1. Vectors and Matrices Solutions

1. Vectors and Matrices Solutions - 1. Vectors and Matrices...

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1. Vectors and Matrices 1A. Vectors 1A-1 a) IAl =&, dir A = A/& b) IAI = 3, dir A =A/3 c) IAI = 7, dir A = A/7 b) A = IAl dir A = 2i + 4j - 4k. Let P be its tail and Q its head. Then OQ = OP + A = - 3 k; therefore Q = (0,4, -3). 1A-4 a) OX = + PX = + ~(PQ) = + $(OQ - OP) = $(OP + OQ) b) OX = s + r OQ; replace 3 by r in above; use 1 - r = s. 1A-5 A = i&i + i j . The condition is not redundant since there are two vectors of length 3 making an angle of 30° with i. 1A-7 a) bi-aj b) -bi+aj c) (3/5)2+(4/5)2=1; jl=-(4/5)i+(3/5)j 1A-8 a) is elementary trigonometry; b) cos a = alda2 + b2 + c2, etc.; dir A = (-1/3,2/3,2/3) c) if t, v, v are direction cosines of some A, then ti +v j +v k = dir A, a unit vector, so t2 + v2 + v2 = 1;conversely, if this relation holds, then ti + v j + v k = u is a unit vector, so dir u = u and t, v, v are the direction cosines of u. 1A-9 Letting A and B be the two sides, the third side is B - A; the line joining the two midpoints is $B - $A, which = $ (B - A), a vector parallel to the third side and half its length. 1A-10 Letting A, B, C,D be the four sides; then if the vectors are suitably oriented, we have A + B = C+D. The vector from the midpoint of A to the midpoint of C is $C - +A; similarly, the vector joining the midpoints of the other two sides is $B - $D, and A+B=C+D C-A=B-D $(C-A)=$(B-D); thus two opposite sides are equal and parallel, which shows the figure is a parallelogram. 1A-11 Letting the four vertices be 0, P,Q, R, with X on PR and Y on OQ, OX = OP+PX = OP++PR = + +(OR - = i(OR+OP) = $OQ = OY; 0 therefore X = Y.
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2 S. 18.02 SOLUTIONS TO EXERCISES 1B. Dot Product 1B-2 A . B = c - 4; therefore (a) orthogonal if c = 4, c-4 b) cos 0 = dm&' the angle 0 is acute if cose > 0, i.e., if c > 4. 1B-3 Place the cube in the first octant so the origin is at one corner P, and i, j , k are three edges. The longest diagonal PQ = i + j + k; a face diagonal PR = i + j. 1B-4 QP = (a, 0, -2), QR = (a, -2,2), therefore a) QP . = a2 - 4; therefore PQR is a right angle if a2 - 4 = 0, i.e., if a = f 2. -4 b) cose = drndrn' the angle is acute if cos0 > 0, i.e., if a2 - 4 > 0, or la1 > 2, i.e., a > 2 or a < -2. 1B-5 a) F . u = -I/& b) u = dir A = A/7, so F - u = -417 1B-6 After dividing by IOPI, the equation says cos0 = c, where 6 is the angle between OP and u; call its solution Oo = cos-' c. Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 200. In particular this cone is a) a plane if Oo = n/2, i.e., if c = 0 b) a ray if Bo = 0, n, i.e., if c = f 1 c) nonexistent if Bo is nonexistent, i.e., c > 1or c < -1. 1B-7 I i'l = I j'l = - Jz = 1;a picture shows the system is right-handed. b) A. il= -1/a; A. jl= -5/a; - if - 5j' since they are perpendicular unit vectors, A = i' - j' . it+ j'. C) Solving, i = - J=- Jz ' ' 2(i1-j') - 3(i1+j')--il- thus A=2i -3j = Jz Jz - , as before. 1B-8 a) Check that each has length 1,and the three dot products . j', i' . k', . k' are 0; make a sketch to check right-handedness. b) A.il=&, A.jl=O, A.kl=&, therefore, A=fii1+&k'. 1B-9 Let u = dir A, then the vector u-component of B is (B .u)u. Subtracting it off gives a vector perpendicular to u (and therefore also to A); thus
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1. VECTORS AND MATRICES or in terms of A, remembering that IAI2 = A .
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1. Vectors and Matrices Solutions - 1. Vectors and Matrices...

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