HW05-solutions

# HW05-solutions - qazi(kaq87 – HW05 – berg –(55290...

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Unformatted text preview: qazi (kaq87) – HW05 – berg – (55290) This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – ﬁnd all choices before answering. 001 to determine a × b. 002 10.0 points Which of the following expressions are welldeﬁned for all vectors a, b, c, and d? 10.0 points Find the cross product of the vectors a = i + 2j + 3k , 1 b = 3i + 3j + 3k . I a × ( b · c) , II (a × b) · (c × d) , III a × (b × c) . 1. a × b = −2i + 7j − 4k 2. a × b = −3i + 6j − 3k correct 1. I and II only 2. I only 3. a × b = −2i + 6j − 4k 3. none of them 4. a × b = −2i + 6j − 3k 4. I and III only 5. a × b = −3i + 6j − 4k 5. III only 6. a × b = −3i + 7j − 3k Explanation: One way of computing the cross product 6. II and III only correct 7. II only (i + 2j + 3k) × (3i + 3j + 3k) 8. all of them is to use the fact that i ×j = k, j× k = i, k ×i = j, j ×j = 0, k× k = 0. while i×i = 0, For then a × b = −3i + 6j − 3k . Alternatively, we can use the deﬁnition ij 12 33 a×b = = 2 3 k 3 3 1 3 i− 3 3 + 1 3 Explanation: The cross product is deﬁned only for two vectors, and its value is a vector; on the other hand, the dot product is deﬁned only for two vectors, and its value is a scalar. For the three given expressions, therefore, we see that I is not well-deﬁned because the second term in the cross product is a dot product, hence not a vector. II is well-deﬁned because both terms in the dot product are cross products, hence vectors. III is well-deﬁned because it is the cross product of two vectors. 3 j 3 2 k 3 keywords: vectors, dot product, cross product, T/F, length, 003 10.0 points qazi (kaq87) – HW05 – berg – (55290) 2 Determine all unit vectors v orthogonal to a = 3 i + 4 j + k, 1. v = b = 3i +6j +2k. 004 2 3 6 i− j+ k 7 7 7 P (2, 2) , 2 6 3 i+ j− k 7 7 7 6. v = −2 i − 6 j + 3 k 3. area = 8 4. area = 7 5. area = 9 Explanation: The non-zero vectors orthogonal to a and b are all of the form λ = 0, with λ a scalar. The only unit vectors orthogonal to a, b are thus v=± a×b . |a × b| Explanation: To use vectors we shall identify a line segment with the corresponding directed line segment. Since the area of the parallelogram having adjacent edges P Q and P R is given by − − →− → |P Q × P R| , ∆P QR has But for the given vectors a and b, a×b = ij 34 36 R(3, −2) . 17 2 15 correct 2. area = 2 2 3 6 i − j + k correct 7 7 7 v = λ(a × b) , Q(−2, 3) , 1. area = 4. v = 2 i − 3 j + 6 k 5. v = ± 10.0 points Find the area of the triangle having vertices 2 6 3 2. v = − i − j + k 7 7 7 3. v = ± keywords: vector product, cross product, unit vector, orthogonal, area = k 1 2 3 31 41 j+ i− = 3 32 62 1− − →− → |P Q × P R| . 2 Now − − → P Q = −4, 1, 0 , 4 k 6 = 2i− 3j+ 6k. − → PR = 1, −4, 0 . But then i − − →− → P Q × P R = −4 1 j 1 −4 k −4 0= 1 0 1 k. −4 Consequently, ∆P QR has In this case, |a × b|2 = 49 . area = Consequently, v=± 2 3 6 i− j+ k 7 7 7 . 15 2 . keywords: vectors, cross product area, triangle, parallelogram qazi (kaq87) – HW05 – berg – (55290) 005 10.0 points 3 3. volume = 31 Find a vector v orthogonal to the plane through the points 4. volume = 28 P (5, 0, 0), Q(0, 2, 0), R(0, 0, 4) . 5. volume = 27 1. v = 4, 20, 10 Explanation: For the parallelopiped determined by vectors a, b, and c its 2. v = 8, 5, 10 volume = |a · (b × c)| . 3. v = 8, 20, 10 correct But 3 −4 4. v = 8, 4, 10 −1 3 5. v = 2, 20, 10 Explanation: Because the plane through P , Q, R con− − → − → tains the vectors P Q and P R, any vector v orthogonal to both of these vectors (such as their cross product) must therefore be orthogonal to the plane. =3 12 14 1 2 3 a · ( b × c) = 1 4 +4 32 34 − 3 1 3 1 . Consequently, Here volume = 30 . − − → P Q = −5, 2, 0 , − → P R = −5, 0, 4 . keywords: determinant, cross product, scalar triple product, parallelopiped, volume, Consequently, − − →− → v = P Q × P R = 8, 20, 10 is othogonal to the plane through P, Q and R. 006 10.0 points Compute the volume of the parallelopiped determined by the vectors a = 3, −4, −1 , 007 10.0 points Which of the following statements are true for all vectors a, b = 0? A. |a × b|2 + |a · b|2 = |a|2|b|2 , B. a × b = b × a, C. If a × b = 0, then a ⊥ b. b = 3, 1, 2 , 1. C only and c = 3, 1, 4 . 2. all of them 1. volume = 30 correct 3. B and C only 2. volume = 29 4. A only correct qazi (kaq87) – HW05 – berg – (55290) 4 so if a = 0 and b = 0, then 5. A and C only |a × b| = 0 6. none of them =⇒ sin θ = 0 . Thus θ = 0, π . In this case, a is parallel to b, not perpendicular. 7. B only keywords: 8. A and B only 008 Explanation: A. TRUE: if θ, 0 ≤ θ ≤ π , is the angle between a, and b, then 10.0 points Compute the volume of the parallelopiped with adjacent edges PQ, |a × b| = |a||b| sin θ , PR, PS determined by vertices while P (1, −1, 1) , Thus |a × b|2 + |a · b|2 Q(3, −5, −1) , R(2, 0, −1) , a · b = |a||b| cos θ . S (2, 1, 0) . 1. volume = 8 correct = |a|2 |b|2(sin2 θ + cos2 θ ) = |a|2 |b|2 . B. FALSE: if 2. volume = 7 3. volume = 5 a = a1 , a2 , a3 , b = b1 , b2 , b3 , 4. volume = 6 then a2 b2 a×b = a3 b3 i− a1 a3 b1 b1 j+ a1 a3 b1 b2 b1 b2 a1 a2 5. volume = 4 k=0 when a, b = 0, while b2 b3 a2 b×a = a3 i+ b1 b3 a1 a3 j+ k=0 when a, b = 0. On the other hand, for a 2 × 2 determinant, a b c d = ad − cb = − c d a b Explanation: The parallelopiped is determined by the vectors − − → a = P Q = 2, −4, −2 , − → b = PR = − → c = PS = 1, 2, −1 . Thus its volume is given in terms of a scalar triple product by . Consequently, 1, 1, −2 , V = | a · ( b × c) | . But 2 −4 a · ( b × c) = C. FALSE: if θ, 0 ≤ θ ≤ π , is the angle between a, and b, then |a × b| = |a||b| sin θ , =2 1 −2 2 −1 1 1 −2 1 a × b = −b × a . −2 2 −1 +4 1 −2 1 −1 −2 1 1 1 2 . qazi (kaq87) – HW05 – berg – (55290) 5 of the third line, so must be scalar multiples of each other. Consequently, volume = 8 . 010 keywords: determinant, cross product, vector product, scalar triple product, parallelopiped, volume, 009 10.0 points Which equation has the surface z 10.0 points Which of the following statements are true for all lines and planes in 3-space? I. two planes parallel to a third plane are parallel, y II. two lines perpendicular to a plane are parallel, III. two lines parallel to a third line are parallel. x as its graph in the ﬁrst octant? 1. 1. II and III only 2. all of them correct 2. 3. III only 3. 4. I and III only 4. 5. I and II only 5. 6. II only 6. 7. none of them 8. I only Explanation: I. TRUE: each of the two planes has a normal vector parallel to the normal vector of the third plane, and so are parallel, hence the planes are parallel. II. TRUE: the two lines will have direction vectors parallel to the normal vector of the plane, and so be parallel, hence the two lines are parallel. III. TRUE: each of the two lines has a direction vector parallel to the direction vector x 3 x 3 x 5 x 4 x 4 x 5 + + + + + + y 4 y 5 y 4 y 3 y 5 y 3 + + + + + + z 5 z 4 z 3 z 5 z 3 z 4 =1 =1 =1 =1 = 1 correct =1 Explanation: As the surface is a plane, it must be the graph of a linear function which can be written in intercept form as xyz + + = 1. ab c But by inspection we see that the x-intercept is x = 4, the y -intercept is y = 5 and the z intercept is z = 3. Consequently, the surface is the graph in the ﬁrst octant of the equation xyz + + =1. 453 qazi (kaq87) – HW05 – berg – (55290) 011 1. Q(5, 0, 1) 10.0 points Find parametric equations for the line passing through the point P (4, −3, 4) and parallel to the vector 2, 4, −1 . 1. x = 4 − 2t, y = 3 − 4t, z = 4 − t 2. x = 2 − 4t, y = −4 + 3t, z = −1 + 4t 3. x = 2 + 4t, y = 4 + 3t, z = 1 − 4t 4. x = 2 + 4t, y = 4 − 3t, z = −1 + 4t 5. x = −4 + 2t, y = 3 + 4t, z = −4 − t 6. x = 4 + 2t, y = −3 + 4t, z = 4 − t correct Explanation: A line passing through a point P (a, b, c) and having direction vector v is given parametrically by r(t) = a + tv , a= a, b, c . 2. Q(3, 0, 5) 3. Q(−1, 0, 3) 4. Q(−1, 1, 0) correct 5. Q(1, −1, 0) 6. Q(5, 3, 0) Explanation: Since the xy -plane is given by z = 0, we have to ﬁnd an equation for and then set z = 0. Now a line passing through a point P (a, b, c) and having direction vector v is given parametrically by r(t) = a + tv , 4, −3, 4 , 3x + y + 2z = 2 . v = 2, 4, −1 . Thus Thus a = 2, 2, 2 , r(t) = 4 + 2t, −3 + 4t, 4 − t . Consequently, are parametric equations for the line. A line passes through the point P (2, 2, 2) and is perpendicular to the plane 3x + y + 2z = 2 . At what point Q does plane? intersect the xy - 3, 1, 2 , 2 + 3t, 2 + t, 2 + 2t . In this case, z = 0 when t = −1. Consequently, intersects the xy -plane at Q(−1, 1, 0) . keywords: line, parametric equations, direction vector, point on line 10.0 points v= and so r(t) = x = 4 + 2t, y = −3 + 4t, z = 4 − t 012 a = a, b, c . But for , its direction vector will be parallel to the normal to the plane Now for the given line, a= 6 keywords: line, parametric equations, direction vector, point on line, intercept, coordinate plane 013 10.0 points Find parametric equations for the line through the point P (9, 5, 1) that is parallel qazi (kaq87) – HW05 – berg – (55290) to the plane x + y + z = 2 and perpendicular to the line x = 1 + t , y = 1 − t , z = 3t . 1. x = 9 − 4t, y = 5 + 2t, z = 1 + t 2. x = 9 + t, y = a + t, z = 1 − 2t 3. x = 9 − 4t, correct y = 5 + 2t, z = 1 + 2t 7 3. P (−7, 9, 6) 4. P (7, 9, 3) 5. P (7, 9, 6) correct Explanation: To determine where the lines intesect it is convenient ﬁrst to convert the lines from equations in symmetric form to ones in parametric form: 4. x = 9 + 4t, y = 5 + 2t, z = 1 + t x = 3 + 4t , y = 4 + 5t , z = 5 + t , 5. x = 9 + 4t, y = 5 + 2t, z = 1 + 2t x = 2 + 5s , y = 6 + 3s , z = 4 + 2s . Explanation: Two vectors which are perpendicular to the required line are the normal, 1, 1, 1 , of the given plane and a direction vector, 1, −1, 3 , for the given line. So a direction vector for the required line is 1, 1, 1 × 1, −1, 3 = −4, 2, 2 . Thus L is given by x, y, z = 9, 5, 1 + t −4, 2, 2 , which can be written in parametric form as For then the lines intersect when the equations 3 + 4t = 2 + 5s , 4 + 5t = 6 + 3s , 5 + t = 4 + 2s , are satisﬁed simultaneously. Solving the ﬁrst two equations gives t = 1, s = 1 and a check shows that these values then satisfy the third equation. Consequently, the lines intersect when s = t = 1, i.e., at the point x = 9 − 4t , y = 5 + 2t , z = 1 + 2t . P (7, 9, 6) . keywords: parametric equations, line, plane, parallel perpendicular, direction vector, cross product 014 10.0 points Find the point of intersection, P , of the lines y−4 z−5 x−3 = = , 4 5 1 x−2 y−6 z−4 = = . 5 3 2 015 10.0 points Find an equation for the plane passing through the point P (1, 2, 5) and perpendicular to the y -axis. 1. z = −5 2. y = −2 3. x = 1 1. P (7, −9, 6) 4. x = −1 2. P (7, −9, 4) 5. z = 5 qazi (kaq87) – HW05 – berg – (55290) 8 lies in the plane, so 6. y = 2 correct Explanation: If the plane is perpendicular to the y -axis, then it is of the form y = d. But if this plane also passes through P (1, 2, 5), then it has equation v ·n = 0. Now i k 3 1 0 −2 y=2 . j 0 3 = 3, −9, 2 . But then 016 10.0 points Find an equation for the plane passing through the points Q(−1, −2, −1) , R(2, −1, −1) , v · n = 3(x + 1) − 9(y + 2) + 2(z + 1) = 3x − 9y + 2z − 13 = 0 . Consequently, the plane S (−3, −2, 2) . 3x − 9y + 2z − 13 = 0 1. 3x + 2y + 9z + 13 = 0 passes through Q, R and S . 2. 3x + 9y − 2z − 13 = 0 keywords: plane, cross product, plane determined by three points, dot product 3. 3x − 9y + 2z + 13 = 0 4. 3x − 9y + 2z − 13 = 0 correct 5. 3x − 2y + 9z + 13 = 0 Explanation: Since the points Q, R, and S lie in the plane, the displacement vectors − − → QR = 3, 1, 0 , 6x + 4y − z = 1 . 1. x + 6y − 4z + 1 = 0 2. 4x − y − 4z + 1 = 0 −2, 0, 3 , lie in the plane. Thus the cross product i j k 3 1 0 −2 0 − − →− → n = QR × QS = 10.0 points Find an equation for the plane passing through the origin and parallel to the plane 6. 3x + 2y + 9z − 13 = 0 − → QS = 017 3 is normal to the plane. On the other hand, if P (x, y, z ) is an arbitrary point on the plane, then the displacement vector − − → v = P Q = x + 1, y + 2, z + 1 3. 6x − 4y + z + 1 = 0 4. x − 6y + 4z = 0 5. 6x + 4y − z = 0 correct 6. 4x + y + 4z = 0 Explanation: The scalar equation for the plane through P (a, b, c) with normal vector n = Ai + B j + C k qazi (kaq87) – HW05 – berg – (55290) is A(x − a) + B (y − b) + C (z − c) = 0 . But P (a, b, c) = (0, 0, 0) if the plane passes through the origin, while n = 6i + 4j − k if the plane is parallel to 6x + 4y − z = 1 since parallel planes have the same normal vector. Consequently, the plane has equation 6x + 4y − z = 0 . 018 10.0 points Find the point at which the line x = 3 + 5t , y = −1 , z = 3t , intersects the plane 4x + 6y − 6z = 12 . 1. P (18, −1, −9) 2. P (18, 1, −9) 3. P (−18, 1, −9) 4. P (−18, 1, 9) 5. P (18, −1, 9) correct 6. P (−18, −1, 9) Explanation: The line x = 3 + 5t , y = −1 z = 3t intersects the plane 4x + 6y − 6z = 12 when 4(3 + 5t) + 6(−1) − 6(3t) = 12 , i.e. when 2t = 6. Thus the point of intersection occurs at t = 3, which corresponds to the point P (18, −1, 9) . 9 ...
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## This note was uploaded on 03/28/2012 for the course M 408 M taught by Professor Gilbert during the Spring '07 term at University of Texas.

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