Starting with an expression L ? , push the projection as far as it can go if L is:
a) b + c->x, c + d->y.
b) a,b,a + d -> z
R X S (cross roduct) : {(a,b), (a, c), (a, d), (a, e), (b, b), (b, c), (b, d), (b, e), (c, b), (c, c), (c, d), (c, e)}
Starting with an expression L ? , push the projection as far as it can go...
Consider the right triangle shown, where , y, and L are lengths. Starting each part of this problem! from the fundamental definition sin θ-y/ we have many choices for how to express t. Our goal is to explore whiclh expressions are most convenient when either y 《 or r 《 (a) Show that sin- Why is this expression, which has the dimensionless ratio x/y inside the square root, a non-ideal choice in the limit y (with r constant)? (b) Construct...
show all work please, thanks Find the projection of vector v onto line L. v = <5,-1,2>, L: x=3, y + 4 22 3 # 1 , 2: Find the projection of vector 1. =(5,-1,2), l: x = onto line f. y+4 3, -1 Z-2 3 y + 4 22 3 # 1 , 2: Find the projection of vector 1. =(5,-1,2), l: x = onto line f. y+4 3, -1 Z-2 3
9. We know from the last homework that the angle between l;(x) = x3 and lt1() basis to construct an orthogonal basis for a polynomial space over the domain 0, 1 not perpendicular. However, we can use this е lo(x). Compute the projection of (a) Let go(r) and call it w(x) (x) onto span{go(ar)}, (b) Сompute q. (") — 6, (ӕ) — w;(»). (c) Compute the projection of l2(x) onto span{go(x), q1 (x)}, and call it w2(x) (d) Compute 2(x)=...
(a) There are two players in this game - A and B. A moves first and can go either left L or right R. B does not see what A did, and can go either up U or down D. If B went D, A moves again, and can choose 1, c, or r. If B went U then A can move again only if he started with going L in the beginning. In this case A can choose either...
Could you show how to do (d) step by step, please? AO = Atomic orbital MO = Molecular orbital (L)GO= (Ligand) Group orbital 2. TaHs has a square pyramidal geometry and a Cev symmetry. (a) Assign x, y, z coordinates on the atoms: - Z axis on central Ta atom is the principal rotation axis (z axes on other atoms are colinear) - X, y axes on central Ta atom match Ta-H bonds (x, y axes on other atoms are...
{ <N> : L(M) contains a string starting with a). Rice's theorem can be F 20, L used to prove that LD. T L(M2) >. Rice's theorem can be used to prove T F 21. L that L D. <M,, M2> L(M,) 22. L-( <M,M> : L(M) = L(M2) }, and R is a mapping reduction function from H to L. It is possible that R retur a TM. T F ns <M#>, where M # is the string encoding...
16. Very far away from the source of the EM wave, you can often treat it as a plane At the location of your detector, which is in vacuum, suppose that you magnitude and direction of the electric field of an EM wave is given by wave. find that the E Eo cos (ot - kz)X (a) Give an exact expression for the magnitude and direction of the magnetic field in the wave: (b) In wh at direction will your...
How can I get the (a) 3*2 matrix A? x 7. [30pts] Let V be the subspace of R consisting of vectors satisfying x- y+z = 0 y (a) Find a 3x2 matrix A whose column space is V and the entries a a1 0 = (b) Find an orthonormal basis for V by applying the Gram-Schmidt procedure (c) Find the projection matrix P projecting onto the left nullspace (not the column space) of A (d) Find an SVD (A...
1. E-far field is in -x direction and H-far field is z direction. Where is the direction of wave propagation? 2. An infinitesimal dipole is located at the origin along the y-axis. Find out the following directivity with both linear scale and dB scale in a) xdirection b) -x direction c) +y direction d) -y direction e) z direction f -z direction
Determine a general expression for the scalar projection of F onto the line directed from B to D. Point M is located at the center of the bottom face of the parallelepiped. Evaluate your expression from d = .3b and d = 1.5b. d= .3 FBD _____ d= 1.5b FBD ______ Determine a general expression for the scalar projection of F onto the line directed from B to D. Point M is located at the center of the bottom face...