Review of Prerequisite Skills, p. 350 C( 2, 0, 1) B( 3, 2, 0) y A(0, 1, 0) D(0, 2, 3) j! k! 2k! Section 7.1, pp

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1 . 5. a. a a b a a b. Case If and are collinear, then b is also collinear with both and. But is perpendicular to and c c c b 9 b c, so a a b b is perpendicular to. Case If b and c b c are not collinear, then b spanning sets, b and c span a plane in R, and is in that plan If a b c is perpendicular to b and c, then it is perpendicular to the plane and all vectors in the plan So, is perpendicular to b c a. Chapter Test, p. 8. Let P be the tail of and let Q be the head of The vector sums a b c a c. and a b c can be depicted as in the diagram below, using the triangle law of addition. We see that a b PQ c a b c. This is the associative propert for vector addition. P. a. (8,, 8) b a (a + b) PQ= (a + b) + c = a + (b + c) a,, b (b + c). 9. a. b a, b 5a a,, 5. a. a and b b 5 c span R, because an vector (, ) in R can be written as a linear combination of a and b. These two vectors are not multiples of each other. p, q. a.,, 9,, 7, r, cannot be written as a linear combination of and In other words, r p q. does not lie in the plane determined b p and q. 7., u.; 7.9 from toward b c Q 8. DE DE CE b a CD Also, BA CA CB BA b a Thus, DE BA Chapter 7 Review of Prerequisite Skills, p. 5. v 8 km> h N 7. E. 5.9 units W. N.. a.,, 7; l ,, ; l.9,, ; l.,, 9; l a. on the -plane on the -plane on the -plane. a. i i 7j 8i j i 7. a. i j j 5i j i j k 8k k j k k Section 7., pp.. a. N is a melon, 5 N is a chair, N is a computer Answers will var.. a. N N C(,, ) B(,, ) A(,, ) D(,, ) N 8. a line along the same direction. For three forces to be in equilibrium, the must form a triangle, which is a planar figur 5. a. The resultant is N at an angle of N. E. The equilibrant is N at an angle of S. W. The resultant is 5 N at an angle of S.9 W. The equilibrant is 5 N at N.9 E.. a. es es no es 7. Arms 9 cm apart will ield a resultant with a smaller magnitude than at cm apart. A resultant with a smaller magnitude means less force to counter our weight, hence a harder chin-up. 8. The resultant would be.7 N at.7 from the N force toward the 8 N forc The equilibrant would be.7 N at 5. from the N force awa from the 8 N forc N 5 from given force,.95 N perpendicular to 9. N force. 9 N directed up the ramp. a. 7 N. approimatel 7. N 5 south of east. a. 7 N The angle between f and the resultant is The angle between f.. and the equilibrant is.7.. a. N 8 N N N N For these three equal forces to be in equilibrium, the must form an equilateral triangl Since the resultant will lie along one of these lines, and since all angles of an equilateral triangle are, the resultant will be at a angle with the other two vectors. Since the equilibrant is directed 7 Answers

2 opposite the resultant, the angle between the equilibrant and the other two vectors is N, 7.5 from f toward f. 5 rope: 75.7 N rope:.8 N 7. cm string: approimatel 9. N, cm string: approimatel 9. N to the starboard side 9. a. magnitude for resultant and equilbrant.75 N u 5N., u 8N 5., u N.7. We know that the resultant of these two forces is equal in magnitude and angle to the diagonal line of the parallelogram formed with f and f as legs and has diagonal f We also know from the cosine rule cos f, where f is the supplement to u in, our parallelogram. Since we know f 8 u, then cos f cos 8 u cos u. Thus, we @ cos @ cos @cos u Section 7., pp a. 8 km> h in the direction of the train s movement 7 km> h in the direction of the train s movement. a. 5 km> h north 7 km> h north.. km> h, W 9.5 S. upstream 5. a. m> s forward m> s in the direction of the car. m> s, N 7. W 7. a. 7.7 km> h, N 5.5 W about 7.7 km 8. a. about 8 km about.5 east of north 9. a. about. south of west h. a. 5 km> h, 5. downstream to the bank. about.7 km min. a. about 8. west of north about 8 km> h. 5 Since her swimming speed is a maimum of km> h, this is her maimum resultant magnitude, which is also the hpotenuse of the triangle formed b her and the river s velocit vector. Since one of these legs is 5 km> h, we have a triangle with a leg larger than its hpotenuse, which is impossibl. a. about. m s. a. about 58.5, upstream about 58. s 5. 5 h Section 7., pp. To be guaranteed that the two vectors are perpendicular, the vectors must be nonero.. a # b is a scalar, and a dot product is onl defined for vectors.. Answers ma var. For eample, let,,, b but. a # a c a a i b j, c i # b # c because c a b b #. a b # c 5.. a a a. a 9 7a # b a.. a. a a a a b b # a a b # a b # a # b a # b # b # b # b # a b b # c, vectors are perpendicular. c, which is the Pthagorean theor.. 5. u v u v # u v u v # u v u v u v v u u v u u # v # v d c b b d c # c a b # a a # a c # a b # a a # a b # a is a scalar c # a b # a a c # a d c # a c # d a d # a c # a # a because because a Section 7., pp An vector of the form is perpendicular to a c, c. Therefore, there are infinitel man vectors perpendicular to a. Answers ma var. For eample:,,,,,.. a. ; 9 7 ; acute ; obtuse. Answer ma var. For eample: a.,, is perpendicular to ever vector in the -plan,, is perpendicular to ever vector in the -plan,, is perpendicular to ever vector in the -plan. a.,, and,, ; no @ a a b # a a # a # b a # a # b a c # b c a # b b b # c c, 5, and a 5,, 5 b Answers 7

3 5. a. a and b are non-collinear and lie on the plan All vectors perpendicular to these are in R. An pair of non-collinear vectors in R line on the plane, All vectors perpendicular to these are in R.. a. about 8 about about about 5 7. a. 8. a. k k (, ) (, ), and, ;, and,, #, or, #, 9. a. 9. a. i. p 8 q ; ii. Answers ma var. For eample, p, q 5. Unique for collinear vectors; not unique for perpendicular vectors. u A 9 ; u B. ; u C.. a. O,,, A 7,,, B 7,,, C,,, D 7,, 5, E,, 5, F,, 5 5. a. Answers ma var. For eample,,,. Answers ma var. For eample,,,.. or 5. a. p q. Answers ma var. For eample,,, and,,.,,,, Let.,, is perpendicular to,, and,,. Let.,, is perpendicular a. a # b Therefore, since the two diagonals are perpendicular, all the sides must be the same length. u ; u 9. a., 8, 87.. a 9.5 or u 7.5 Mid-Chapter Review, pp a cm cord: 7. N; cm cord: 88. N.. a. about 575. km> h at S 7. E about.7 h 5. a. about. N about 9.9 N a. 8. a. 5 5i j k 9 9. a. or no value. a. i j k or.. about. F.9.8 N, 5. from the N force towards the N forc E.8 N, from the N force awa from the N forc. a. about 9. about a. about N. E about.7 h 5. a or,, b a,, b. a. about. m about 8.9 s 7. a. when and have the same length Vectors a and b determine a parallelogram. Their sum a b is one diagonal of the parallelogram formed, with its tail in the same location as the tails of a and Their difference a b b. is the other diagonal of the parallelogram. 8. about 8. N Section 7.5, pp. 98. a. scalar projection, vector projection i scalar projection, vector projection j. Using the formula would cause a division b. Generall the has an direction and magnitud You cannot project onto nothing.. You are projecting a onto the tail of b, which is a point with magnitude. Therefore, it is the projections of onto the tail of a ; are also and b.. Answers ma var. For eample, p A E, q A B D E p A C q B scalar projection p on C@, vector projection p on scalar projection q q A C, on p A D, vector projection on p q 5. scalar projection of on vector projection of a ii A D a, on i, scalar projection of a on, vector projection of a on jj j, scalar projection of a on vector projection of a k 5, on k 5k ; Without having to use formulae, a projection of,, 5 on i, j, or k is the same as a projection of on i,,, on j,,, and,, 5 on k, which intuitivel ields the same result. p # q. a. scalar 58 vector projection:, 5, about 8.5, about 7.9, about. 7 Answers

4 7. a. scalar projection:, vector projection: scalar projection:, vector projection: i 5 scalar projection:, 5 vector projection: 5, 9 8. a. The scalar projection of a on the -ais X,, is ; The vector projection of a on the -ais is The scalar projection of a i ; on the -ais, Y, is ; The vector projection of a on the -ais is The scalar projection of a j ; on the -ais,, Z is ; The vector projection of a on the -ais is The scalar projection of ma k. on the -ais X,, is The vector projection of on the -ais is mi ma m; ; The scalar projection of ma on the -ais, Y, is m; The vector projection on the -ais is The scalar projection of ma mj ma, Y, ; on the -ais,, Z is m; The vector projection of ma on the -ais is mk. 9. a. a vector projection: a scalar projection: a cos u a a a cos a. The vector projection is the scalar a projection multiplied b a a a a a,.. a. B a O a A a # a a a a a So, the vector projection is a a a a b a.. a. scalar projection of AB on the -ais is ; vector projection of AB on the -ais is scalar projection of AB i ; on the -ais is ; vector projection of on the -ais is j AB ; scalar projection of AB on the -ais is ; vector projection of AB on the -ais is k. C a b B u u D b a C b B u u D c A In an isosceles triangle, CD is a median and a right bisector of BA. Yes. a. 7.7, 8.9 b Q OQ is the vector projection of on OP b is the vector projection of a a on b. a. The scalar projection of BC on OD 9 is 9. The scalar projection of AC on OD is. Same lengths and both are in the direction of OD. Add to get one vector. 5. a. cos a cos b cos g a a a b c b b a a b c b c a a b c b a a b c b a b c c O 5 a b c a b c a b c Answers ma var. For eample:,,,,, If two angles add to 9, then all three will add to 8. P a. a. about 5.7 about cos sin cos sin cos a cos b cos g sin a sin b sin g 8. Answers ma var. For eample: Section 7., pp a. sin a sin b sin a sin b sin g a b is perpendicular to a. Thus, their dot product must equal. The same applies to the second cas sin g ab ab a b a ab a b is still in the same plane formed b and thus is perpendicular to a b a a b, b making the dot product again. Once again, a b is still in the same plane formed b a and thus a b is perpendicular to a b, b making the dot product again.. a b produces a vector, not a scalar. Thus, the equalit is meaningless. b B (, c, d) A (a, b, ) Answers 7

5 . a. It s possible because there is a vector crossed with a vector, then dotted with another vector, producing a scalar. This is meaningless because a # b produces a scalar. This results in a scalar crossed with a vector, which is meaningless. This is possibl produces a vector, and c a b d also produces a vector. The result is a vector dotted with a vector producing a scalar. This is possibl # b produces a scalar, and c d a produces a vector. The product of a scalar and vector produces a vector. This is possibl produces a vector, and c a b d produces a vector. The cross product of a vector and vector produces a vector. This is possibl a b produces a vector. When added to another vector, it produces another vector.. a. 5.. a. 7, 8,, 5,,, 9,, 7,, 8,,,, Vectors of the form (, b, c) are in the -plan Thus, the onl vectors perpendicular to the -plane are those of the form (a,, ) because the are parallel to the -ais. 7. a.,,,,,,,, a, b, c ka, kb, kc bkc ckb, cka akc, akb bka Using the associative law of multiplication, we can rearrange this: bck bck, ack ack, abk abk 8. a. p, p q, p q r p q p 9. a. i j j q r r, 7,, 7, p r,, 5,, 5 j i,,,, k k k,,,, j,,,, i k,,,, i k i i k,,,, j,,,, j. ka b a b, a b a b, a b a b a, a, a ka a b a a b a a b a a b a a b a a b k a is perpendicular to ka b.. a.,,,,,,, All the vectors are in the -plan Thus, their cross product in part is between vectors parallel to the -ais and so parallel to each other. The cross product of parallel vectors is. Let,, and.,,,,, Then,,,,,,,,,,,,,,,, Thus,,,.. B the distributive propert of cross product: a b a a b b B the distributive propert again: a a a b b b a b A vector crossed with itself equals, thus: b a a b a a b b a b a a b b Section 7.7, pp. 5. B pushing as far awa from the hinge as possible, r is increased, making the cross product bigger. B pushing at right angles, sine is its largest value,, making the cross product larger.. a. This makes sense because the vectors lie on the same lin Thus, the parallelogram would just be a line making its area.. a. 5 J about 78.9 J about 889. J J. a. k k j j 5. a. square units square units., a. square units 5 square units. An two sides of a triangle can be used to calculate its area. 8. about.99 J 9. or about.8. a. p q,, p q 9, r, 5 5, 5, 9 5, 5, a,, b,, 5, 5, Looking at -components: a b 5 a 5 b -components: a b 5 Substitute a: b b 5 5b 5 b Substitute b back into the -components: a 5 a Check in -components: a b 9 p # r q # r p # r q q # r p,,,,,,,, 9,, 9 5, 5, Review Eercise, pp. 8. a.,,,, The cross products are parallel, so the original vectors are in the same plan 7 Answers

6 . a a about a. about rope at 5 : about 87.8 N, rope at : about 7.7 N 7.. km> h, W 9. N 8. a. OB approimatel , 5 5, 5. a. about 77. is the largest angle.5. cm string: 78. N; cm string: 58.8 N. a. N OA 5 N N 5 N The resultant is N in a direction N. W. The equilibrant is N in a direction S. E.. a. Let D be the origin, then: A,,, B,,, C,,, D,,, E,,, F,,, G,,, H,, about. about a. about 8. about 8 min s Such a situation would have resulted in a right triangle where one of the legs is longer than the. a. hpotenuse, which is impossibl OA OA OB OB, 8, 8, 9,, about a. a and b a b 9 a 5,, 5 b 8. a. about 7. about ,, about a. special not special. a.,,,, 5,,. about.55 N..., 8, 5 or 7 5. about.. a. C,, 5, F,,,, 5 about. 7. a. about 7. about.8 about.8 8. a. scalar:, vector: i scalar:, vector: j scalar:, vector: 9. a. b k j a, c ; When dotted with d, it equals J. a. a # b 5 a with the -ais: a 5 cos a a with the -ais: cos b 5 a with the -ais: cos 9 with the -ais: cos a b with the -ais: cos b b with the -ais: cos g S S m j 9j k 8. a cos a, cos b cos g ; acute case: 9., obtuse case:.7 5 a # b a a b 8 b 8 b 98 c c b 8 b a b a a a b AB a b a b a # b a b 5 Answers 75

7 cos A cos B cos C area of triangle ABC Chapter 7 Test, p.. a.,,,, (,, ). a. scalar projection:, vector projection:,,. 9 -ais: 8. ; -ais: 9.5 ; -ais:.8 or 5.. Both forces have a magnitude of 78. N. The resultant makes an angle.7 to the N force and. to the 5 N forc The equilibrant makes an angle. to the N force and 5.7 to the 5 N forc..99 km> h, N 5.7 W 5. a. 9 m downstream 8.7 upstream..5 square units. 7. cord at 5 : about 5. N; cord at 8. a. # 7 : about 9. N So, the equation holds for these vectors. So, the right side of the equation is # # # # # # # # # # # # # # # # Chapter 8 Review of Prerequisite Skills, pp. 5. a., 9,,,. a. es es es no. es. 5. t 8 a. (, ) (5, ), 7, a., 8,,, 8. C 9. a. 7,,, 8, 5, 5,. a. 7,,, 8, 5, 5,. a. slope: ; -intercept: slope: -intercept: ; slope: -intercept: 5 ; 5 slope: ; -intercept: = = 8 =8 5 =5 5. Answers ma var. For eample: a. (8, ) 5, i j, 9 i k j 8k B A D. a. 77, 8, 8, 8, 8 55,,. The dot product of two vectors ields a real number, while the cross product of two vectors gives another vector. Section 8., pp.. Direction vectors for a line are unique onl up to scalar multiplication. So, since each of the given vectors is just a scalar multiple of Q each is an, R, acceptable direction vector for the lin. a. Answers ma var. For eample,, 7, (, 5), and (, ). t 5 If t 5, then and 5. So P, 5 is a point on the lin. Answers ma var. For eample: a. direction vector: (, ); point: (, ) direction vector:, 7; point: (, ) direction vector: (, ); point: (, ) direction vector: 5, ; point: (, ). Answers ma var. For eample: r q, t5,, tr, 5 s5,, sr 5. a. R9, 8 is a point on the lin When t 7, 9and 8. Answers ma var. For eample: r 9, 8 t,, tr Answers ma var. For eample: r, t,, tr. Answers ma var. For eample: a., (, ), and (, ) r, t,, tr This describes the same line as part a. 7. One can multipl a direction vector b a constant to keep the same line, but multipling the point ields a different lin 8. a. 8 R(, 9) Q(, 7) r, 7 t,, tr;, 7 t, tr 7 Answers

8 9. a. r, 5 t5,, tr;. a. r 5t, 5, tr, t5,, tr,.. 9. First, all the relevant vectors are foun AB, AC, AD a. AC, 9 AD,, AB, 9, AB AC,, 9 AD. a. A(5, ); B, or about.. In the parametric form, the second equation becomes t, t, tr. If t is solved for in this equation, we obtain t and t. Setting these two epressions equal to each other, the line is described b, or b simplifing, So, the. second equation describes a line with a slope of. If is solved for in the first epression, we see that 5. (, ) is on the second line but not the first. Hence, both equations are lines with slope of and must be parallel. Section 8., pp.. a. m n, 5 5, (, 9) r 7, 9 t, 5, tr; r 7 t, 9 5t, tr, t5,, tr; 5t, t, tr. a. q It produces a different lin M(, 5) r N(9, 5). a. r, t8, 7, tr; r 8t, 7t, tr, 5 t,, tr; r t, 5 t, tr, t,, tr; r t,, tr, t,, tr;, t, tr. If the two lines have direction vectors that are collinear and share a point in common, then the two lines are coincident. In this eample, both have (, ) as a parallel direction vector and both have, as a point on the lin Hence, the two lines are coincident. 5. a. The normal vectors for the lines are, and,, which are collinear. Since in two dimensions, an two direction vectors perpendicular to, are collinear, the lines have collinear direction vectors. Hence, the lines are parallel. k a.. a a. acute: 5, obtuse: 5. a., 8 A(, ) C(, ) B(8, ) CA # CB 8 Since the dot product of the vectors is, the vectors are perpendicular, and ACB 9.. The sum of the interior angles of a quadrilateral is. The normals make 9 angles with their respective lines at A and C. The angle of the quadrilateral at B is 8 u. Let represent the measure of the interior angle of the quadrilateral at O u u u Therefore, the angle between the normals is the same as the angle between the lines.. ; Section 8., pp a.. a.. a. CA,, CB 8, 8,,, 8,,,,,,,, a,, 5 b,, 9,,,,,,,,, r,,, t, 5,, q tr;,, 5 s, 5,, sr t, 5t, t, tr; s,. a. r 5s, 5 s, sr, 5, t,,, tr t, 5,, tr Since two of the coordinates in the direction vector are ero, a smmetric equation cannot eist. 5. a. r,, t,,, tr; t, t, t, tr; Answers 77

9 r,, t,,, tr;, t, t, tr; r,,, t,,, tr;, t, t, tr; r,,, t,,, tr;, t,, tr; Since two of the coordinates in the direction vector are ero, there is no smmetric equation for this lin r t,,, tr; t, t,, tr; r,,, t,,, tr;,, t, tr; Since two of the coordinates in the direction vector are ero, there is no smmetric equation for this lin. a. t, t, 7 t, tr; 7 s, s, 5, sr about The directional vector of the first line is 8,,,,. So,,, is a directional vector for the first line as well. Since,, is also the directional vector of the second line, the lines are the same if the lines share a point. (,, ) is a point on the second lin Since 7 5 (,, ) 8, is a point on the first line as well. Hence, the lines are the sam 8. a. The line that passes through (,, ) with a directional vector of,, is given b the parametric equation is t, t, t, tr. So, the -coordinate is equal to onl when t. At t, and 5. So, A,, 5 is a point on the lin So, the -coordinate is equal to 5 onl when t 5. At t 5, 5 5 and 5 7. So, B5, 5, 7 is a point on the lin t, t, t, t a. 8,,,, 8,,,, 5 9,, 5, (,, ),,, 9,,, (,, ),,,,,,, 5, 8,,. a. t, t, 5 8t, tr; 5 8 r,, 9 s5,,, r 9 sr; 5,, t,,, tr; t, t, t, tr r,, t,, 5, tr;, t, 5t, tr. t, 5 5t, t, tr.,,,,,.. P,, and P,, 5. about 7 Mid-Chapter Review, pp a. 7,, 5,,,, 5, (, ), (5, ) Q, Q, 8, 5 R, 5 R,,,,,,,,, 5. a. Q 8 (, ) 5, R; Q, R;,. approimatel ais: about 5 ; -ais: about t, 9t, 5 t, tr approimatel 79., 7.7, and , -ais: t,,, tr; -ais:, t,, tr; -ais:,, t, tr. a units,. a. r, t,, tr t, t, tr about r.9 t,, tr. r ;, t,, tr; t, t, tr 5. Q 5, 5 R. a. 5 t, t, tr t, t, tr, t, tr 7. a. -plane at (, 8, ); -plane at (,, ); -plane at (,, ) -ais at (,, ) (,, ) (, 8, ) 8. a. r,, 8 t5,,, tr; 5t, t, 8 t, tr; 8 r 5,, 9 t,,, tr; t, t, 9 t, tr; r 9,, t, 5,, tr; t, 5t, t, tr; r 5,, t,,, tr;,, t, tr; There is no smmetric equation for this lin 9. r t5, 5,, tr.. t, 8 t,, tr,, 5,,, Since 8, the point 7,, 8 lies on the lin Section 8. pp. 59. a. plane; This is a vector equation of a plane in R. line; This is a vector equation of a line in R. line; This is a parametric equation for a line in R. plane; This is a parametric equation of a plane in R using (,, ) as r.. a.,, 9, r, 5,, s,, 9 t,, 5, t, sr. a.,,,, and, 5,, 7, m and n For the point B, 5, 8, the first two parametric equations are the same, ielding m and n ; however, the third equation would then give: 8 m n which is not tru So, there can be no solution. 78 Answers

10 . a. r,, t,,, t, r s,, sr,, t,,, t, 5. a. r s,, sr,, s,, t,, 8, t, sr, does not represent a plane because the direction vectors are the sam We can rewrite the second direction vector as,,. And so we can rewrite the equation as: r,, s,, t,,,, s t,,,, n,,, nr This is an equation of a line in.. a. r R,, 7 t,, s,,, t, sr; t s, t s, t, r 7 s, sr,, t,, s,,, t, sr; t s, t, t, r s, sr,, t,, s7,,, t, sr; t 7s, t s, t s, t, sr 7. a. s and t, 5,,, s,, t,, gives the following parametric equations: s t t s 5 s t 5 s s s s t a b t The third equation then sas: s t 8. a. 7 which is a false, statement. So, the point A, 5, is not on the plan l p, 5, s,,, sr;, 5, t,,, tr, 5, 9. (,, 5). r,, s,, 5 t,,, t, sr. r m,, 7 n,,, m, nr. a. (,, ), (,, ) and (,, ),, r, s,, t,,, t, sr; s, t, t,. a. r, sr s,, 5 t,, 7, t, r sr,, s,, 5 t,, 7, t, sr The planes are parallel since the have the same direction vectors.., 7,,,, 5,, 7 7,,, 7, 5. r, 5, 7,, t,,, tr. The fact that the plane r OP sa tb contains both of the given lines is easil seen when letting s and t, respectivel. Section 8.5 pp a. n A, B, C, 7, 8 In the Cartesian equation: A B C D If D, the plane passes through the origin. (,, ),. a. n,,,,, A, B, C, 5, In the Cartesian equation: D. So, the plane passes through the origin.,. a. n,, 5,, 5,, A, B, C,, In the Cartesian equation: D. So, the plane passes through the origin.,,,,,,,. a Method : Let A(,, ) be a point on the plan Then, PA,, 5 is a vector on the plan n # PA Method : n., 7, 5 so the Cartesian equation is 7 5 D We know the point,, 5 is on the plane and must satisf the equation, so 7 55 D D D This also gives the equation: a a. There is onl one simplified Cartesian equation that satisfies the given information, so the equations must be the sam a,, b a,, b a. First determine their normal vectors, n and n. Then the angle between the two planes can be determined from the formula: cos u n # n n n. a a. 8 5 No, the planes cannot ever be coincident. If the were, then the would also be parallel, so k 8, and we would have the two equations: Here all of the coefficients are equal ecept for the D-values, which means that the don t coincid a,, b Section 8., pp a. A plane parallel to the -ais, but two units awa, in the negative direction. A plane parallel to the -ais, but three units awa, in the positive direction. A plane parallel to the -ais, but units awa, in the positive direction. Answers 79

11 .,,. P must lie on plane p since the point has an -coordinate of 5, and doesn t have a -coordinate of.. 8. a.. a. 5. a. i. -intercept 9, -intercept, no -intercept ii. -intercept, -intercept, -intercept iii. no -intercept, -intercept, -intercept 9 i. (,, ),,, ii. (,, ), 5,, iii. (,, ), (,, ). a. i. (,, ), (,, ), (, 5, ) ii. 9. a. 7. -plane, -plane, -plane. a Answers

12 Review Eercise, pp Answers ma var. For eample, r,, s,, t,, tr; s t, s, s t. AC BC,, 5 r,, b c,, s,, 5 t,,, s, tr b c,,,, 5,, A B C D D D D Both Cartesian equations are the same regardless of which vectors are use. a. Answers ma var. For eample, r,, 9 t7,,, tr; 7t, t, 9 t, tr; 9 7 Answers ma var. For eample, r,, 9 t7,, s,,, t, sr; 7t s, t s, 9 t s, t, sr There are two parameters.. r 7,, t,,, tr; 7 t, t, t; 7 5. a r 7 8,, t,, s,, t, sr;, t, s Answers ma var. For eample, r,, s,,, sr; s, s, s;. Answers ma var. For eample, r,, s,, t, 5,, s, tr; s t, 5t, s t. Answers ma var. For eample, r,, 7 t,,, tr; t,, 7 t; 7,. r,, s,, 5 t,,, s, tr; s t, s t 5s t, s, tr; a. 5. a. Answers ma var. For eample, r,, t,, s,,, t, sr; t s, t s, t s; 5 8 Answers ma var. For eample, BC,, D 8 8 a,, b s a,, b t,,, s, tr 8 Answers ma var. For eample, r,, t,, 5 s,,, t, sr; t, t, 5t s; Answers ma var. For eample, r,, 5 t,, 9 s, 9,, t, sr; t s, t 9s, 5 9t s; The are in the same plane because both planes have the same normal vectors and Cartesian equations. L : r,, s, 5, L : r t,,, s, tr,, u,, v, 5,, u, vr, 5,,,, 9,,,,,, 5,, 9,,, A B C D D D D D D 7. a,, b 8. a. The plane is parallel to the -ais through the points (,, ) and,,. The plane is parallel to the -ais through the points (,, ) and,,. The plane is parallel to the -ais through the points (,, ) and,,. 9. a. A a 8, b. a a a. i. no ii. es iii. no i. es ii. no iii. no.,,,, p,, q,,,,,,,,,,,,, 5, 8, 5,. s t, t, s t, s, tr Answers 8

13 5. A plane has two parameters, because a plane goes in two different directions, unlike a line that goes onl in one direction.. This equation will alwas pass through the origin, because ou can alwas set s and t to obtain (,, ). 7. a. The do not form a plane, because these three points are collinear. r,, t,, The do not form a plane, because the point lies on the lin r r, 9, t,,, 9,,, 8, 7, 5 8. bc ac ab abc a., r,, t, 7, s5,, t, sr; t 5s, 7t s, t s 9 no. a a. These lines are coincident. The angle between them is. Q,. a. r 8.8, 5R,, 5 t,,, tr; t, t, 5 t; 5 r,, 5 t8,,, tr; 8t, t, 5 t; 5 r 8,, 5 t,,, tr; t, t, 5 t; 5 r,, 5 t,,, tr; t,, 5, t, 5 t; tr r r,, 5 t,,, tr,, 5 t,, ;, t, 5 t. a Chapter 8 Test, p. 8. a. i. r,, s,, t,,, s, tr; s t, s t, s, s, tr ii. 8 8 no. a. (,, ). a. r s,, t,, 5, s, tr. a. r 7,, 5 s,, t5,,, s, tr 5. a. 5. a. about 7.5 i. ii. 5 The -intercepts are different and the planes are different. 7. a a, 5, b The equation for the plane can be written as A B. For an real number t, A B t, so (,, t) is on the plan Since this is true for all real numbers, the -ais is on the plan Chapter 9 Review of Prerequisite Skills, p. 87. a. es es no no. Answers ma var. For eample: a. r, 5 t5,, tr; r 5t, 5 t, tr, 7 t7,, tr; r 7t, 7 t, tr, t,, tr; r t, t, tr,, 5 t5,, 5, tr; 5t, t, 5 5t, tr r,, t, 5,, tr; t, 5t, t, tr r, 5, t,,, tr; t, 5 t, t, tr. a L is not parallel to the plan L is on the plan L is parallel to the plan L is not parallel to the plan. a. 7. r,, t,, s,,, s, tr 8. Section 9., pp a. p: r,,, s5,, sr This line lies on the plan. a. A line and a plane can intersect in three was: () The line and the plane have ero points of intersection. This occurs when the lines are not incidental, meaning the do not intersect. () The line and the plane have onl one point of intersection. This occurs when the line crosses the plane at a single point. () The line and the plane have an infinite number of intersections. This occurs when the line is 8 Answers

14 coincident with the plane, meaning the line lies on the plan Assume that the line and the plane have more than one intersection, but not an infinite number. For simplicit, assume two intersections. At the first intersection, the line crosses the plan In order for the line to continue on, it must have the same direction vector. If the line has alread crossed the plane, then it continues to move awa from the plane, and can not intersect again. So, the line and the plane can onl intersect ero, one, or infinitel man times.. a. The line r s,, is the -ais. The plane is parallel to the -plane, but just one unit awa to the right. There are no intersections between the line and the plan. a. For, if we substitute our parametric equations, we have t t t All values of t give a solution to the equation, so all points on the line are also on the plan For the plane, we can substitute the parametric equations derived from r, 5, t,, : t 5 t t All values of t give a solution to this equation, so all points on the line are also on the plan 5. a. s s s 5 Since there are no values of s such that 5, this line and plane do not intersect. t 5t t Since there are no values of t such that, there are no solutions, and the plane and the line do not intersect.. a. The direction vector is m,, and the normal is m # n,,, n. So the line is parallel to the plane, but 5. So, the point on the line is not on the plan The direction vector is m and the normal is m # n, 5,,,, n,, so the line is parallel to the plan and So, the point on the line is not on the plan 7. a. 9,,,, 8. a. There is no intersection and the lines are skew.,, 9. a. not skew not skew not skew skew. 8. a. Comparing components results in the equation s t for each component. From L, we see that at,,, s. When this occurs, t. Substituting this into L, we get,, 7,,,,. Since both of these lines have the same direction vector and a common point, the lines are coincidental.. a. a, 5, b.. a.,, 5. a. (,, ) r,, t, 55,, t R. a. L (,, ) If p and q, the intersection occurs at (,, ). L 7. a. Represent the lines parametricall, and then substitute into the equation for the plan For the first equation, t, 7 8t, t. Substituting into the plane equation, t 7 8t t. Simplifing, t. So, the line lies on the plan For the second line, s,, s. Substituting into the plane equation, 8 s s. Simplifing, s. This line also lies on the plan,, 8. Answers ma var. For eample, r,, p,,, pr. Section 9., pp a. linear not linear linear not linear. Answers ma var. For eample: a. 5,, 8. a. es no. a.,, The two sstems are equivalent because the have the same solution. 5. a.. a. These two lines are parallel, and therefore cannot have an intersection. The second equation is five times the first; therefore, the lines are coincident. 7. a. 8. a. 9. a.,, 5, t, t, t R t, s, s t, t R t t 7 t t 7 k not possible k Answers 8

15 . a infinitel man t, t R t, This equation will not have an integer solutions because the left side is an even function and the right side is an odd function.. a. a b, b a Since the have different direction vectors, these two equations are not parallel or coincident and will intersect somewher. a.,,,,,, 8,, 8,,,,. Answers ma var. For eample: a. Three lines parallel Two lines coincident and the third parallel L L L L L L Two parallel lines cut b the third line L L L The lines form a triangle Lines meet in a point Three coincident lines L L L L. a c, a b c, a b 5. a. k k k ; Section 9., pp L. a. The two equations represent planes that are parallel and not coincident. Answers ma var. For eample:,. a. s, t; s t, s, tr; the two planes are coincident. Answers ma var. For eample:,. a. s, s,, sr; the two planes intersect in a lin Answers ma var. For eample:,. a. m p q, q, and p ;, The value for m is unique, but p just has to be twice q and arbitrar values can be chosen. L L L L m q, and p ;, The value for m is unique, but p and q can be arbitraril chosen as long as p q. m ; This value is unique, since onl one value was found to satisf the given conditions. m, p, q ; The value for m is unique from the solution to, but the values for p and q can be arbitrar since the onl value which can change the angle between the planes is m. 5. a. 9s, s, s, sr t, t, tr t, Since t is an arbitrar real number, we can epress t as part t s, sr.. a. es; plane no es; line es; line es; line es; line 7. a. s t, s, t, s, tr no solution s,, s, sr s 5, s, s, sr 5 s, sr s, s, s 8, s,, sr 8. a. The sstem will have an infinite number of solutions for an value of k. No, there is no value of k for which the sstem will not have a solution. 9. r,, s5, 8,, sr. The line of intersection of the two planes, s, s, s; sr; 5 5 s s s 5 s s s Since this is true, the line is contained in the plan. a. s, s, s, sr about Answers

16 Mid-Chapter Review, pp a.,,,,,, 5. a, Answers ma var. For eample: t, t, t, tr; t,, 5 tr; 8 7t, 5 t, 7 t, tr,, 5 C: 8 7t, 5 t, 7 t, tr t 8 7, 5, 7. a.,, 5,, 5, r, 5 7,, t,,, tr r a 9 7,, b t,, 7, 7 tr 7,,. a. t t 7 5, 5, t, tr 7 s 5 s 9 5 5, 5, s, tr The lines found in.a. and. do not intersect, because the are in parallel and distinct planes. 5. a. a a a ;, ar. Since there is no t-value that satisfies the equations, there is no intersection, and these lines are skew. 7. a. no intersection The lines are skew. 8.,, 9. a.,, These lines are the same, so either one of these lines can be used as their intersection.. a. Answers ma var. For eample: i. coincident ii. parallel and distinct iii. skew iv. intersect in a point L L i. When lines are the same, the are a multiple of each other. ii. When lines are parallel, one equation is a multiple of the other equation, ecept for the constant term. iii. When lines are skew, there are no common solutions to make each equation consistent. iv. When the solution meets in a point, there is onl one unique solution for the sstem. L L L L L L. a. when the line lies in the plane Answers ma var. For eample: r ; r t, 5,, tr t, 5, s,,, t, sr. a., 8 no solution,,. a. The two lines intersect at a point. The two planes are parallel and do not meet.. a. a,, b u 9 Section 9., pp a. 9, 5, This solution is the point at which all three planes meet.. a. Answers ma var. For eample, and 8. These three planes are intersecting in one single plane because all three equations can be changed into one equivalent equation. The are coincident planes. t, s, s t, s, tr t, s, t s, s, tr. a. Answers ma var. For eample,,, and 7 ;,, and 8 5. no solutions. a. a,, b This solution is the point at which all three planes meet. 5. a. Since equation equation, equation and equation are consistent or lie in the same plan Equation meets this plane in a lin, t, and t, tr. If ou multipl equation b 5, ou obtain a new equation, , which is inconsistent with equation. 7. a. Yes, when this equation is alone, this is tru Answers ma var. For eample: Answers 85

17 8. a.,, is the point at which the three planes meet.,, is the point at which the three planes meet. 99,, is the point at which the three planes meet.,, is the point at which the three planes meet. 9. a. 5 and 7 7 t 9 7, 7 t, t, tr; the planes intersect in a lin no solution t,, and t, tr; the planes intersect in a lin. a., t, and t, tr t s, t, and s, s, tr. a. Equation equation Equation or Equation equation Equation 5 or Since the -variable is different in Equation and Equation 5, the sstem is inconsistent and has no solution. Answers ma var. For eample: n,, n,, n,, m n n,, m n n,, m n n,, The three lines of intersection are parallel and coplanar, so the form a triangular prism. Since n n # n, a triangular prism forms.. a. Equation and equation have the same set of coefficients and variables; however, equation equals, while equation equals, which means there is no possible solution. All three equations equal different numbers, so there is no possible solution. Equation equals 8, while equation equals 7, which means there is no possible solution. The coefficients of equation are half the coefficients of equation, but the constant term is not half the other constant term.. a.,, 5 5t 5 t, t, tr,, t, t, tr no solution t,, t, tr,,. a. p q 5 t, t, t, tr 5. a. m m ;, mr m.,, Section 9.5, pp a. 5 5 or. or a. 5 or. 5 5 or. 5. a.. about.9 about.88. a. d A B C A B If ou substitute the coordinates,, the formula changes to A B C d, A B which reduces to d 5 the answers are the same 5 ; 5. a. 7 or. 5 or. or 8.. a. about.8 about.8 about. 7. a. about.8 about.8 8. a. a 7, 7, b about.5 C A B. 9. about.;. a 8,, 7 b. a. about.75 D and G about. units Section 9., pp a. Yes, the calculations are correct. Point A lies in the plan The answer means that the point lies in the plan. a. or. 5 7 or.8. a. 5 8 Answers ma var. For eample: a,, b a, 5, b. a. 5. or about.5 8. a. about. U,, is the point on the first line that produces the minimal distance to the second line at point V,,. Review Eercise, pp no solution. a. no solution 99,,. a. All four points lie on the plane about.9 5. a. or.8. r,, t,,, tr 7. a. no solution no solution no solution 8 Answers

18 8. a t, tr 7 t, 7 t,,, t s 7, t, s, s, tr 9. a. 5 t, t, t, tr 9 t, 8 t, t, tr. a. These three planes meet at the point, 5,. The planes do not intersect. Geometricall, the planes form a triangular prism. The planes meet in a line through the origin, with equation t, 7t, 5t, tr..9. a. r m,. n, 5 s,,, sr,,,, Since the line s direction vector is perpendicular to the normal of the plane and the point,,5 lies on both the line and the plane, the line is in the plan,, 5 5 The point,, 5 is on the plane since it satisfies the equation of the plan 7 5. a. 5.8,,. a. r,,.,, t,,, tr 5. a. 9 8 about.5. a. r,, t,, 7, tr 7. a.,, 7 t, t, t, tr a, b, c a, 7, 7 b a 5, 8, b. a. r a 5,, b t,, 5, tr;. r a 7 5,, b ; r t,, 5, tr 7,, t,, 5, tr; 5t, tr All three lines of intersection found in part a. have direction vector,, 5, and so the are all parallel. Since no pair of normal vectors for these three planes is parallel, no pair of these planes is coincident. a,, b, a,, b, a,, b, a,, b, a,, b, a a and,, b,, b, a,, b. 7. a 9 7, 7, 7 b 5. A. a. r 5, B, C,, t 5,,, tr a,, b about. units Chapter 9 Test, p. 55. a.,, a. or.8 or.. a. t t t, tr 5, 5,,, 5. a., 5, The three planes intersect at the point, 5,. 5. a. t t, t,, tr The three planes intersect at this lin. a. m, n, t, t, tr 7.. Cumulative Review of Vectors, pp a. about. scalar projection:, vector projection: a 5 9, 5 9, 8 9 b scalar projection:, vector projection: a 8 9, 9, 8 9 b. a. 8 t, t, t, tr about 5.9. a.. a. 7i 9j k ais: about., -ais: about.8, -ais: about.9. a. 7, 5,,, about 8. square units 7. a and a,,, b, b 8. a. vector equation: Answers ma var. r,, t, 5,, tr; parametric equation: t, 5t, t, tr If the -coordinate of a point on the line is, then t, or t. At t, the point on the line is,,, 5,,,. Hence, C,, is a point on the lin 9. The direction vector of the first line is, 5, and of the second line is, 5,, 5,. So the are collinear and hence parallel. The lines coincide if and onl if for an point on the first line and second line, the vector connecting the two points is a multiple of the direction vector for the lines. (,, 9) is a point on the first line and, 5, is a point on the second lin,, 9, 5,, 5, k, 5, for kr. Hence, the lines are parallel and distinct. Answers 87

19 . vector equation: r,, t,,, tr; parametric equation:, t, t, tr. (,, ) (,, ). a,, b. a. (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) Answers ma var. For eample,,, and,,. (,, ) (,, ) (,, ) Answers ma var. For eample,,, and,,. Answers ma var. For eample,,, and,,.. 7, 5. q,,, t, 7,, tr. a. 9 about.9 units 7. a km h, N. W 9. a. r >,, s,, t,,, s, tr. To verif, find the Cartesian equation corresponding to the above vector equation and see if it is equivalent to the Cartesian equation given in the problem. A normal vector to this plane is the cross product of the two directional vectors. n,,,,,,,, So the plane has the form D, for some constant D. To find D, we know that (,, ) is a point on the plane, so D. So, D, or D. So, the Cartesian equation for the plane is. Since this is the same as the initial Cartesian equation, the vector equation for the plane is correct.. a. The two planes are perpendicular if and onl if their normal vectors are also perpendicular. A normal vector for the first plane is,, and a normal vector for the second plane is,, 7. The two vectors are perpendicular if and onl if their dot product is ero.,, #,, 7 7 Hence, the normal vectors are perpendicular. Thus, the planes are perpendicular. The two planes are parallel if and onl if their normal vectors are also parallel. A normal vector for the first plane is,, and a normal vector for the second plane is,,. Since both normal vectors are the same, the planes are parallel. Since, the point,, is on the second plan Yet since,,, is not on the first plan Thus, the two planes are parallel but not coincident.. resultant: about 5.79 N, 7. from the 5 N force toward the N force, equilibrant: about 5.79 N,. from the 5 N force awa from the N force. a. b a a b 88 Answers

20 . a. a + b a 7, 7, 7 b a a 7, 7, 7 b b. a. OC BD 8, 9,, 5 about 7.9 about a. t, t,, tr,,, t, t, tr s t, t, s, s, tr. a. es;, t, t, tr no es; t, t, t, tr a r t,,, tr,.,,. a..8 km min. a. Answers ma var. r,, t,,, tr The line found in part a will lie in the plane if and onl if both points A,, and B,, lie in this plan We verif this b substituting these points into the equation of the plane, and checking for consisten For A: For B: Since both points lie on the plane, so does the line found in part a.. km>h at N 5. E. parallel: 9 N, perpendicular: about 9.8 N 5. a. True; all non-parallel pairs of lines intersect in eactl one point in R. However, this is not the case for lines in R (skew lines provide a countereample). True; all non-parallel pairs of planes intersect in a line in R. True; the line has direction vector (,, ), which is not perpendicular to the normal vector,, to the plane k, k is an constant. Since these vectors are not perpendicular, the line is not parallel to the plane, and so the will intersect in eactl one point. False; a direction vector for the line is (,, ). A direction vector for the line is,,, or (,, ) (which is parallel to,,. Since (,, ) and (,, ) are obviousl not parallel, these two lines are not parallel.. a. A direction vector for L :, is (,, ), and a direction vector for L : k is (,, k). k But (,, ) is not a nonero scalar multiple of (,, k) for an k, since the first component of (,, ) is. This means that the direction vectors for L and L are never parallel, which means that these lines are never parallel for an k. ;,, Calculus Appendi Implicit Differentiation, p. 5. The chain rule states that if is a composite function, then d d du To differentiate an d du d. equation implicitl, first differentiate both sides of the equation with respect to, using the chain rule for terms d involving, then solve for. a. d a (, ) 5. a a. a and a 5, 5 b 5, 5 b 8. a. 9 ;. a. 7 d d 8 d 8Q d R one tangent one tangent Answers 89

21 ... one tangent two tangents a b d d a d b d = k Let Pa, b be the point of intersection where a and b. For k, d d d d At Pa, b, d d a b d d d d d as require d, and = p Multipl b : d d a a b d b d 5 d d d d 5 d d Q 5 5 R d d = p P(a, b) = k For P, d # d P d d At Pa, b, d d b a At point Pa, b, the slope of the tangent line of P is the negative reciprocal of the slope of the tangent line of k. Therefore, the tangent lines intersect at right angles, and thus, the two curves intersect orthogonall for all values of the constants k and P. 5. d d d d Let Pa, b be the point of tangen d b d a Equation on tangent line l and P is b a b a. -intercept is found when. b a b a ba b ab ab ba b Therefore, the -intercept is ab ba. b For the -intercept, let, b a b a. ab -intercept is a The sum of the intercepts is ab ba ab ba b a a b ab b a a b a b a ab b a b a ab b a b Since Pa, b is on the curve, then or a a b k, b k. Therefore, the sum of the intercepts is k k, as require and 8 Related Rates, pp a. da m>s dt ds dt m >min ds 7 km>h, when t.5 dt d d dt dt du rad> s dt p. a. decreasing at 5.9 C> s about.58 m Solve T.. area increasing at cm > s; perimeter increasing at cm> s. a. increasing at cm > s increasing at cm > s 5. increasing at cm > s. a. 5 p km>h 5 p m>s p km>h m>s 8 m>min m>s 5 km>h. a. 7p cm>s cm>s or about. cm>s 9p cm>s or about. cm>s 8p.. 5 cm>min; 9.5 min or about.5 h p Answers ma var. For eample: a. The diameter of a right-circular cone is epanding at a rate of cm> min. Its height remains constant at cm. Find its radius when the volume is increasing at a rate of 8p cm > min. Water is being poured into a right-circular tank at the rate of p m > min. Its height is m and its radius is m. At what rate is the water level rising? The volume of a right-circular cone is epanding because its radius is increasing at cm> min and its height is increasing at cm> min. Find the rate at which its volume is changing when its radius is cm and its height is cm. 5..5p m >ear 9 Answers

22 p cm>min m>min m>min.8 km>h. a. 5p cm>s 8 5p cm>s. a. a l b k l k The Natural Logarithm and its Derivative, p A natural logarithm has base e; a common logarithm has base.. Since e lim h n, let h n. hs Therefore, e lim a n. n S n b But as n Sq. n S, Therefore, e lim nsq n n. If e a n, b..7 8 Tr n, et. a t t t t t 5. a. ln e t ln t et t e e e te t ln t e t tln t eu a eu ln u u b 5. a. e. The value shown is approimatel e, which matches the calculation in part a. This value matches the calculation in part. a. no solution, ;e 7. a. 8. The equation on the calculator is in a different form, but is equivalent to the equation in part a. ln 9. a. a and (, ) e, e b The solution in part a is more precise and efficient.. ln. a. 9 km>h 9 t about.8 km>h>s. s.. a. ln The function s domain is 5R 7. The domain of the derivative is 5R 7 and. The Derivatives of General Logarithmic Functions, p a. ln 5 ln ln ln 7 ln ln. a. ln ln 8 ln 5 ln ln 8 ln 8 ln 7. a. 5 5 ln 8 log 8ln ln. a. ln ln ln 5 ln ln ln ln ln Answers 9

23 ln ln ln ln log a k d f d f ln a k k ln a ln a Since the derivative is positive at t 5, the distance is increasing at that point. 9. a vertical asmptote at The tangent line will intersect this asmptote because it is defined for.. D 5R or 7 ; critical number at,, and ; function is decreasing for and increasing for 7. a. point of inflection at is a possible point of inflection. Since the graph is alwas concave up, there is no point of inflection.. The slope of log at (, ) is ln. Since ln 7, the slope of at (, ) is greater than the slope of log at,. Logarithmic Differentiation, p. 58. a. 5 pt p e e e. a. ln ln a b Q ln R a t t b a ln t b. a. e e e e # e 7. ln 8 ln e, e e 7., and, ln ln 8 ln 8 8 a sin b sin cos a sin b cos a sin ln cos b. a. vt at t t ln t ln t t t ln t t ae e e t e;. Using a calculator, e p. and p e.. So, e p 7 p e. Vector Appendi Gaussian Elimination, pp a.. Answers ma var. For eample: c 5.5 d. Answers ma var. For eample: 7. a. Answers ma var. For eample:, 8, a.. a. t t ln t a t b, c d 9,, 9, no solution 9,, 5 9 Answers

24 t s, s, t, s, tr, 8, 7. a. It satisfies both properties of a matri in row-echelon form.. All rows that consist entirel of eros must be written at the bottom of the matri.. In an two successive rows not consisting entirel of eros, the first nonero number in the lower row must occur further to the right that the first nonero number in the row directl abov A solution does not eist to this sstem, because the second row has no variables, but is still equal to a nonero number, which is not possibl Answers ma var. For eample: 8. a. no; Answers ma var. For eample: no; Answers ma var. For eample: no; Answers ma var. For eample: es 9. a. i. 5,, ii. 7,, iii. t, t,, tr iv. 9t, t, t, tr i. The solution is the point at which the three planes meet. ii. The solution is the point at which the three planes meet. iii. The solution is the line at which the three planes meet. iv. The solution is the line at which the three planes meet.. a.,, The three planes meet at the point,,. t, t, t, tr The three planes meet at this lin, t, t, tr The three planes meet at this lin,, The three planes meet at the point,,., t, t, tr The three planes meet at this lin 5,, 5 The three planes meet at the point 5,, 5.. 7a c 5b, c b 5a, c b a.. 7 p, q 9, r 9. a. a a a or a Gauss-Jordan Method for Solving Sstems of Equations, pp a. c 7 d 8. a. 7, (,, ) (,, ) 8,,. a.,,, 5, 7,,,,,,,,. a.,, 8, 8, 5. a. k k Z, kr The matri cannot be put in reduced row-echelon form.. a. Ever homogeneous sstem has at least one solution, because (,, ) satisfies each equation. 7.,, The reduced row-echelon form shows that the intersection of these planes is a line that goes through the point,,.,, t, tr t t Answers 9

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