# symbolic notation and fractions

Daniel Targonski Math233 1293 El-Achkar Su21 Assignment Lesson01 12.1 Vectors in the plane due 06/06/2021 at 11:59pm EDT 1. (1 point) 4. (1 point) Find the vector of length 2 making an angle of 30◦ with the x-axis. −→ Find the components of PQ, P = (−6, −4), Q = (2, −8). Solution: Solution: The desired vector is *√ + √ 3 1 ◦ ◦ 2hcos 30 , sin 30 i = 2 , = h 3, 1i. 2 2 (Use symbolic notation and fractions where needed. Give your answer as the point’s coordinates in the form (*,*),(*,*)…)) −→ PQ = Correct Answers: • Solution: Using the definition of the components of a vec−→ tor we have PQ = (2 − (−6), −8 − (−4)) = (8, −4) 5. (1 point) Find the components and length of the following vectors: −3i + 5j , Components: Length: Correct Answers: • (8,-4) 2. (1 point) ~ has compoLet R = (−1, −2). Find the point P such that PR nents h3, 3i. Solution: Solution: Denoting P = (x0 , y0 ), we have: ~ = h−1 − x0 , −2 − y0 i = h3, 3i PR Equating corresponding components yields: −1 − x0 −2 − y0 = 3 ⇒ = 3 x0 = −4, y0 = −5 ⇒ P = (−4, −5) Correct Answers: • (-4,-5) −3i + 3j Components: Length: , −4i − 1j Components: Length: , 1i + 2j Components: Length: , Solution: Solution: Since i = h1, 0i and j = h0, 1i, using vector algebra we have: 3. (1 point) Find the unit vector in the direction opposite to v = h5, 2i. −3i + 5j Solution: Solution: We first compute the unit vector ev in the direction of v and then multiply by −1 to obtain a unit vector in the opposite direction. This gives: 1 1 1 5 2 h5, 2i = √ h5, 2i = √ , √ ev = v= p kvk 29 29 29 (5)2 + (2)2 = −3h1, 0i + 5h0, 1i = h−3, 0i + h0, 5i = h−3 + 0, 0 + 5i = h−3, 5i The length of the vector is: k − 3i + 5jk = q √ (−3)2 + (5)2 = 34 We use vector algebra and the definition of the standard basis vector to compute the components of the vector −3i + 3j: The desired vector is thus 5 2 −5 −2 −ev = − √ , √ = √ ,√ . 29 29 29 29 −3i + 3j = −3h1, 0i + 3h0, 1i = h−3, 0i + h0, 3i = h−3 + 0, 0 + 3i = h−3, 3i The length of this vector is: Correct Answers: • k − 3i + 3jk = 1 q (−3)2 + (3)2 = √ 18 We find the components of the vector −4i − 1j: • −3 , 5 −4i − 1j = −4h1, 0i − 1h0, 1i = h−4, 0i + h0, −1i = h−4 + 0, 0 − 1i = h−4, −1i • 5.83095 • −3 , 3 The length of this vector is: q √ k − 4i − 1jk = (−4)2 + (−1)2 = 17 • 4.24264 • −4 , −1 We find the components of the vector 1i + 2j, using vector algebra: 1i + 2j = 1h1, 0i + 2h0, 1i = h1, 0i + h0, 2i = h1 + 0, 0 + 2i = h1, 2i • 4.12311 • 1, 2 The length of this vector is k1i + 2jk = q √ (1)2 + (2)2 = 5 • 2.23607 Correct Answers: Generated by c WeBWorK, http://webwork.maa.org, Mathematical Association of America 2 Daniel Targonski Math233 1293 El-Achkar Su21 Assignment Lesson02 12.2 Vectors in three dimensions due 06/06/2021 at 11:59pm EDT Solution: The line through Q = (−13, 24, 5) and R = −→ (−5, 3, 7) has direction vector QR = h8, −21, 2i. The parallel line through P = (−4, 15, 9) has vector parametrization −→ −→ r(t) = OP + t QR = h−4, 15, 9i + th8, −21, 2i . The corresponding parametric equations are 1. (1 point) v = h10, 16, 23i Which of the following vectors is parallel to v? • • • • • A. < 23, 16, 10 > B. < 10, 8, 5.75 > C. < 10, −16, −23 > D. < −10, −16, −23 > E. < −16, 23, −10 > x(t) = −4 + 8t y(t) = 15 − 21t z(t) = 9 + 2t Which of the following points in the same direction as v? • • • • • Note : there are many sets of parametric equations for this line (each of those sets which is also linear in the variable t would be scored as correct). A. < 10, −16, −23 > B. < 20, 32, 46 > C. < −10, −16, −23 > D. < −16, 23, −10 > E. < 16, 23, 10 > Correct Answers: • -4+8*t; 15-21*t; 9+2*t 4. (1 point) r1 (t) = (−6, −2, 15) + t h−2, 2, 2i r2 (t) = (−2, −9, 5) + t h2, 0, 2i Find the point of intersection, P, of the lines r1 and r2 . Correct Answers: P= Solution: • D Solution: The two lines intersect if there exist parameter • B values t1 and t2 such that r1 (t1 ) = r2 (t2 ). 2. (1 point) That is Find a unit vector ev where v = h10, 7, 4i. h−6, −2, 15i + t1 h−2, 2, 2i = h−2, −9, 5i + t2 h2, 0, 2i. ev = This is equivalent to the three equations for the components: Solution: x = −6 − 2t1 = −2 + 2t2 , √ √ Solution: kvk = v· v = 10 ∗ 10 + 7 ∗ 7 + 4 ∗ 4 = y = −2 + 2t1 = −9 + 0t2 , 12.8452. z = 15 + 2t1 = 5 + 2t2 . 1 10 7 4 ev = kvk v = 12.8452 , 12.8452 , 12.8452 = h0.778499, 0.544949, 0.3114iSolving the first two equations for t and t in turn yields 1 2 Correct Answers: t1 = −3.5 and t2 = 1.5 • Now, substituting these values into the third equation yields 15 − 3.5 ∗ 2 = 5 + 1.5 ∗ 2 3. (1 point) 8=8 Find parametric equations for the line through P = (−4, 15, 9) Since this does indeed come out equal, the lines intersect at and parallel to the line through Q = (−13, 24, 5) and R = the point (−5, 3, 7) h−6, −2, 15i − 3.5 ∗ h−2, 2, 2i = h1, −9, 8i x(t) = Note: If the third equation could not be reconciled, that y(t) = would mean that the lines do not intersect. z(t) = Correct Answers: Solution: Solution: h−10, −16, −23i is −1v, so it is parallel to v (in the opposite direction). h20, 32, 46i is 2v, so it points in the same direction as v. Solution: • (1,-9,8) Generated by c WeBWorK, http://webwork.maa.org, Mathematical Association of America 1 Daniel Targonski Assignment Lesson03 12.3 Dot product due 06/06/2021 at 11:59pm EDT (b) We follow the identical procedure as in part a, with the − → − → two vectors BA= h−2, −5i and BC = h2, −2i. h−2,−5i·h2,−2i = 1.1659. θb = cos−1 kh−2,−5ikkh2,−2ik (c) We follow the identical procedure as in parts a and b, with − → − → the two vectorsCA = h−4, −3i and CB = h−2, 2i. h−4,−3i·h−2,2i θc = cos−1 kh−4,−3ikkh−2,2ik = 1.4289. 1. (1 point) If θ is the angle between h−1, 2, 2i and h−2, 5, −2i, then cos(θ) = Solution: Solution: q √ k h−1, 2, 2i k = (−1)2 + 22 + 22 = 9 q √ k h−2, 5, −2i k = (−2)2 + 52 + (−2)2 = 33 Correct Answers: • 0.546789 • 1.1659 • 1.4289 h−1, 2, 2i · h−2, 5, −2i = (−1) · (−2) + 2 · 5 + 2 · (−2) = 8 imply cos(θ) = Math233 1293 El-Achkar Su21 4. (1 point) The projection of u = h−3, 2, 1i along v = h5, 3, 0i is h−1, 2, 2i · h−2, 5, −2i 8 =√ k h−1, 2, 2i k k h−2, 5, −2i k 9 · 33 Correct Answers: u|| = • 0.464207 help (vectors) Solution: Solution: 2. (1 point) Assume that u· v = 9, kuk = 6, and kvk = 10. What is the value of 1u· (8u − 4v)? u · v = h−3, 2, 1i · h5, 3, 0i = (−3) · 5 + 2 · 3 + 1 · 0 = −9 v · v = h5, 3, 0i · h5, 3, 0i = 52 + 32 + 02 = 34 u·v −9 27 h5, 3, 0i = − 45 u|| = v= 34 , − 34 , 0 v·v 34 Solution: Solution: Using the distributive property of dot products and the fact that u· u = kuk2 we can evaluate this expression. 1u· (8u − 4v) 8(u· u) − 4(u· v) 8kuk2 − 4(u· v) Now substitute in our known values: 8(62 ) − 4(9) And finally we arrive at the answer 252. Correct Answers: • 5. (1 point) If u = h4, 3, 2i and v = h0, 0, 5i, then the component of u along v is Correct Answers: Hint: Hint: The correct answer is not a vector. Solution: Solution: • 252 3. (1 point) A triangle is defined by the three points: A = (6, 5) B = (8, 10) C = (10, 8). Determine all three angles in the triangle (in radians). θa = θb = θc = Solution: Solution: − → − → (a) First we find the two vectors AB = h2, 5i and AC = h4, 3i. Then, from the identity v· w = kvkkwk cos θ we solve for θ which yields v·w θ = cos−1 kvkkwk . h2,5i·h4,3i Therefore, θa = cos−1 kh2,5ikkh4,3ik = 0.546789. u · v = h4, 3, 2i · h0, 0, 5i = 4 · 0 + 3 · 0 + 2 · 5 = 10 v · v = h0, 0, 5i · h0, 0, 5i = 02 + 02 + 52 = 25 The component of u along v is u·v 10 =√ kvk 25 Correct Answers: • 2 6. (1 point) Let a = h−1, 3, 3i and b = h−5, 5, 5i, then find the decomposition a = a|| + a⊥ of a with respect to b a|| = help (vectors) a⊥ = Solution: 1 Solution: a · b = h−1, 3, 3i · h−5, 5, 5i = (−1) · (−5) + 3 · 5 + 3 · 5 = 35 Correct Answers: • • b · b = h−5, 5, 5i · h−5, 5, 5i = (−5)2 + 52 + 52 = 75 a·b 35 h−5, 5, 5i = − 37 , 73 , 37 a|| = b= b·b 75 a⊥ = a − a|| = h−1, 3, 3i − − 73 , 37 , 37 = 4 2 2 3, 3, 3 Generated by c WeBWorK, http://webwork.maa.org, Mathematical Association of America 2 Daniel Targonski Assignment Lesson04-05 12.4 Cross product due 06/06/2021 at 11:59pm EDT Math233 1293 El-Achkar Su21 4. (1 point) 1. (1point) 1 4 , −10, and w = h−2, −6, −12i, then If v = 3 2 v×w = help (vectors) Let v = −4, −4, 2 Calculate: i v×i = h , , i v×j = h , , i v×k = h , , Solution: Solution: i 4/3 −2 j k 4/3 −10 4/3 1/2 −10 1/2 −10 1/2 = kSolution: j+ i− −2 −6 −2 −12 −6 −12 −6 −12 Solution: 4j + 2k and use the distributive We law: 1 4 1 write v = −4i − 4 = (−10) · (−12) − · (−6) i − · (−12) − v· × (−2) j + · (−6) − (−10) · (−2) k4j × i + 2k × i = i = (−4i − 4j + 2k) × i = −4i × i − 2 3 2 3 −4 · 0 + 4k + 2j = 0, 2, 4 = 123i + 15 j − 28k = h123, 15, −28i v × j = (−4i − 4j + 2k) × j = −4i × j − 4j × j + 2k × j = −4k − 4 · 0 − 2i = −2, 0, −4 Correct Answers: v × k = (−4i − 4j + 2k) × k = −4i × k − 4j × k + 2k × k = • +4j − 4i − 2 · 0 = −4, 4, 0 2. (1 point) (i − 9 j + 3k) × (− j + k) = help (vectors) Correct Answers: Solution: • 0 Solution: • 2 • 4 = [(i × k) − 9( j × k) + 3(k × k)] − [(i × j) − 9( j ×• j)-2+ 3(k × j)] • 0 = [− j − 9(i) + 3(0)] − [k − 9(0) + 3(−i)] • -4 = [−9i − j] − [−3i + k] • -4 • 4 = −6i − j − k • 0 (i − 9 j + 3k) × (− j + k) = [(i − 9 j + 3k) × k] − [(i − 9 j + 3k) × j] Correct Answers: 5. (1 point) Sketch the triangle with vertices O, P = (7, 0, 7) and Q = (3, 6, 0) and compute its area using cross products. Area= Solution: Solution: The area of the triangle is half of the area of the parallelogram ~ = h7, 0, 7i and OQ ~ = h3, 6, 0i. determined by the vectors OP Thus, ~ × OQ|| ~ S = 12 ||OP We compute the cross product: i j k 0 7 7 7 ~ × OQ ~ = 7 0 7 OP = i − j + 6 0 3 0 3 6 0 7 0 k 3 6 = −42i + 21 j + 42k p S = 21 || − 42i + 21 j + 42k|| = 21 (−42)2 + (21)2 + (42)2 ≈ 31.5. • (i-9*j+3*k) >< (-j+k) 3. (1 point) Calculate the cross product assuming that u × w = −5, −1, −7 i (−u − 4w) × w = h , , Solution: Solution: Using the properties of the cross product we obtain: (−u − 4w) × w = −u × w − 4w × w = 5, 1, 7 Correct Answers: • 5 • 1 • 7 1 Correct Answers: • 1/2*sqrt(3969) Generated by c WeBWorK, http://webwork.maa.org, Mathematical Association of America 2 Daniel Targonski Math233 1293 El-Achkar Su21 Assignment Lesson06-07 12.5 Planes in three-space due 06/13/2021 at 11:59pm EDT 5. (1 point) 1. (1 point) Write an equation of the plane with normal vector n= h−9, 2, −1i passing through the point (5, 0, −5) in scalar form = −200 Determine whether or not the lines have a single point of intersection. If they do, give an equation of a plane containing them. Answer(s) submitted: • -45x+10y-5z (correct) r1 (t) = h19t, 9t − 1, 9t − 9i and r2 (t) = ht − 19, −t + 18,t − 28i 2. (1 point) The plane whose points satisfy the equation −6(x − 15) − 4(y + 13) − 3z = 0 has many vectors which are normal to it. help (vectors) One of those normal vectors is n = (Use symbolic notation and fractions where needed. ”DNE” if two lines do not intersect.) Enter Equation of a plane: Answer(s) submitted: • (correct) Answer(s) submitted: • 18x-10y-28z=262 3. (1 point) Find the equation of the plane which passes through O and is parallel to z − (2x + y) = 9 x+ y+z = (correct) 6. (1 point) Answer(s) submitted: • -2 • -1 • 0 Determine whether or not the point lies on the line. If it does not, give an equation of a plane containing the point and the line. (correct) (−2, 7, −2) and r(t) = h1 − 3t, 7t − 3,t − 3i 4. (1 point) Find an equation of the plane through the three points given: P = (0, 4, 0) , Q = (4, 7, −4) , R = (4, 6, −3) = −16 (Use symbolic notation and fractions where needed. ”DNE” if the point doesn’t lie on the line.) Equation of a plane: Answer(s) submitted: • -x-4y-4z Answer(s) submitted: • x+3z=-8 (correct) (correct) Generated by c WeBWorK, http://webwork.maa.org, Mathematical Association of America 1 Enter Find an arc length parametrization of r(t)=etcos(t)i +7etj+etsin(t)K Find the speed at the given value of t: r(t)=(1+2+1)1 – 3e+-1; +t in(t)k at t=1 State the type of the given quadric surface and describe the trace obtained by intersecting with the given plane: P+C)** +25z2=1; y=2 State the type of the given quadric surface and describe the trace obtained by intersecting with the given plane: P+C)** +25z2=1; y=2 Question 4 Evaluate the limit problem: x2-y2= lim II (x, y) + (0,0) XY Find an equation of the tangent plane to the surface f(x, y)= ln x2 + y2 at (3, 4, 1n5) 3×2 Use the linear approximation of flx,y)=- at (-2, 1) to: 5y +1 a. estimate f(-2.01, 1.01) b. Compute the percentage error Question 7 Find the solution r(t) of the differential equation with the given initial condition: 1 r'(t)=cos(2t)ĩ – 2sin(t)j +. 5k and r(@)=31 -21 +k =+h 1+t2 Question 8 Consider the points A(2, -3,1), B(6.5.-1) and C(7.2.2): a. Verify that the points A, B and c are not colinear. b. Calculate the area of the triangle with vertices A, B and C Question 9 Given the Function f(x, y)=in(x3+2y2), show that fxy(x,y)=fyx(x,y) Question 9 Given the Function f(x, y)=in(x3+2y2), show that fxy(x,y)=fyx(x,y) a. Calculate the directional derivative of f(x, y)=tan-+(x2+3y) at the point P(1, -1) in the direction of the vector V = -81 +67 b. Find the rate of change of f in the direction fp. Question 11 Describe and sketch the domain of the function f(x,y)=\n(3-x-y) Question 12 10 points Save Answer Determine whether or not the point P(-3.2.-1) lies on the line r(t)=, if it does not find the equation of the plane containing the point and the line Question 12 10 points Save Answer Determine whether or not the point P(-3.2.-1) lies on the line r(t)=, if it does not find the equation of the plane containing the point and the line

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