Homework Assignment 1 Solutions


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1 Dept. of Mth. Sci., WPI MA 1034 Anlysis 4 Bogdn Doytchinov, Term D01 Homework Assignment 1 Solutions 1. Find n eqution of sphere tht hs center t the point (5, 3, 6) nd touches the yzplne. Solution. The rdius connecting the center of the sphere nd the point where it touches the yzplne is perpendiculr to the yzplne. In other words, the point A where the sphere touches the yzplne is the orthogonl projection of the center C(5, 3, 6) onto the yzplne. This mens tht A hs coordintes (0, 3, 6). The rdius is r = CA = 5. The eqution of the sphere is then (x 5) + (y 3) + (z + 6) = 5.. Find the eqution of the sphere with dimeter determined y its endpoints, (1,, 7) nd (9, 0, 1). Solution. The length of the dimeter is (1 9) + ( 0) + (7 1) = 104 = 6. The rdius is then r = 6. The center is the midpoint of the line segment with endpoints (1,, 7) nd (9, 0, 1), i.e. the center hs coordintes ( 1 + 9, + 0, ) = (5, 1, 4). Putting this informtion together, we get the eqution (x 5) + (y + 1) + (z 4) = Find the center nd the rdius of sphere with n eqution x + y + z + 4y z = 1. Solution. We rerrnge the terms nd complete the squres: x + y + z + 4y z = 1 x + y + z + y z = 1 x + (y + y) + (z z) = 1 x + (y + 1) + (z 1 ) = ( 1 ) x + (y + 1) + (z 1 ) = 7 4 Which gives us the center (0, 1, 1 ) nd rdius 7 1
2 4. Two spheres with equtions nd x + y + z 4x y 6z + 1 = 0 x + y + z x + y 10z + 4 = 0 respectively, intersect. Their intersection is circle in IR 3. Find the center nd the rdius of this cirle. Solution. As in the previous prolem, we rewrite the equtions in form which gives us the centers nd rdii: nd (x ) + (y 1) + (z 3) = (x 1) + (y + 1) + (z 5) = 3. The first sphere hs center A(, 1, 3) nd rdius. The second sphere hs center B(1, 1, 5) nd rdius 3. Let C e point on the intersection of the two spheres. Then AB = = 3, AC =, nd BC = 3. Since three points lwys lie in plne, we cn put A, B, nd C on plnr picture s follows: C r 3 u v A O B Let O e the orthogonl projection of C onto AB. Then O is the center of the circle nd OC is its rdius. Denoting AO = u, OB = v, OC = r, we get u + v = 3 u + r = v + r = 3 Solving this system, we get r = /3, u = 4/3, nd v = 5/3. Thus, the rdius of the circle is r = /3, nd the center is the point O which divides the line segment AB into rtio 4 : 5. Therefore the center O hs coordintes 5 9 (, 1, 3) + 4 ( 14 9 (1, 1, 5) = 9, 1 9, Write the vector u = 1, 0, s liner comintion of the three vectors 1, 1, 0, 1, 1,, nd 0, 1,. Solution. We re looking for rel numers λ 1, λ, λ 3, such tht or, in components, ). 1, 0, = λ 1 1, 1, 0 + λ 1, 1, + λ 3 0, 1,, λ 1 λ = 1 λ 1 + λ + λ 3 = 0 λ λ 3 =
3 Solving this liner system, we get λ 1 = 1, λ =, λ 3 = 3, so u = 1, 0, = 1, 1, 0 1, 1, + 3 0, 1,. 6. Let + =. Find the ngle etween nd. Solution. We hve: i.e., nd re perpendiculr. 7. The Prllelorm Lw sttes tht + = + = ( + ) ( + ) = ( ) ( ) + + = + 4 = 0 = 0, + + = +. Give geometric interprettion of this lw. Prove it. Solution. Let ABCD e prllelogrm nd denote AB =, AD =. Then DC =, BC =, AC = +, DB =. The prllelogrm lw sttes tht AC + BD = AB + BC + CD + AD, i.e. the sum of the squres of the digonls in prllelogrm is equl to the sum of the squres of the sides. To prove the equlity, oserve tht + + = ( + ) ( + ) + ( ) ( ) = = If =, show tht = ( ). Solution. = = = ( ) ( ) = + 0 = = = ( ) 3
4 9. If c = +, where,, nd c re ll nonzero vectors, show tht c isects the ngle etween nd. Solution. Since c is liner comintion ov nd, it lies in the sme plne s nd. It suffices to show tht the ngle etween nd c is the sme s the ngle etween nd c. The cosine of the ngle etween nd c is c c = ( ) + c nd the cosine of the ngle etween nd c is the sme thing. c c = + ( ) c = +, c = +, c 10. In the tringle ABC, points A 1, B 1, nd C 1 re chosen on the sides BC, AC, nd AB respectively in such wy tht AA 1, BB 1, nd CC 1 re the three (ngulr) isectors of the tringle ABC. Show tht, if AA 1 + BB 1 + CC 1 = 0, (1) then the tringle ABC must e equilterl. Solution. Let s define the vectors BC =, CA =, AB = c, nd let s denote their lengths y,, nd c respectively. Then, we see immeditely tht + + c = 0, nd hence c =. () Since the point C 1 is on the line segment AB, we must hve CC 1 = λ CA + (1 λ) CB = λ + (1 λ)( ) for some rel λ etween 0 nd 1, s shown in clss. On the other hnd, ccording to the result of the previous prolem, CC 1 must e proportionl to CB CA + CA CB = + ( ). Comining these two oservtions, we conclude tht Similrly, we hve nd CC 1 = AA 1 = CC 1 = + + c c c + c +. c + c + c c. 4
5 Adding these three equlities, nd using (1), we get ( c + c ) ( c ) + c Comining this with () we get, fter simplifiction, ( + c + ) ( + 1 = + + ( + + c ) + c 1. ) c = 0. + c Since the vectors nd re not colliner, this is possile only if oth coefficients re zeros, i.e. + c c = 1 = 1 To solve this system, we get rid of the denomintors, y multiplying the first eqution y ( + c)( + ), nd the second y ( + )( + c). After simplifiction we get: or, equivlenly, = c = c = c = c. In other words, we hve which implies = c = c, = = c. 5
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