i need help with these, thank you so much

2.1 & Right-handed, orthogonal, unit vectors. (Section 4.1) Draw a set of right-handed orthogonal (mutually perpendicular) unit vectors consisting of Ix, ny, n2. In other words, draw fix, Ry, n, so that ny is perpendicular (orthogonal) to fix and Riz = Rx x Ry- 2.2 & Adding and subtracting vectors. (Sections 2.6, 2.8) Given: Vectors p and q expressed in terms of unit vectors p = ai+ bj+ ek i, j. k. Form the vector sums and differences below. q = ri+ yj + zk Result: p t q = (atz)it it(k p- q= (a- r)it ()i+ 2.3 & Words: Physical vectors and column matrices. (Section 2.1, Hw 1.2) True/False As defined by Gibbs and for F = ma, physical vectors have magnitude and direction. True/False In math (linear algebra), a column matrix is called a "vector". True/False The physical vector a, + 2a, + 3a, can be written (as a, a.]* Note: as, ay, a, are the orthogonal unit vectors shown below. True/False The physical vector a, + 2a, + 3a, is equal to the column matrix True/False ax + 2a, + 3a, + 4b, + 5by + 6bz 49 19 - + (Ax, Ay, as and bx, by, be are shown right). . Complete the following statement with one equal sign =] and one not-equal sign # ax + 2ay + 3as a a, a, ] 2.4 & Dot products with orthogonal unit vectors. (Sections 2.9, 2.9.4) Given: Vectors v and w expressed in terms of right-handed orthogonal unit vectors i, j. k, with: V . w = (ai + bj + ck) . (ri + yj+ zk . Use the distributive property for dot products to write v . w in terms of i. i, i. j, etc. Next, use the definition of the dot product to calculate i. i, i . j, etc. (below-right). Result: vow= ariit ayi.j i. i=lij= i . k = + brjit by Pi=o jj= j . k = + er kit cy 6. i=0 k. j= E . K = . Simplify V . w and use its special dot-product formula for the calculations that follow. Result: V- w = or+ by+ Use this special dot-product formula to calculate v - w when i, j, k are orthogonal unit vectors. Given Calculate p = 2i + 3j + 4k p . q =2x + 3y + 2 P . p = 29 p = V29 q = xi+ vi + zk p . r = 4.q=2'+ + b F = 51 - 6j + 7k 4 . F = F - F = Copyright (5 1990-2021 Pand Mitigay. All rights reserved. 141 Homework 2: Vectors with basis2.5 & Perpendicular vectors. (Note: 1].; are orthogonal unit vectors]. {Section 2.9) | I Draw two vectors v and\" w that are perpendicular Hence i'r - i? = . : : WhenvI1+2_]+3k lsperpendlculartow41+5j tilt, at: I | 2.6 Dot products to calculate distance and angles. {Sections 2.9, 3.4) / The gure to the right shows a block with sides of length 2 n1, 3 n1. 4m and points A, B, C located at corners. Righthanded orthogonal unit vectors x. y, g are directed with x from B to Cand y frotho A. (3.} Express i: (position. from A to C} in terms of Ex, y. 3,. and nd a numerical value for |F| 2. Next calculate the distance d between A to C [magnitude of F}. Result: f 2 x _ a], lit"? {2:} f . f = m2 d = \\fm .4 (b) Calculate the unit vector directed from A to C and the unit vector ? directed From A to D. Result: \":74 (c) Calculate LBHC {angle between iinc 313 and line TC] and LCAD{ {angle between line TC and line AD}. Result 15.40 = 40.43 = m 2.7 & Construct a unit vector ii in the direction of each vector given below. (Section 2.9.2] Note: 1 j E are orthogonal unit vectors. _ 3i 4} 33 43+12E Ensure your last answer agrees with your ci i or signfi c is a \":31 nonzero number rst two answers: I:_g_, if c : 3 or c : -3_ 2.8 II. Vector components: Sine and cosine. {Section 1.4] '.' I Replace each ? in the gure to the right with 5111(3) or 005(6). E 0 Use vector addition to express 3 and E in terms of stow}. oos], 13 1 I? j if? a = i " 3 i 9 1 '-' mm or: A A A '- SohCahToa b = i __ 005(3) j I B ,2 _|j 2.9 & Vector components for a crane-boom. (Section 1.4) Shown right is a crane whose cab A supports a boom B that swings a wrecking ball Co. Draw position vectors L Right-handed orthogonal unit vectors i, j, k are directed with i horizontally-right, j vertically- upward, and k outward-normal to the plane containing points No, AB, Bo, Co. (N Draw each position vector listed below and No X then use your knowledge of sine/cosine to re- (A AB solve these vectors into i and j components. Position from No to Ap NopAs = + Position from As to Be ABFBC = i + Position from Be to Co BoCo = i + Position from N. to Bc No BC = 0+ + Position from N. to Co NoCo = i + [LB sin(GB) - LC cos(ec)] j 2.10 Dot products and distance calculations. Show work. (Section 2.9) BC Shown right is a crane whose cab A supports a boom B that swings a wrecking ball Co. To prevent the wrecking ball from hitting a car, the distance between No and point Bo (the tip Any of the boom) must be controlled. To start this problem, express I (the position vector from No to Be) in terms of r, LB, nix, bx. Result: r = nx + bx . Without resolving r into ix and ny components, use [r| = vr . r [from equation (3.1)] and the distributive property to calculate the distance between No and Bo in terms of r, LB, OB. Result: (if stumped, hint below). Optional: Calculate [F| when x = 20 m, Lg = 10 m, 0g = 30". Distance between N. and Be: If| = VO' + + 2LB cos(0B) = 29.1m . Homework 2.9 showed I can be expressed as r = [r + LB cos(0B)] nix + LB sin(0B)ny. Use this expression to verify your previous result for |r| = vr . F. Result: (F] simplifies to the previous result but uses an inefficient process and sin () + cos'(0B) = 1