v = 7 cos(β) = b v = 6 7 cos(γ) = c v = 3 7 α = cos 1 ( 7 ) β = cos 1 ( 6 7 ) γ = cos 1 ( 3 7 ) α β 31.0 γ 6.6 Now, plug into the formula: v = v [cos(α)i + cos(β)j + cos(γ)k] = 7[cos(106.6 )i + cos(31.0 )j + cos(6.6 )k]