a a. θ = cos 1 a b ) b For non-zero vectors a and b, then the component of b along a is given as comp

Transcription

1 Textbook Assignment 4 Your Name: LAST NAME, FIRST NAME (YOUR STUDENT ID: XXXX) Your Instructors Name: Prof. FIRST NAME LAST NAME YOUR SECTION: MATH 0300 XX Due Date: NAME OF DAY, MONTH DAY, YEAR. SECTION 13.4: do problems 7, 16,, 4, 6, 3, 33, 36, 37, 39, 40, 4. Formulas: If a = a 1, a, a 3 and b = b 1, b, b 3, then i) a + b = a 1 + b 1, a + b, a 3 + b 3 and α a = α a 1, α a, α a 3 for any scalar α R. ii) = a 1 + a + a 3. iii) If a is any non-zero vector, then a unit vector û in the direction of a is given as û = 1 a. iv) a b = a 1 b 1 + a b + a 3 b 3 for any a and b. v) a b = a b 3 a 3 b, a 3 b 1 a 1 b 3, a 1 b a b 1 vi) If θ is the angle between two vectors a and b, then ( θ = cos 1 a b ). b vii) For non-zero vectors a and b, then the component of b along a is given as comp a b a b = Theorems and ideas: The vector a b is orthogonal to both a and b. If θ is the angle between a and b (so 0 θ π), then a b = b sin θ. Two nonzero vectors a and b are parallel if and only if a b = 0 The length of the cross product a b is equal to the area of the parallelogram determined by a and b. The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product: a ( b c). If a, b, and c are vectors and α is a scalar, then i) a b = b a ii) (α a) b = α( a b) = a (α b) iii) a ( b + c) = a b + a c iv) ( a + b) c = a c + b c. v) a ( b c) = ( a b) c (The Scalar Triple Product or Box Product). vi) a ( b c) = ( a c) b ( a b) c (The Vector Triple Product) -START SOLVING PROBLEMS ON A NEW PAGE (WRITE OUT THE QUESTION!!!!) -

2 Exercise 7: Find the cross product a b and verify that it is orthogonal to both a and b where a = t, t, t 3 and b = 1, t, 3t. The Plan: A straight forward computation using the definitions. a b = t, t, t 3 1, t, 3t = t 4, t 3, t a ( a b) = t, t, t 3 t 4, t 3, t = t 5 t 5 + t 5 = 0 which shows that a b is orthogonal to a. b ( a b) = 1, t, 3t t 4, t 3, t = t 4 4t 4 + 3t 4 = 0 which shows that a b is orthogonal to b. Exercise 16: Find two unit vectors orthogonal to both i + j + k and i + k. THE PLAN We know that a cross product is orthogonal to both vectors involved in calculating the cross product. So this is a good starting point. Let a = 1, 1, 1 and b = 1, 0, 1. Now the vector a b = 1, 1, 1 1, 0, 1 = 1, 0, 1 is orthogonal to both a and b. We can now define a unit vector û in the direction of a b which is orthogonal to both a and b as û = 1 a b ( a b) = 1 1, 0, 1. We need a second unit vector ˆv in the opposite direction of a b, i.e., a unit vector in the opposite direction of û. We define ˆv as ˆv = û = 1 1, 0, 1 Exercise : Prove property 4 of Theorem 8. THE PLAN: A straight forward proof of Property 4 of Theorem 8 (page 854 of your Textbook) which says that for vectors a, b and c it is always true that ( a + b) c = a c + b c. Let a = a 1, a, a 3 and b = b 1, b, b 3 and c = c 1, c, c 3. Now we have by direct computation that ( a + b) c = ( a 1, a, a 3 + b 1, b, b 3 ) c 1, c, c 3 = a 1 + b 1, a + b, a 3 + b 3 c 1, c, c 3 = (a + b )c 3 (a 3 + b 3 )c, (a 3 + b 3 )c 1 (a 1 + b 1 )c 3, (a 1 + b 1 )c (a + b )c 1 = a c 3 + b c 3 a 3 c b 3 c, a 3 c 1 + b 3 c 1 a 1 c 3 b 1 c 3, a 1 c + b 1 c a c 1 b c 1 = a c 3 a 3 c + b c 3 b 3 c, a 3 c 1 a 1 c 3 + b 3 c 1 b 1 c 3, a 1 c a c 1 + b 1 c b c 1 = a c 3 a 3 c, a 3 c 1 a 1 c 3, a 1 c a c 1 + b c 3 b 3 c, b 3 c 1 b 1 c 3, b 1 c b c 1 = a c + b c. Exercise 4: Find the area of the parallelogram with vertices K(1,, 3), L(1, 3, 6), M(3, 8, 6), and N(3, 7, 3). THE PLAN: Form the two pairs of parallel vectors ( LM, LM NM which is the area of the parallelogram we seek. LM =, 5, 0 and NM = 0, 1, 3 and so the area of the parallelogram is KN), and ( LK,, 5, 0 0, 1, 3 = 15, 6, = = 65 units NM) and calculate Exercise 6: (a) Find a vector orthogonal to the plane through the points P(, 1, 5), Q( 1, 3, 4), R(3, 0, 6) and find the area of triangle PQR.

3 THE PLAN: Form the vectors PQ and PR and PQ PR. Next 1 PQ PR is the area of the triangle with vertices at P, Q, and R. Now PQ = 3,, 1 and PR = 1, 1, 1 and so PQ PR = 1,, 1 is a vector orthogonal to plane spanned by PQ and PR. Next, PQ PR = 3,, 1 1, 1, 1 = 1,, 1 = 6 so the area of triangle PQR is 1 6 units. Exercise 3: Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS where P(0, 1, ), Q(, 4, 5), R( 1, 0, 1), and S(6, 1, 4) are its vertices. THE PLAN: Calculate the Scalar Triple Product and then take its absolute value and this should do the job. Let a = PQ =, 3, 3 and b = PR = 1, 1, 1 and c = PS = 6,, and so the scalar triple product is a ( b c) = ( a b) c = (, 3, 3 1, 1, 1 ) 6,, = 0, 1, 1 6,, = 4 and so the volume of the parallelepiped with edges PQ, PR, and PS is a ( b c) = 4 = 4 units 3 Exercise 33: Use the scalar triple product to verify that the vectors a = i + 3 j + k, b = i j, and c = 7 i + 3 j + k are coplanar. THE PLAN: Clearly if the scalar triple product yields 0, then it must be that the vectors involved are coplanar, i.e., they all lie in the same plane. Otherwise all are not in the same plane. ( a b) c = (, 3, 1 1, 1, 0 ) 7, 3, =,, 10 7, 3, = = 0 which tells us that a, b and c are coplanar. This means that c = α a + β b for scalars α, β R. In other words, c lies in the plane spanned by a and b. Note that α = and β = 3 does the job. Exercise 36: Find the magnitude of the torque about P if a 36-lb force is applied as shown. P 4 ft r 30 4 ft C F line-of-action of the force F Notice that the angle θ between r and F is = 105 and that r = 4 since r is the diagonal of a square. Let τ = r F denote the torque about P in the figure. Now we know that τ = r F = r F sin 105 = (4 ft)(36 lb) sin ft-lbs

4 Exercise 37: A wrench 30 cm long lies along the positive y-axis and grips a bolt at the origin. A force is applied in the direction 0, 3, 4 at the end of the wrench. Find the magnitude of the force needed to supply 100 J of torque to the bolt. Let O(0, 0, 0) denote the origin (where the bolt is) and let P(0, 0.3, 0) denote the position of the end of the wrench. Then r = 0, 0.3, 0 is the position vector emanating from the origin to the end of the wrench. Let û = 1 5 0, 3, 4 denote a unit vector in the direction of 0, 3, 4. The angle θ between r and û is given by ( 1 ) θ = cos 1 5 ( 0, 0.3, 0 0, 3, 4 ) = cos 1 (0.6). r Now the magnitude of the torque τ at the origin is given by r F sin θ 100 = 0.3 F sin(cos 1 (0.6)) or F = 100 Nm (0.3 m) sin(cos 1 (0.6)) = N 417 N. Exercise 39 (Distance from a point to a line): (a) Let P be a point not on the line L that pass through the points Q and R. Show that the distance d from the point P to the line L is where a = QR and b = QP. d = a b Use the formula in part (a) to find the distance from the point P(1, 1, 1) to the line through Q(0, 6, 8) and R( 1, 4, 7). (a) Notice that d = b sin θ where θ is the angle between a and b. Thus d = b sinθ = b sin θ = a b With a = QR = 1,, 1 and b = QP = 1, 5, 7 and a b = 9, 8, 7. Then by the formula in part (a) we have d = 9 + ( 8) + 7 ( 1) + ( ) + ( 1) = 97 3 Exercise 40 (Distance from a point to a plane): (a) Let P be a point not on the plane that passes through the points Q, R, and S. Show that the distance d from P to the plane is where a = QR, b = QS, and c = QP. d = ( a b) c a b Use the formula in part (a) to find the distance from the point P(, 1, 4) to the plane through the points Q(1, 0, 0), R(0,, 0), and S(0, 0, 3). (a) Consider the parallelepiped formed with the vectors a, b, and c with the parallelogram formed with a and b as its base. Then the distance d is the height h of this parallelepiped which is c cosθ where θ is the angle between the vectors a b and c. Let V denote the volume of this parallelepiped and let A denote the area of the parallelogram formed by a and b, then we have d = c cosθ = h = V A = a ( b c) a = ( a b) c b a b We use the above formula with a = QR = 1,, 0, b = QS = 1, 0, 3, and c = QP = 1, 1, 4 we have a b = 6, 3, and ( a b) c = 6(1)+3(1)+(4) = 17 and a b = = 49 = 7. Thus d = ( a b) c a b = 17 7 = 17 7

5 Exercise 4 (Vector Triple Product) Prove part 6 of Theorem 8, that is, a ( b c) = ( a c) b ( a b) c. Consider vectors a = a 1, a, a 3, b = b 1, b, b 3, and c = c 1, c, c 3. Now we have a ( b c) = a 1, a, a 3 ( b 1, b, b 3 c 1, c, c 3 ) = a 1, a, a 3 b c 3 b 3 c, b 3 c 1 b 1 c 3, b 1 c b c 1 = a (b 1 c b c 1 ) a 3 (b 3 c 1 b 1 c 3 ), a 3 (b c 3 b 3 c ) a 1 (b 1 c b c 1 ), a 1 (b 3 c 1 b 1 c 3 ) a (b c 3 b 3 c ) = a b 1 c a b c 1 + a 3 b 1 c 3 a 3 b 3 c 1, a 3 b c 3 a 3 b 3 c + a 1 b c 1 a 1 b 1 c, a 1 b 3 c 1 a 1 b 1 c 3 + a b 3 c a b c 3 = a b 1 c + a 3 b 1 c 3, a 3 b c 3 + a 1 b c 1, a 1 b 3 c 1 + a b 3 c a b c 1 + a 3 b 3 c 1, a 3 b 3 c + a 1 b 1 c, a 1 b 1 c 3 + a b c 3 = (a 1 c 1 + a c + a 3 c 3 ) b 1, b, b 3 (a 1 b 1 + a b + a 3 b 3 ) c 1, c, c 3 = ( a c) b ( a b) c