United Arab Emirates University College of Sciences Department of Mathematical Sciences HOMEWORK 1 SOLUTION. Section 10.1 Vectors in the Plane

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1 United Arab Emirates University College of Sciences Deartment of Mathematical Sciences HOMEWORK 1 SOLUTION Section 10.1 Vectors in the Plane Calculus II for Engineering MATH 110 SECTION 0 CRN 510 :00 :00 on Monday & Wednesday Due Date: Wednesday, October 6, 010

2 Calculus II for Engineering HOMEWORK 1 SOLUTION Fall, Comute a + b for a = h ; i and b = h ; 1 i. a + b = h ; i + h ; 1 i = h 9; 6 i + h 6; 6 i = h 9 + 6; 6 6 i = h ; i :. Determine whether the vectors a = h 1; i and b = h ; 1 i are arallel. We recall that a and b are arallel if and only if there is a scalar s such that a = sb. If a = sb, then h 1; i = s h ; 1 i = h s; s i ; 1 = s and = s: There is no such a scalar s satisfying both s = 1 and s = arallel.. Therefore, given two vectors cannot be. Find the vector with initial oint A(; ) and terminal oint B(5; ).! AB = h 5 ; i = h ; 1 i :. Find a vector with the magnitude in the same direction as the vector v = i j. Let u be such a vector, i.e., the one with the magnitude in the same direction as the vector v = i j. Since u is arallel to v, there should be a ositive scalar s (negative scalar gives the oosite direction) such that u = sv. Since u has the magnitude, we have i:e:; jsj = = kuk = ksvk = jsjkvk = jsjki jk = jsj 5; 5 = Therefore, we deduce such a vector u = ; i:e:; s = 5 5 : 5 (i j) : Page 1 of

3 Calculus II for Engineering HOMEWORK 1 SOLUTION Fall, The thrust of an airlane s engines roduces a seed of 600 mh in still air. The wind velocity is given by h 0; 60 i. In what direction should the airlane head to fly due east? Let v = h x; y i and w = h 0; 60 i be the velocities of the airlane and the wind, resectively. Since we want the airlane to move due east, the sum of two vectors v and w should satisfy v + w = h c; 0 i, where the vector h c; 0 i oints due east with the ositive constant c. The equation imlies h c; 0 i = v + w = h x; y i + h 0; 60 i = h x 0; y + 60 i ; i:e:; c = x 0; and y = 60: The airlane roduces a seed of 600 mh and so we get kvk = 600, which imlies q 600 = kvk = h x; y i = x = = 180 x + y ; 600 = x + y = x + ( 60) ; Because of the condition that x 0 = c > 0, we choose x = , hence, D v = h x; y i = 11: This oints right and down at an angle of tan 1 1 east. E D E ; 60 = 60 11; 1 : 11! 5:7917 0: (radian) south of 6. Suose a vector a has magnitude kak = and vector b has magnitude kbk =. (6.1) What is the largest ossible magnitude for the vector a + b? The largest magnitude of a + b is 7 (if the vectors oint in the same direction). (6.) What is the smallest ossible magnitude for the vector a + b? The smallest magnitude is 1 (if the vectors oint in the oosite directions). (6.) What will be the magnitude of a + b if a and b are erendicular? If the vectors are erendicular, then a + b can be viewed as the hyotenuse of a right triangle with sides a and b, so it has length + = 5. Page of

4 Calculus II for Engineering HOMEWORK 1 SOLUTION Fall, A vector i j is given. Here i = h 1; 0 i and j = h 0; 1 i. (7.1) Find two unit vectors arallel to the given vector. Let v = i j = (i j). Then it has the magnitude q kvk = k (i j) k = k (i j) k = + ( 1) = 5: The unit vector in the same direction as v can be found by v kvk (i j) = 5 = 5 1 (i j) : The unit vector in the oosite direction as v is v kvk = 1 (i j) : 5 Thus, two unit vectors arallel to v are 5 1 (i j). (7.) Write the given vector as the roduct of its magnitude and a unit vector. Using the results above, we have v = kvk v kvk = 5! i 1 j : If v V lies in the first quadrant of the xy lane and makes the angle = = with the ositive x axis and the magnitude, then find v in the comonent form. The unit vector making the angle = = with the ositive x axis is given by h cos(=); sin(=) i. (Why?) Hence, the desired vector v V is obtained by v = h cos(=); sin(=) i = * 1 ; + D = E ; : Page of

5 Calculus II for Engineering HOMEWORK 1 SOLUTION Fall, (Think!) Draw the vectors a = h ; i, b = h ; 1 i and c = h 7; 1 i. By using the sketch involved with those vectors, show that there exist scalars s and t such that c = sa + tb. Can you rove the existence of such s and t algebraically? Justify your answer. The equation c = sa+tb imlies h 7; 1 i = s h ; i + t h ; 1 i, i.e., 7 = s+t and 1 = s t. Solving the equations for s and t, we get s = 9 7 and t = Find the correct figure of the sum a+b with a = i j and b = i+j, where i and j are standard basis vectors of V. A simle comutation shows a + b = i j i + j = 5i + j = h 5; 1 i : That is, a + b should oint in the direction h 5; 1 i. It is easy to see that the red colored vector in () is the vector h 5; 1 i. Hence, the answer is () Page of

6 United Arab Emirates University College of Sciences Deartment of Mathematical Sciences HOMEWORK SOLUTION Section 10. Vectors in the Sace Calculus II for Engineering MATH 110 SECTION 0 CRN 510 :00 :00 on Monday & Wednesday Due Date: Wednesday, October 1, 010

7 Calculus II for Engineering HOMEWORK SOLUTION Fall, Find two unit vectors arallel to the vector i j + k. Let v = i j + k = (i j + k). Then it has the magnitude kvk = k (i j + k) k = ki j + kk = The unit vector in the same direction as v can be found by q + ( 1) + = 9 = 6: v kvk = (i j + k) 6 = 1 (i j + k) : The unit vector in the oosite direction as v is v kvk = 1 (i j + k) : Thus, two unit vectors arallel to v are 1 (i j + k).. Identify the lane y = as arallel to the xy lane, xz lane or yz lane and sketch a grah. The lane y = is arallel to the xz lane and asses through (0; ; 0).. Find the dislacement vectors! P Q and! QR and determine whether the oints P (; ; 1), Q(0; ; ) and R(; 1; ) are colinear (on the same line). The dislacement vectors are! P Q = h 0 ; ; 1 i = h ; 1; 1 i ;! QR = h 0; 1 ; i = h ; ; i : There does not exist any scalar s such that!! P Q = s QR, because no scalar s satisfies simultaneously = s; 1 = s; 1 = s: It imlies that two vectors! P Q and! QR are not arallel. That is, the oints P, Q and R are not colinear. Page 1 of

8 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010. Use vectors to determine whether the oints (; 1; 0), (5; 1; ), (0; ; ) and (; 1; 5) form a square. We recall that if a square has the side length s, then it has the diagonal of length s by the Pythagorean Theorem. Let P = (; 1; 0), Q = (5; 1; ), R = (0; ; ) and S = (; 1; 5). There are six airs of vectors (! P Q,!!!!!!! P R, P S, QR, QS, RS) and four of them will corresond to sides. Further P Q, P R and! P S should form two sides and one diagonal of the square, because they have the same initial oint. When we comute the lengths, we get k! P Qk = k h ; ; i k = 17; k! P Rk = k h ; ; i k = k! P Sk = k h 1; 0; 5 i k = 6: 17; It imlies! P S should be the diagonal. However, since 6 6= 17, i.e., k! P Sk 6= k! P Qk, so! P S cannot be the diagonal of a square. That is, those given oints cannot form a square. 5. In the accomanying figure, two roes are attached to a 00 ound crate. Roe A exerts a force of h 10; 10; 00 i ounds on the crate, and roe B exerts a force of h 0; 180; 160 i ounds on the crate. (5.1) If no further roes are added, find the net force on the crate and the direction it will move. Let the force due to roe A be a = h 10; 10; 00 i, the force due to roe B be b = h 0; 180; 160 i, and write the force due to gravity as w = h 0; 0; 00 i. Then the net force is a + b + w = h 10; 10; 00 i + h 0; 180; 160 i + h 0; 0; 00 i = h 10; 50; 60 i : (5.) If a third roe C is added to balance the crate, what force must this roe exert on the crate? In order to comensate, roe C must exert a force of h 10; 50; 60 i or 78.7 (= k h 10; 50; 60 i k) ounds in direction h 1; 5; 6 i. Page of

9 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010 (5.) We want to move the crate u and to the right with a constant force of h 0; 0; 0 i ounds. If a third roe C is added to accomlish this, what force must the roe exert on the crate? Let the force due to roe C be c. We want the net force to be a + b + c + w = h 0; 0; 0 i ; i:e:; c + h 10; 50; 60 i = h 0; 0; 0 i ; where a, b and w defined in (1) above are used. That is, c = h 0; 0; 0 i h 10; 50; 60 i = h 10; 0; 0 i : So roe C must exert a force of h 10; 0; 0 i or 5.8 (= k h 10; 0; 0 i k) ounds in direction h 1; ; i. Page of

10 United Arab Emirates University Faculty of Science Deartment of Mathematical Sciences HOMEWORK SOLUTION Section 10. Dot Product and Section 10. Cross Product Calculus II for Engineering MATH 110 SECTION 0 CRN 510 :00 :00 on Monday & Wednesday Due Date: Wednesday, October 0, 010

11 Calculus II for Engineering HOMEWORK SOLUTION Fall, Comute a b. (1.1) a = h ; ; 0 i and b = h ; ; i. Section 10. Dot Product a b = h ; ; 0 i h ; ; i = ( ) + () + 0() = : (1.) a = i k and b = j k. a b = (i k) (j k) = (0) + 0() + ( 1)( 1) = 1:. Comute the angle between the vectors. (.1) a = h ; 0; i and b = h 0; ; i. cos = a b kakkbk = cos ; h ; 0; i h 0; ; i = k h ; 0; i kk h 0; ; i k =! :555 (.) a = i + j k and b = i + j + k. cos = a b kakkbk = cos ; (i + j k) ( i + j + k) = ki + j kk i + j + kk = 8! 6 :111. Determine whether the vectors are orthogonal. (.1) a = h ; 1; 1 i and b = h ; ; i. So a and b are not orthogonal. a b = h ; 1; 1 i h ; ; i = 8 6= 0: (.) a = 6i + j and b = i + j. So a and b are orthogonal. a b = (6i + j) ( i + j) = 0:. Find a vector erendicular to the given vector. Page 1 of 9

12 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010 (.1) h ; 1; 1 i. Let v = h a; b; c i be a vector erendicular to u = h ; v u = 0, i.e., 1; 1 i. Then we should have 0 = v u = h a; b; c i h ; 1; 1 i = a b + c; i:e:; a b + c = 0: There are so many numbers, a, b, and c satisfying the equation. We choose just one, a = 1, b = and c = 0. That is, v = h 1; ; 0 i is one vector erendicular to u = h ; 1; 1 i. (.) 6i + j k. Let v = h a; b; c i = ai + bj + ck be a vector erendicular to w = 6i + j by the same argument as above, we deduce k. Then 0 = v w = (ai + bj + ck) (6i + j k) = 6a + b c; i:e:; 6a + b c = 0: There are so many numbers, a, b, and c satisfying the equation. We choose just one, a = 1, b = 1 and c =. That is, v = h 1; 1; i = i j + k is one vector erendicular to w = 6i + j k. 5. Find Com b a and Proj b a. (5.1) a = i + j and b = i j. Com b a = a b kbk Proj b a = (Com b a) (5.) a = h ; ; 0 i and b = h ; ; 1 i. Com b a = a b kbk Proj b a = (Com b a) (i + j) (i j) = ki jk b kbk = 9 5 h ; ; 0 i h ; ; 1 i = k h ; ; 1 i k b kbk = = 9 5 i j 9 ki jk = = (i j) 5 h ; ; 1 i k h ; ; 1 i k = h ; ; 1 i 9 6. A constant force of h 60; 0 i ounds moves an object in a straight line from the oint (0; 0) to the oint (10; 10). Comute the work done. The dislacement vector is d = h 10 0; 10 0 i = h 10; 10 i. The force is F = h 60; 0 i. By the formula, the work W done is obtained by W = F d = h 10; 10 i h 60; 0 i = 10(60) + ( 10)( 0) = 900: Page of 9

13 Calculus II for Engineering HOMEWORK SOLUTION Fall, Label each statement as true or false. If it is true, briefly exlain why; if it is false, give a counterexamle. (7.1) If a b = a c, then b = c F a = h 1; 0; 0 i, b = h 0; 1; 0 i and c = h 0; 0; 1 i satisfies a b = 0 = a c. But, obviously, b 6= c. (7.) If b = c, then a b = a c T If b = c, then b c = 0 and so a (b c) = a 0 = 0; i:e:; a b a c = 0; i:e:; a b = a c: (7.) a a = kak T The formula a b = kakkbk cos, where is the angle between a and b, imlies a a = kakkak cos 0 = kak ; i:e:; a a = kak : One may comute the dot roduct with a = h a; b; c i and rove the equality. (7.) If kak > kbk, then a c > b c F We observe a = h ; 0 i and b = h 0; 1 i satisfy the inequality kak = > 1 = kbk. However, with c = h 0; i, we get a c = 0 < = b c. (7.5) If kak = kbk, then a = b F We observe i = h 1; 0; 0 i, j = h 0; 1; 0 i and k = h 0; 0; 1 i have kik = 1 = kjk = kkk. However, obviously, i 6= j 6= k. 8. By the Cauchy Schwartz Inequality, ja bj kakkbk. What relationshi must exist between a and b to have the equality ja bj = kakkbk? (1) a = 0 or b = 0. () The cosine of the angle between the vectors is 1. This haens exactly when the vectors oint in the same or oosite directions. In other words, when a = sb for some scalar s. 9. By the Triangle Inequality, ka + bk kak + kbk. What relationshi must exist between a and b to have the equality ka + bk = kak + kbk? (1) a = 0 or b = 0. () Vectors a and b must be arallel with the same direction so that a = sb for some ositive scalar s. 10. The orthogonal rojection of vector a along vector b is defined as Orth b a = a Proj b a. Sketch a icture showing vectors a, b, Proj b a and Orth b a, and exlain what is orthogonal about Orth b a. Page of 9

14 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010 Orth b a is the comonent of a that is orthogonal to b: b Orth b a = b (a Proj b a) = b a b Proj b a = b a b (a b)b kbk = b a (a b)(b b) kbk = b a a b = 0; where b b = kbk is used above. 11. A car makes a turn on a banked road. If the road is banked at 15, show that a vector arallel to the road is h cos 15 ; sin 15 i. If the car has weight 500 ounds, find the comonent of the weight vector along the road vector. This comonent of weight rovides a force that hels the car turn. The vector b = h cos 15 ; sin 15 i reresents the direction of the banked road. The weight of the car is w = h 0; 500 i. The comonent of the weight in the direction of the bank is Com b w = w b kbk = 500 sin 15 67:0 lbs toward the inside of the curve. Page of 9

15 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010 Section 10. Cross Product 1. Comute the determinant: = 0( ) ( ) (1 + 1) = : 1. Comute the cross roduct a b. (1.1) a = h ; ; 0 i and b = h ; 0; 1 i. a b = h ; ; 0 i h ; 0; 1 i = i j k = i j + 6k = h 1; 1; i (1.) a = i + j k and b = j k. a b = h ; 1; i h 0; ; 1 i = i j k = 5i j k = h 5; ; i 1. Find two unit vectors orthogonal to the two given vectors. (1.1) a = h 0; ; 1 i and b = h 1; 0; 1 i. We recall Orthogonal Vector a b is orthogonal to both a and b. Thus, a b are two unit vectors orthogonal to both k a b k a and b. a b = h 0; ; 1 i h 1; 0; 1 i = h ; 1; i ; a b k a b k = 1 h ; 1; i (1.) a = i + j k and b = i k. a b = h ; ; i h ; 0; 1 i = h ; 8; 6 i ; a b k a b k = 1 h ; 8; i Page 5 of 9

16 Calculus II for Engineering HOMEWORK SOLUTION Fall, Use the cross roduct to determine the angle between the vectors, assuming that 0 =. (15.1) a = h ; ; 1 i and b = h 0; 0; i. We recall Angle Between Two Vectors k a b k = k a k k b k sin ; i:e:; sin k a b k! = k a k k b k ; i:e:; k a b k = sin 1 k a k k b k where 0 is the angle between a and b. a b = h 1; 1; 0 i ; k a k = ; k b k = ; k a b k = ;! = sin 1 6 1:096 rad 70:588 (15.) a = i + j + k and b = i + j. a b = h ; 6; 5 i ; k a k =! = sin ; k b k = 5; k a b k = 70; 1:01 rad 59: Find the distance from the oint Q(1; ; 1) to the line through (1; ; ) and (1; 0; ). The distance d from the oint Q to the line through the oints P and R is obtained by the following formula. Distance from a Point to a Line P! Q! P R d = P! R Letting P = (1; ; ) and R = (1; 0; ), we have! P Q = h 0; 0; 1 i ;! P R = h 0; 1; 0 i ;! P Q P! R = 9 h 1; 0; 0 i ; d = 9 k h 1; 0; 0 i k k h 0; 1; 0 i k = : 17. If you aly a force of magnitude 0 ounds at the end of an 8 inch long wrench at an angle of = to the wrench, find the magnitude of the torque alied to the bolt. We recall Page 6 of 9

17 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010 Torque Torque τ is defined to be the cross roduct of the osition vector r and force vector F, i.e., τ = r F ; k τ k = k r k k F k sin where is the angle between r and F. Given information: k F k = 0 and = = and k r k = 8 inch = 8=1 = = feet. Thus we have k τ k = (0) sin = 10 ft lbs 18. Label each statement as TRUE or FALSE. If it is true, briefly exlain why. If it is false, give a counterexamle. (18.1) If a b = a c, then b = c F We observe that for a = h 1; 0; 0 i, b = h 1; 1; 1 i and c = h ; 1; 1 i, a b = h 0; 1; 1 i = a c; but b 6= c: (18.) a b = b a T We observe that for a = h a 1 ; a ; a i and b = h b 1 ; b ; b i, a b = h a b a b ; a 1 b a b 1 ; a 1 b a b 1 i = b a: (18.) a a = kak F a a = h 0; 0; 0 i is a vector, while k a k is a scalar. As a counterexamle, for a = i, we get i i = 0 = h 0; 0; 0 i 6= 1 = k i k. (18.) a (b c) = (a b) c F (a b) c is not ossible because a b is a scalar. A cross roduct must involve two vectors. (18.5) If the force is doubled, the torque doubles T Torque is the cross roduct of direction and force. r (F ) = (r F ) = τ : 19. Find the area of the arallelogram with two adjacent sides formed by h ; 1 i and h 1; i. We recall Page 7 of 9

18 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010 Area The area of the arallelogram with two adjacent sides formed by a and b is the magnitude of their cross roduct: A = k a b k = k a k k b k sin where 0 is the angle between a and b. The vectors h ; 1 i and h 1; i are in the lane (R ). The cross roduct is defined only for the vectors in the sace (R ). But we observe h ; 1; 0 i and h 1; ; 0 i in the sace can corresond to those vectors in the lane, resectively. So we comute the area with these vectors: A = k h ; 1; 0 i h 1; ; 0 i k = k 5 h 0; 0; 1 i k = 5: 0. Find the area of the triangle with vertices (0; 0; 0), (0; ; 1) and (1; ; 0). Letting P (0; 0; 0), Q(0; ; 1) and R(1; ; 0), by the formula, the area of the arallelogram with two adjacent sides formed by P! Q and P! R is the magnitude of their cross roduct: A arallelogram =! P Q! P R = k h 0; ; 1 i h 1; ; 0 i k = k h ; ; 1 i k = 1: Since the area of the triangle is the half of the area of the arallelogram, hence the desired area is A triangle =! P Q! P R = 1 : 1. Find the volume of the aralleleied with three adjacent edges formed by h 0; 1; 0 i, h 0; ; 1 i, and h 1; 0; i. We recall Volume The volume of the aralleleied determined by the vectors a, b, and c is the magnitude of their scalar trile roduct: V = jc (a b)j Letting a = h 0; 1; 0 i, b = h 0; ; 1 i and c = h 1; 0; i, the formula imlies the volume V = 1.. Use geometry to identify the cross roduct. (Do not comute!) (.1) j (j k). Geometry imlies Page 8 of 9

19 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010 Cross Product on Standard Basis Vectors i j = k j k = i k i = j j i = k k j = i i k = j Hence, we deduce j (j k) = j i = k (.) (j i) k. By the same argument as above, we get (j i) k = k k = 0. Use the aralleleied volume formula to determine whether the vectors h 1; 1; i and h 0; 1; 0 i and h ; ; i are colanar. Letting a = h 1; 1; i, b = h 0; 1; 0 i and c = h ; ; i, the formula above imlies the volume V =. Since the volume of the aralleleied is not zero, it means those three vectors do not lie on the same lane. That is, they are not colanar. Page 9 of 9

20 United Arab Emirates University Faculty of Science Deartment of Mathematical Sciences HOMEWORK SOLUTION Section 10.5 Lines and Planes in Sace Calculus II for Engineering MATH 110 SECTION 0 CRN 510 :00 :00 on Monday & Wednesday Due Date: Wednesday, October 7, 010

21 Calculus II for Engineering HOMEWORK SOLUTION Fall, Find (a) arametric equations and (b) symmetric equations of the line. (1.1) The line through (; ; ) and arallel to h ; ; 1 i. (a) x = + t; y = + t; z = t (b) x = y + = z 1 (1.) The line through ( 1; 0; 0) and arallel to the line x + 1 y = = z. The direction comes from the given line, i.e., h ; ; 1 i. (a) x = 1 t; y = 0 + t; z = 0 + t (b) x + 1 = y = z 1 (1.) The line through ( ; 1; 0) and erendicular to both h 0; ; 1 i and h ; ; 1 i. h 0; ; 1 i h ; ; 1 i = h 1; ; 1 i is in the direction erendicular to both vectors. (a) x = + t; y = 1 + t; z = 0 + 1t (b) x + 1 = y 1 = z 1 (1.) The line through (0; ; 1) and normal to the lane y + z =. h 0; 1; i is normal to the lane. (a) x = 0 + 0t = 0; y = + t; z = 1 + t (b) x = 0; y + 1 = z 1. State whether the lines are arallel or erendicular and find the angle between the lines: L : x = t; y = t; z = 1 + t M : x = + s; y = s; z = 1 + s: The vectors arallel to lines L and M are resectively v L = h ; ; i and v M = h 1; ; i. Since there is no constant c satisfying v L = h ; ; i = c h 1; ; i = cv M, so the lines L and M are not arallel. By the formula on the dot roduct, we have cos = v L v M k v L k k v M k = = ; 8 = cos 1 8! 1:7 rad; which is not =. Thus, the lines L and M are not erendicular and the angle between the lines L and M is about 1.7 rad.. Determine whether the lines are arallel, skew or intersect: L : x = + t; y = + t; z = t Page 1 of 5

22 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010 M : x = s; y = 1 s; z = 6 + s: The vectors arallel to lines L and M are resectively v L = h 1; ; 1 i and v M = h 1; ; i. Since there is no constant c satisfying v L = h 1; ; 1 i = c h 1; ; i = cv M, so the lines L and M are not arallel. To determine whether or not the lines intersect, we set the x, y and z values equal: + t = s; + t = 1 s; t = 6 + s; simly; (i) t + s = 1; (ii) t + s = ; (iii) t + s = : From (i) and (ii), we get t = 0 and s = 1. Putting t = 0 and s = 1 into (iii), the equation (iii) holds. This imlies that the lines L and M intersect when t = 0 and s = 1: x = ; y = ; z = : Hence, the lines intersect at the oint (; ; ).. Find an equation of the given lane. (.1) The lane containing the oint ( ; 1; 0) with normal vector h ; 0; i. (x + ) + 0(y 1) + (z 0) = 0; simly; x z + 6 = 0: (.) The lane containing the oints ( ; ; 0), ( ; ; ) and (1; ; ). Letting P ( ; ; 0), Q( ; ; ) and R(1; ; ), we deduce P! Q = h 0; 1; i and P! R = h ; 0; i and P! Q P! R = h ; 6; i is normal to the lane. Hence, using the oint P ( ; ; 0) (one can use Q or R), the equation of the lane is obtained as (x + ) + 6(y ) (z 0) = 0; simly; x + 6y z 8 = 0: (.) The lane containing the oint (; ; 1) and arallel to the lane x + y z =. The vector normal to the lane is h 1; ; i. Hence, the equation of the lane is (x ) + (y + ) (z 1) = 0; simly; x + y z + 7 = 0: (.) The lane containing the oint (; 0; 1) and erendicular to the lanes x + y z = and x z = 1. Let v 1 and v be the vectors normal to the lanes: v 1 = h 1; ; 1 i and v = h ; 0; 1 i. Normal vector must be erendicular to the normal vectors of both lanes. So, the lane that we are looking for has the normal vector v 1 v = h ; 1; i. Hence, the equation of the lane is (x ) (y 0) (z + 1) = 0; simly; x + y + z = : Page of 5

23 Calculus II for Engineering HOMEWORK SOLUTION Fall, Sketch the given lane. (5.1) x y + z =. The equation reresents a lane which has the normal vector h ; 1; i and asses through the oint P (; 0; 0), Q(0; roughly. ; 0) and R(0; 0; 1). We connect those three oints and sketch the lane Y axis 0 Normal Vector R Z axis 0 Q P 0 X axis (5.) x + y = 1. The equation reresents a lane such that (i) it is arallel to the z axis and erendicular to the xy lane, (ii) it has the normal vector h 1; 1; 0 i, (iii) it asses through the oint P (1; 0; 0), Q(0; 1; 0). Y axis 0 Z axis 0 Q P Normal Vector 0 X axis 6. Find the intersection of the lanes x + y z = and x y + z = 1. Solve the equations for z and equate: x + y = z = x y 1; i:e:; x + y = x y 1; Page of 5

24 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010 i:e:; 5x + y = 1; i:e:; y 5x = + 1 : Putting it into the first equation, we get = x + 5x + 1 z; i:e:; z 7x = 7 : Using x = t as a arameter, we deduce the line: x = t; y = 1 5 t; z = t: If one use y = s as a arameter, one can get the line: x = s; y = s; z = s: One can use z = u as a arameter and get the equation of the line. 7. Find the distance between the given objects. (7.1) The oint (1; ; 0) and the lane x + y 5z =. The distance formula imlies d = j(1) + 1() 5(0) j q ( 5) = 5 : (7.) The lanes x + y z = and x + y z = 1. Two lanes are arallel. We choose a oint P (1; 0; 0) on the lane x + y z = 1. Alying the distance formula to the oint P and the other lane, we get d = j1(1) + (0) (0) j q ( ) = 1 : 8. Find an equation of the lane containing the lines: L : x = + t; y = ; z = + t M : x = + s; y = s; z = 1 + s: The vectors arallel to the lines are resectively v L = h 1; 0; i and v M = h ; ; i. The normal vector to the lane is v L v M = h ; 0; i. We choose one oint (; ; ) from L. (One can choose any oint from either line.) Hence, the equation of the lane is (x ) + 0(y ) + (z ) = 0; simly; x z = 5: 9. State whether the statement is true or false (not always true). Page of 5

25 Calculus II for Engineering HOMEWORK SOLUTION Fall, 010 (9.1) Two lanes either are arallel or intersect T Two lanes can be even both arallel and intersect if the lanes coincide. (9.) The intersection of two lanes is a line F It can be a lane if the lanes coincide, or can be emty. (9.) The intersection of three lanes is a oint F It can be a oint, a line, or a lane, or can be emty. (9.) Lines that lie in arallel lanes are always skew T It is true, unless the arallel lanes coincide. (9.5) The set of all lines erendicular to a given line forms a lane F It is false. However, it is true if we take all lines erendicular to a given line through a given oint. (9.6) There is one line erendicular to a given lane F It is false. There is one line erendicular to a given lane through each oint of the lane. (9.7) The set of all oints equidistant from two given oints forms a lane T 10. Determine whether the given lines are the same: L : x = 1 + t; y = t; z = + 6t M : x = 9 s; y = + s; z = 8 s: The vectors arallel to the lines are v L = h ; ; 6 i and v M = h ; 1; i, resectively. Since v L = v M, the lines L and M are arallel. The oint (9; ; 8) lies on the line M when s = 0. We solve for t in the x coordinate of the line L to see that 1 + t = 9 imlies t =. It means when t = and s = 0, the lines L and M have the same x coordinate 9. What about the y and z coordinates when t = and s = 0? At this time t = and s = 0, the line L has the y and z coordinates, y = () = and z = + 6() = 1, resectively. However, the line M has the y and z coordinates, y = and z = 8. That is, t = & s = 0 =) L : (x; y; z) = (9; ; 1); M : (x; y; z) = (9; ; 8): It imlies that these lines L and M are not the same. Page 5 of 5