91Ó°ÊÓ

The dot product is a way to multiply two vectors to get a scalar (a single number). It helps in measuring how much one vector goes in the direction of another. For two vectors \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) and \(\mathbf{u} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\), their dot product is calculated as:

• \( \mathbf{v} \cdot \mathbf{u} = ax + by + cz \).

This formula adds the products of corresponding components of the vectors.
In our exercise, \( \mathbf{v}=10 \mathbf{i}+11 \mathbf{j}-2 \mathbf{k}\) and \( \mathbf{u}=0 \mathbf{i}+3 \mathbf{j}+4 \mathbf{k}\), the dot product is \( 10 \times 0 + 11 \times 3 - 2 \times 4 = 25 \).
The result, 25, shows the degree of parallelism between \(\mathbf{v}\) and \(\mathbf{u}\): a higher result means the vectors are more aligned.

Understanding vector magnitude is like understanding the length of a vector in space. It tells you how long that vector stretches. For any vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), its magnitude is found using:

• \( |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \)

Let's calculate the magnitudes in the example:
For vector \( \mathbf{v} = 10\mathbf{i} + 11\mathbf{j} - 2\mathbf{k} \):
\(|\mathbf{v}| = \sqrt{10^2 + 11^2 + (-2)^2} = 15 \)
For vector \( \mathbf{u} = 0\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} \):
\(|\mathbf{u}| = \sqrt{0^2 + 3^2 + 4^2} = 5 \)
Magnitudes help us understand how "big" or "strong" each vector is.

Projection of a vector breaks down its components along another vector's direction. Think of it as shadowing one vector on another. It helps in understanding how much of one vector is in the direction of another.
The formula for the projection of vector \( \mathbf{u} \) on vector \( \mathbf{v} \) is:

• \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left(\frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|^2}\right) \mathbf{v} \)

In our example, substituting the values gives:
\( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{25}{15^2}(10\mathbf{i} + 11\mathbf{j} - 2\mathbf{k}) = \left(\frac{1}{9}\right)(10\mathbf{i} + 11\mathbf{j} - 2\mathbf{k}) \).
This results in the vector \( \frac{10}{9}\mathbf{i} + \frac{11}{9}\mathbf{j} - \frac{2}{9}\mathbf{k} \).
It shows how much \( \mathbf{u} \) points along \( \mathbf{v} \).

Finding the angle between two vectors can tell you how much they point in the same or opposite directions. The relationship is given by the formula:

• \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}||\mathbf{u}|} \)

The \( \cos \theta \) can give us a direct insight into the vector alignment.
If \( \cos \theta \) is 1, vectors point in the same direction;
-1 means opposite directions, and 0 shows perpendicular vectors.
In this exercise, the computation gives \( \cos \theta = \frac{1}{3} \), which is a positive value between 0 and 1.
It implies the vectors \( \mathbf{v} \) and \( \mathbf{u} \) are not completely aligned, but they do point somewhat in the same direction.