vectr_mtrics_sol

# vectr_mtrics_sol - MIT OpenCourseWare http/ocw.mit.edu...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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; therefore X = Y. 0
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 = the angle 0 is acute if cose > 0, i.e., if c > 4. dm&' 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 = the angle is acute if cos0 > 0, i.e., if a2 drndrn' - 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 = - Jz J=- ' ' 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
1. VECTORS AND MATRICES or in terms of A, remembering that IAI2 = A .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 05/04/2011 for the course MATH 18.02 taught by Professor Auroux during the Spring '08 term at MIT.

### Page1 / 14

vectr_mtrics_sol - MIT OpenCourseWare http/ocw.mit.edu...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online