6. Compute the dot product u ∙ v.

To compute the dot product between two vectors, u and v, we need to multiply their corresponding components and then sum them up.

Given that u = (-1, 3, 3) and v = (9, 2, -3), we will multiply their corresponding components and add the results:

u ∙ v = (-1 9) + (3 2) + (3 * -3)

Now let's simplify this expression step by step:

u ∙ v = -9 + 6 - 9

u ∙ v = -12

Therefore, the dot product of u ∙ v is -12.

So, the correct answer is D) -12.

To find the distance between two vectors, u and v, we can use the Euclidean distance formula. This formula calculates the straight-line distance between two points in a three-dimensional space.

Given that u = (-4, 6, -5) and v = (-3, 10, 4), the distance between these two vectors can be calculated as follows:

Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)

Where (x1, y1, z1) are the components of vector u and (x2, y2, z2) are the components of vector v.

Now let's plug in the values:

Distance = sqrt((-3 - (-4))^2 + (10 - 6)^2 + (4 - (-5))^2)

Distance = sqrt(1^2 + 4^2 + 9^2)

Distance = sqrt(1 + 16 + 81)

Distance = sqrt(98)

Therefore, the distance between the vectors u and v is sqrt(98), which is approximately equal to 9.9.

So, the correct answer is D) 98.

Express the vector x as a linear combination of the u's.

To express the vector x as a linear combination of the vectors u1, u2, and u3, we need to multiply each vector u by its corresponding scalar coefficient and then add them together.

Given that u1 = (-2, 0, 1), u2 = (3, 5, 6), u3 = (-2, 6, -4), and x = (6, 50, 2), we can express x as:

x = a*u1 + b*u2 + c*u3

where a, b, and c are scalar coefficients.

Now let's find the values of a, b, and c:

6 = a*(-2) + b*3 + c*(-2)

50 = a*0 + b*5 + c*6

2 = a*1 + b*(-4) + c*(-4)

To solve this system of equations, we can use various methods such as substitution or elimination. I will use the elimination method.

Multiplying the first equation by -1 and adding it to the third equation eliminates variable a:

2 = -a - 4b - 4c   (Modified third equation)

Now we have a system of two equations with two variables:

-3 = 7b + 5c     (Second modified equation)

2 = -a - 4b - 4c   (First equation)

Solving this system of equations, we find that b = 1 and c = -1.

Substituting these values back into the original second equation, we can solve for a:

50 = a*0 + 1*5 + (-1)*6

50 = 5 - 6

a = 51

Therefore, the scalar coefficients are a = 51, b = 1, and c = -1.

Now, we can express vector x as a linear combination of the vectors u1, u2, and u3:

x = 51*u1 + 1*u2 - 1*u3

So, the correct answer is A) x = 2u1 - 4u2 - 5u3.

Find the orthogonal projection of y onto u.

C)

4

-2

9)

D)

4

5

2

5

To find the orthogonal projection of vector y onto vector u, we can use the formula:

Proj_u(y) = (y dot u) / (u dot u) * u

Let's calculate step by step:

1. Calculate the dot product of y and u:

(y dot u) = (10 4) + (-30 -2)

          = 40 + 60

          = 100

2. Calculate the dot product of u and itself:

(u dot u) = (4 4) + (-2 -2)

          = 16 + 4

          = 20

3. Divide the dot product of y and u by the dot product of u and itself:

(100) / (20) = 5

4. Multiply the scalar value (5) by u:

(5) u = (5) (4, -2) = (20, -10)

Therefore, the orthogonal projection of vector y onto vector u is (20, -10).

So, the orthogonal projection of vector y onto vector u is (20, -10). Answer is B

Let W be the subspace spanned by the u's. Write y as the sum of a vector in W and a vector orthogonal to W.

C)

10)

D)

1         -11

y =  21  +       7

17            -8

To write vector y as the sum of a vector in the subspace W spanned by the u's and a vector orthogonal to W, we can use the orthogonal projection formula.

1. Calculate the orthogonal projection of y onto the subspace W:

   Proj_W(y) = [(y dot u1) / (u1 dot u1)] u1 + [(y dot u2) / (u2 dot u2)] u2

   Let's calculate this step by step:

   - Calculate the dot product of y with u1:

     (y dot u1) = (12 2) + (14 2) + (25 * -1)

                = 24 + 28 - 25

                = 27

   - Calculate the dot product of u1 with itself:

     (u1 dot u1) = (2 2) + (2 2) + (-1 * -1)

                 = 4 + 4 + 1

                 = 9

   - Calculate the dot product of y with u2:

     (y dot u2) = (12 -1) + (14 3) + (25 * 4)

                = -12 + 42 + 100

                = 130

   - Calculate the dot product of u2 with itself:

     (u2 dot u2) = (-1 -1) + (3 3) + (4 * 4)

                 = 1 + 9 + 16

                 = 26

   - Plug the values into the projection formula:

     Proj_W(y) = [(27 / 9)] (2, 2, -1) + [(130 / 26)] (-1, 3, 4)

     Simplifying, we have:

     Proj_W(y) = (3) (2, 2, -1) + (5) (-1, 3, 4)

               = (6, 6, -3) + (-5, 15, 20)

               = (1, 21, 17)

2. Calculate the orthogonal complement of the subspace W:

   This is the vector that is orthogonal to every vector in W. If we take any vector orthogonal to W and add it to a vector in W, the resulting vector will be vector y.

   We can find the orthogonal complement by subtracting the projection from y:

   Orthogonal complement = y - Proj_W(y)

                        = (12, 14, 25) - (1, 21, 17)

                        = (12 - 1, 14 - 21, 25 - 17)

                        = (11, -7, 8)

Therefore, y can be written as the sum of a vector in W and a vector orthogonal to W as follows:

y = (1, 21, 17) + (11, -7, 8)

So, the vector (12, 14, 25) can be written as the sum of a vector in the subspace spanned by u1 and u2, which is (1, 21, 17), and a vector orthogonal to the subspace, which is (11, -7, 8). Answer is C