# 10.1 Three Dimensional Space

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1 Math 172 Chapter 10A notes Page 1 of Three Dimensional Space 2D space 0 xx.. xx-, 0 yy yy-, PP(xx, yy) [Fig. 1] Point PP represented by (xx, yy), an ordered pair of real nos. Set of all ordered pairs R 2 = {(xx, yy) xx, yy R} Identify 2D space with R 2 3D space 0 xx xx-, 0 yy yy-, 0 zz zz-, PP(xx, yy, zz) [Fig. 2] Point PP represented by (xx, yy, zz) an ordered triple of real nos. Set of all ordered triples R 3 = {(xx, yy, zz) xx, yy, xx R} Identity 3D space with R 3 2D space is divided into quadrants axes, 1 st or positive quadrant [Fig. 3] 3D space is divided into octants 3D axes, xxxx-plane, yyyy-plane, xxxx-plane, 1 st or positive octant [Fig. 4] Example. Plot PP(4,3,0) and QQ(1,6, 4) axes, points plotted in perspective [Fig. 5] A single equation in R 2 represents a curve.

2 Math 172 Chapter 10A notes Page 2 of 12 Example yy = xx 2 0 xx-, 0. yy-, parabola, {(xx, xx 2 ) xx R} [Fig. 6] A single equation in R 3 represents a surface. Example yy = xx 2 0. xx-, 0 yy-, 0 zz-, parabolic cylinder, {(xx, xx 2, zz) xx, zz R} [Fig. 7] Distance in R 2 2D axes, 1 st quadrant, PP(xx 1, yy 1 ), QQ(xx 2, yy 2 ), segment PPPP, complete right triangle with sides of length xx 2 xx 1 and yy 2 yy 1 [Fig. 8] Pythagorean theorem PPPP 2 = xx 2 xx yy 2 yy 1 2 = (xx 2 xx 1 ) 2 + (yy 2 yy 1 ) 2 Distance in R 3 3D axes, show plane zz = zz 1 as xxxx-plane, PP(xx 1, yy 1, zz 1 ), QQ(xx 2, yy 2, zz 1 ), RR(xx 2, yy 2. zz 2 ), sides of right triangle [Fig. 9] PPPP 2 = (xx 2 xx 1 ) 2 + (yy 2 yy 1 ) 2 PPPP 2 = PPPP 2 + QQQQ 2 = (xx 2 xx 1 ) 2 + (yy 2 yy 1 ) 2 + (zz 2 zz 1 ) 2 PPPP = (xx 2 xx 1 ) 2 + (yy 2 yy 1 ) 2 + (zz 2 zz 1 ) 2 Distance between PP 1 (xx 1, yy 1, zz 1 ) and PP 2 (xx 2, yy 2, zz 2 ) PP 1 PP 2 2 = (xx 2 xx 1 ) 2 + (yy 2 yy 1 ) 2 + (zz 2 zz 1 ) 2 PP 1 PP 2 = (xx 2 xx 1 ) 2 + (yy 2 yy 1 ) 2 + (zz 2 zz 1 ) 2 Example. Find the distance between PP(4,3,0) and QQ(1,6, 4) PPPP 2 = (4 1) 2 + (3 6) ( 4) 2 = = 34

3 Math 172 Chapter 10A notes Page 3 of 12 PPPP = 34 Circle of radius rr centered at (aa, bb) in 2D 2D axes and 1 st quadrant, (aa, bb), circle, radius rr to (xx, yy) on circle [Fig. 10] Distance between (xx, yy) and (aa, bb) is rr (xx aa) 2 + (yy bb) 2 = rr (xx aa) 2 + (yy bb) 2 = rr 2 Equation for circle in 2D Sphere of radius rr centered at (aa, bb, cc) in 3D 3D axes and 1 st octant, (aa, bb, cc), sphere, radius rr to (xx, yy, zz) on sphere [Fig. 11] Distance between (xx, yy, zz) and (aa, bb, cc) is rr (xx aa) 2 + (yy bb) 2 + (zz cc) 2 = rr (xx aa) 2 + (yy bb) 2 + (zz cc) 2 = rr 2 Equation for sphere of radius rr and center (aa, bb, cc) in 3D Example. Show that the given equation represents a sphere. Find its center and radius. 2xx 2 + 2yy 2 + 2zz 2 + 4yy 2zz = 1 xx 2 + yy 2 + zz 2 + 2yy zz = 1 2 xx 2 + (yy 2 + 2yy + 1) 1 + zz 2 zz = 1 2 xx 2 + (yy + 1) 2 + zz = = 7 4 Center (0, 1, 1 2 ) Radius 7/2

4 Math 172 Chapter 10A notes Page 4 of Vectors Motivation Quantities with magnitude and direction Examples. Position, velocity, force. Algebraic Properties of Vectors In 2D space, vectors are ordered pairs of real numbers. aa = aa 1, aa 2 In 3D space, vectors are ordered triples aa = aa 1, aa 2, aa 3 aa 1 is the xx-component aa 2 is the yy-component aa 3 is the zz-component Addition of vectors aa + bb = aa 1, aa 2, aa 3 + bb 1, bb 2, bb 3 = aa 1 + bb 1, aa 2 + bb 2, aa 3 + bb 3 aa bb = aa 1, aa 2, aa 3 bb 1, bb 2, bb 3 = aa 1 bb 1, aa 2 bb 2, aa 3 bb 3 Zero Vector 00 = 0,0,0 aa + 00 = aa

5 Math 172 Chapter 10A notes Page 5 of 12 Multiplication by Scalars scalar kk ( a real number) kkkk = kk aa 1, aa 2, aa 3 = kkaa 1, kkaa 2, kkaa 3 Three Algebraic Properties (1) commutative law aa + bb = bb + aa (2) associative law aa + (bb + cc) = (aa + bb) + cc (3) distributive law kk(aa + bb) = kkaa + kkbb Example. Find aa + bb, aa bb, 3aa + 4bb, where aa = 3,2, 1, bb = 0,6,7?? aa + bb = 3,8,6?? aa bb = 3, 4, 8?? 3aa + 4bb = 9,30,25 Geometric Properties of Vectors In 2D, consider 2,1 0 6 xx-,0 2 yy-, PP(2,1), AA(4,1), BB(6,2) [Fig. 12] OOOO = 2,1 position vector of point PP(2,1) add AAAA = 2,1 These are different representations of the vector 2,1. Geometrically, vectors add tip-to-tail 0, AA(aa 1, aa 2 ), BB(bb 1, bb 2 ),, 0AA, 0BB AAAA [Fig. 13]

6 Math 172 Chapter 10A notes Page 6 of 12 0AA + AAAA = 0BB AAAA = 0BB 0AA = bb 1, bb 2 aa 1, aa 2 = bb 1 aa 1, bb 2 aa 2 consistent with 0AA + AAAA = 0BB aa 1, aa 2 + bb 1 aa 1, bb 2 aa 2 = bb 1, bb 2 Three Dimensions 0AA = aa 1, aa 2, aa 3 0BB = bb 1, bb 2, bb 3 AAAA = bb 1 aa 1, bb 2 aa 2, bb 3 aa 3 Example xx-, 0 2 yy-, aa, bb, cc [Fig. 14] bb = 1,2, aa = 3,1 what is cc? aa + cc = bb add tip-to-tail cc = bb aa = 1,2 3,1 = 4,1 Magnitude or length of a vector In 2D aa = aa 1, aa 2 aa = aa aa 2 2 In 3D aa = aa 1, aa 2, aa 3 aa = aa aa aa 3 2

7 Math 172 Chapter 10A notes Page 7 of 12 Two notations for components Vectors: components denoted by subscript: aa = aa 1, aa 2, aa 3 components denoted by base: xx = xx 0, yy 0, zz 0 Points: components denoted by subscript: PP(xx 1, yy 1, zz 1 ) components denoted by base: AA(aa 1, aa 2, aa 3 ) Both notations are frequently seen Example. Length of vector between two points in space AA(xx 1, yy 1, zz 1 ), BB(xx 2, yy 2, zz 2 ), AAAA Let aa = AAAA = xx 2 xx 1, yy 2 yy 1, zz 2 zz 1 aa = AAAA = (xx 2 xx 1 ) 2 + (yy 2 yy 1 ) 2 + (zz 2 zz 1 ) 2 Multiplication by a scalar xx-, yy- [Fig. 15] aa = 2,1 add aa aa = 2, 1 add aa 2aa = 4,2 add 2aa aa, aa, 2aa are parallel. Properties of Lengths Let aa, bb be any two vectors and let kk be a scalar

8 Math 172 Chapter 10A notes Page 8 of aa 0 aa = 0 if and only if aa = kkaa = kk aa 3. Triangle Inequality aa + bb aa + bb sketch 0, aa, bb, aa + bb [Fig. 16] Unit vectors A unit vector is a vector of length xx-, 0 1 yy-, ıı = 1,0, ȷȷ = 0,1, 1, 1 [Fig. 17] , 1 2 = = = 1 For any vector vv 00 vv vv is a unit vector vv vv = 1 vv vv = 1 Standard basis vectors In 2D ıı = 1,0 ȷȷ = 0,1 In 3D ıı = 1,0,0 ȷȷ = 0,1,0 kk = 0,0,1 Any vector can be represented in terms of ıı, ȷȷ, kk 2,1 = 2,0 + 0,1 = 2 1,0 + 0,1 = 2 ıı + ȷȷ

9 Math 172 Chapter 10A notes Page 9 of 12 5,2, 4 = 5 ıı + 2 ȷȷ 4 kk 10.3 The Dot Product The dot product is also known as the scalar or inner product 2D aa = aa 1, aa 2 bb = bb 1, bb 2 aa bb = aa 1 bb 1 + aa 2 cc 2 3D aa = aa 1, aa 2, aa 3 bb = bb 1, bb 2, bb 3 aa bb = aa 1 bb 1 + aa 2 bb 2 + aa 3 bb 3 Example. aa = 1, 2, 2 bb = 2,8, 6 aa bb = = 6 Properties of the dot product 1. aa aa = aa 2 2. aa bb = bb aa 3. aa (bb + cc) = aa bb + aa cc 4. kk(aa bb) = kkaa bb = aa kkbb 5. aa 00 = 0 Prove 1. aa = aa 1, aa 2, aa 3 = aa aa aa 3

10 Math 172 Chapter 10A notes Page 10 of 12 aa aa = aa 1, aa 2, aa 3 aa 1, aa 2, aa 3 = aa aa aa 3 2 = aa 2 The angle between two vectors is the smallest positive angle between the vectors drawn with their tails at the same point. sketch aa, bb, θθ [Fig. 18] Theorem. If θθ is the angle between aa and bb then aa bb = aa bb cos(θθ) Proof sketch aa, bb, θθ, bb aa [Fig. 19] Law of cosines aa bb 2 = aa 2 + bb 2 2 aa bb cos(θθ) Properties of dot product (aa bb) (aa bb) = aa 2 + bb 2 2aa bb Thus aa 2 + bb 2 2 aa bb cos(θθ) = aa 2 + bb 2 2aa bb 2 aa bb cos(θθ) = 2aa bb aa bb = aa bb cos(θθ) Example. Find the angle between the vectors 3,1 and 2,4. sketch 0 3 xx-, 0 4 yy-, aa = 3,1, bb = 2,4, θθ [Fig. 20] cos(θθ) = aa bb aa bb 3,1 2,4 = 10 3,1 = 10 2,4 = 20 cos(θθ) = = 1 2

11 Math 172 Chapter 10A notes Page 11 of 12 θθ = ππ/4 Two vectors are orthogonal or perpendicular if the angle between them is ππ/2. sketch vectors, indicate right angle [Fig. 21] cos ππ 2 = 0 If aa 00 and bb 00 then aa and bb are orthogonal iff aa bb = 0 Example. Determine whether the following vectors are orthogonal: aa = 1,5,2 and bb = 4,2, 3. aa bb = = 0 therefore the vectors are orthogonal. Define the component or scalar projection of a vector along another vector. sketch vector aa long and horizontal, vector bb at angle θθ upwards, projection from tip of bb to aa, comp aa bb [Fig. 22] comp aa bb = bb cos (θθ) Define the projection or vector projection of a vector along another vector. sketch vector aa long and horizontal, vector bb at angle θθ upwards, projection from tip of bb to aa, proj aa bb [Fig. 23] proj aa bb = bb cos(θθ) aa aa indicate length of projection and unit vector in direction of aa below the formula above Example. Find the scalar and vector projection of bb on to aa: aa = 3,1, bb = 2,3. Recall cos(θθ) = aa bb aa bb comp aa bb = bb cos(θθ) = aa bb aa = 9 10 proj aa bb = comp aa bb aa = 9 3,1 = 9 3,1 aa Example Let

12 Math 172 Chapter 10A notes Page 12 of 12 aa = 2ii 3jj + kk = 2, 3,1 bb = ii + 6jj 2kk = 1,6, 2 Then comp aa bb = bb cos(θ) = aa bb aa = proj aa bb = comp aa bb aa = 18 2, 3,1 aa = 18 2, 3,1 14 Application to Work For force FF in the same direction as displacement DD WW = FF DD Example. Work sliding a crate on a flat surface surface, crate, FF = 50N, DD = 10m [Fig. 24] WW = FF DD = (50N)(10m) = 500J In US customary units surface, crate, FF = 10 lbs, DD = 30ft [Fig. 25] WW = 300 ft lbs What if displacement is not in the same direction as the force? wagon with handle up at angle θθ, horizontal displacement vector DD and force FF in direction of handle, projection of FF on DD of magnitude FF cos (θθ) [Fig. 26] Work done is the product of the component of FF along DD and the distanced moved WW = ( FF cos(θθ)) DD = FF DD cos (θθ) = FF DD Example. Suppose FF = 50N, θθ = 45 = ππ, DD = 10m 4 Then WW = (50N)(10m)(cos ππ 2 ) = 500 J 354J 4 2

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