Solutions

These solutions collect the exercise reasoning in one place.

MATH-C04-C-001

(1,4) + [2,-3] = (3,1)

The reached point is (3, 1).

First compute the difference:

b - a = [8 - 2, 9 - 1] = [6, 8]

Then compute its length:

\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10

The distance is 10.

u \cdot v = 3(1) + 1(-3) = 3 - 3 = 0

The vectors are orthogonal.

Since u = [1, 0] is a unit vector:

\operatorname{proj}_u(a) = (a \cdot u)u

Now:

a \cdot u = 6(1) + 2(0) = 6

So:

\operatorname{proj}_u(a) = 6[1, 0] = [6, 0]

w \cdot x + b = 1(3) + 2(1) - 5 = 3 + 2 - 5 = 0

The point lies on the boundary.

The shifted line is:

y = 2x + 1

At [0,0], the left side is 0, but the right side is 1. The zero vector is not on the shifted line.

The vectors [1,0] and [2,0] both lie on the horizontal axis.

They span a line, not the full plane, so they do not form a basis for two-dimensional space.

Nearby embeddings are useful because they show similarity under the model's learned representation.

They are not a complete explanation because the model may use other directions, layers, or context to make its final decision.

Low-dimensional drawings are useful models.

They should still be checked against formulas or code when reasoning about the actual high-dimensional space.

A good question to keep is:

What is this operation
doing to the space?

That question connects vector and matrix computation to geometry.

The residual is original minus projection:

[5,2] - [5,0] = [0,2]

The projection kept the horizontal component. The residual is the vertical part that was not kept.

The first boundary is:

s(x) = 0

The second boundary is:

2s(x) = 0

These have the same zero-score points. Multiplying the whole score by a positive constant changes raw score size, but not the boundary.

Use the basis vectors:

2b_1 + 3b_2 = 2[1,1] + 3[0,1]

So:

[2,2] + [0,3] = [2,5]

The standard vector is [2, 5].