Rather than use the standard definitions of addition and scalar multiplication in R³, suppose these two operations are defined as follows. With these new definitions, is R³ a vector space? Justify your answers. (a) (X₁, X₁, Z₁) + (x2, Y2, 2₂) = (X₁ + X2, Y₁ + Y2, Z1 + Z₂) c(x, y, z)= (0, cy, cz) O The set is a vector space. O The set is not a vector space because the associative property of addition is not satisfied. vector space because it is not closed under scalar multiplication. O The set is not a O The set is not a O The set is not a vector space because the associative property of multiplication is not satisfied. vector space because the multiplicative identity property is not satisfied. (b) (x1, ₁, 2₁) + (x2, Y2, 2₂) = (0, 0, 0) c(x, y, z) = (cx, cy, cz) O The set is a vector space. O The set is not a vector space because the commutative property of addition is not satisfied. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the multiplicative identity property is not satisfied. (c) (X₁, ₁, 2₁) + (x2, Y2, 2₂) = (x₁ + x₂ + 9, Y₁ + y₂ + 9, Z₁ + Z₂ + 9) c(x, y, z) = (cx, cy, cz) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because the additive inverse property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. (d) (x₁, Y₁, 2₁) + (x₂, Y/2, 2₂) = (X₁ + X₂ + 8, Y₁ + y₂ + 8, Z₁ + Z₂ + 8) c(x, y, z) = (cx + 8c-8, cy + 8c-8, cz + 8c-8) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied.

Calculate the triple scalar products w. (v × u) and u ⋅ (w × v), where u = (7, 2, −1), v = w. (vx u) = 330 = (2, 8, -3), and w = u. (w x v) = 350 = (9, 8, -10).

1. Find the scalar product (a.b) and the vector product ( a × b ), when (a)   a = i + 2j – k, b = 2i + 3j + k (b)  a = 2i + 3j + 4k, b = 5i – 2j + k   2.

For the vectors A = ux + 3uy + 5uz, B = 2ux + 4uy + 6uz, and C = 3ux + 4uy + 5uz, show that A × ( B × C ) = B ( A ⋅ C ) - C ( A ⋅ B)

[3] respectively. Let y and u = -8 [] Write y as the sum of two orthogonal vectors, y, in the span of u, and y, which is orthogonal to u. Enter them as 2x1 arrays into y_u and y_o

1. Let x = i +4j + 2k and y = i-3j - k. In this question, write all vectors in i, j, k notation. (a) ( Find |x||. (b) M Find 2x + 3y. (c) Find x. y. (d) Find x × y. Answer: Answer: Answer:

Let x, y, z be (non-zero) vectors and suppose that z = -1x + 1y and w = −1x + 1y - 3z. Are the following statements true or false?

For the vectors A = Ux + 3uy + 5uz , B = 2ux, 4uy + 6uz, and C = 3ux + 4uy + 5uz. Show that A x (B x C) = B (A x C) - C(A x B)

The scalar triple product of vectors a, b, c is a⋅(b×c) . Write a program in Scilab that includes a function scalarTripleProduct(a,b,c) where a,b,c are any three-dimensional vectors. The program should calculate and display the value of a⋅(b×c) for a= (1 2 3) b= (−2 1 3) c= (3 −2 −1) It should also calculate and display the values of c⋅(a×b) and b⋅(c×a) to verify the identity a⋅(b×c)=c⋅(a×b)=b⋅(c×a) need help with the scilab code for this plz.

Let V = R?. For (u1, u2), (v1, v2) E V and a E R define vector addition by (u1, u2) # (v1, v2) := (u1+ v1 + 3, u2 + v2 a O (u1, u2) - 1) and scalar multiplication by := (au1 + 3a – 3, au2 – a + 1). It can be shown that (V, H, D) is a vector space. Find the following: the sum: (2, —9) в (0, —8) %3D( 5 -18 the scalar multiple: -70 (2, –9) =( -38 57 the zero vector: " 0 "=( -1 22 3 the additive inverse "-v" of v = (x,y): " -v "=( 3 )(Must be in terms of x and y)