Question

What linear combination of (1, 2, -1) and (1, 0, 1) is closest to b = (2, 1, 1 )

Answer 1

To find the linear combination of (1, 2, -1) and (1, 0, 1) that is closest to b = (2, 1, 1), we need to minimize the distance between the linear combination and b.

Step-by-step explanation:

To find the linear combination of (1, 2, -1) and (1, 0, 1) that is closest to b = (2, 1, 1), we need to find coefficients a and c such that a(1, 2, -1) + c(1, 0, 1) is closest to b. We can do this by minimizing the distance between a(1, 2, -1) + c(1, 0, 1) and b, which is equivalent to minimizing the quadratic function (a + c - 2)² + (2a - c - 1)² + (-a + c - 1)². By expanding and simplifying this expression, we can find the values of a and c that minimize the distance.

Answer 2

To find the linear combination of (1, 2, -1) and (1, 0, 1) that is closest to b = (2, 1, 1), we can use the concept of projection. The projections of b onto each of the given vectors can be found using the formula proj_v(b) = (dot_product(b, v) / dot_product(v, v)) * v. By calculating the projections and adding them together, the closest linear combination to b is (5/6, 2/3, 2/3).

Step-by-step explanation:

In order to find the linear combination of (1, 2, -1) and (1, 0, 1) that is closest to b = (2, 1, 1), we can use the concept of projection. The projection of b onto a vector v is given by the formula proj_v(b) = (dot_product(b, v) / dot_product(v, v)) * v. Here, the dot_product is the scalar product or dot product between two vectors.

So, the linear combination of (1, 2, -1) and (1, 0, 1) that is closest to b = (2, 1, 1) can be found by finding the projections of b onto each of the given vectors, and then adding them together. Let's calculate step-by-step:

First, let's calculate the projection of b onto the vector (1, 2, -1):

proj_v1(b) = (dot_product(b, v1) / dot_product(v1, v1)) * v1

dot_product(b, v1) = (2 * 1) + (1 * 2) + (1 * -1) = 1 + 2 - 1 = 2

dot_product(v1, v1) = (1 * 1) + (2 * 2) + (-1 * -1) = 1 + 4 + 1 = 6

proj_v1(b) = (2 / 6) * (1, 2, -1) = (1/3) * (1, 2, -1) = (1/3, 2/3, -1/3)

Next, let's calculate the projection of b onto the vector (1, 0, 1):

proj_v2(b) = (dot_product(b, v2) / dot_product(v2, v2)) * v2

dot_product(b, v2) = (2 * 1) + (1 * 0) + (1 * 1) = 2 + 0 + 1 = 3

dot_product(v2, v2) = (1 * 1) + (0 * 0) + (1 * 1) = 1 + 0 + 1 = 2

proj_v2(b) = (3 / 2) * (1, 0, 1) = (3/2) * (1, 0, 1) = (3/2, 0, 3/2)

Finally, let's add the two projections together to get the linear combination closest to b:

(1/3, 2/3, -1/3) + (3/2, 0, 3/2) = (1/3 + 3/2, 2/3 + 0, -1/3 + 3/2) = (5/6, 2/3, 4/6) = (5/6, 2/3, 2/3)