Need number 3,4,5 thanks! And explain too!! As shown in Figure 1-1, consider the

Question

Need number 3,4,5 thanks! And explain too!! As shown in Figure 1-1, consider the orthogonal projection to the Let theta be the angle vector "in the direction of the straight line * of arbitrary vectors v and w. Answer the following questions. (1) Find the unit vector in the vector direction. Jo) (2) Represent with v and . (3) From (1) and (2), find the matrix A which is (4) Next, as shown in Figure 1-2, consider a new XY orthogonal coordinate system with the X axis in the direction of the line as shown in Fig. 1-2 Find a transformation matrix P such that transforms the pointin xy coordinate to ' In X coordinate system 5.Express the matrix A obtained in (3) with the matrix B and the transformation matrix P Figure 1-1 Figure 1-2

Explanation / Answer

1. w is a vector in direction y = x/3

so, w passes though (0,0) and (1,1/3)

hence a unit vector in direction of w is ew

ew = [(1 - 0)i + (1/3 - 0)j]/sqroot(1^2 + (1/3)^2)

ew = (3i + j)/sqroot(10) = 0.94868i + 0.316j

2. now given f(v) is orthogonal projection of xi + yj on vector w

hence

f(v) = vcos(theta)ew

3. f(v) = Av

hence A = cos(theta)ew = cos(theta)[0.94868]

[0.316 ]

4. for the new axis system

[X Y]T = P[x y]T

now as we can see

if I and J are unit vectors along X and Y

I =0.94868i + 0.316j

J = ai + bj

I.J = 0

so, 0.94868a + 0.316b = 0

a = -0.333094b

alos sqroot(a^2 + b^2) = 1

0.1109b^2 + b^2 = 1

b = 0.94868, a = - 0.316

hence

I =0.94868i + 0.316j

J = -0.316i + 0.94868j

so P = [ 0.94868 0.316 ]

[-0.316 0.94868]

now, A = PB

cos(theta)[0.94868]

[0.316 ] = [ 0.94868 0.316 ]*B

[-0.316 0.94868]

B = cos(theta)[ 0.94868 -0.316 ][0.94868]

[0.316 0.94868][0.316]

B = cos(theta)[0.8001377424 0.59956576]T ( where T stands for trnaspoose)

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.