91Ó°ÊÓ

In vector calculus, a **direction vector** plays a crucial role, as it describes the direction in which a line or vector extends. Imagine it as an arrow indicating the line's path. To identify a direction vector from a line's parametric equations, look at the coefficients of the parameter—usually noted as 't' or 's.' For instance, if you encounter parametric form equations, such as:

• For line \( \mathscr{L}_{a} \): \( x = 4 - t \), \( y = 3 + 2t \), \( z = -2t \), gives us the direction vector \( \mathbf{a} = \langle -1, 2, -2 \rangle \).
• For line \( \mathscr{L}_{b} \): \( x = 5 + 2s \), \( y = 1 + 3s \), \( z = 5 - 6s \), resulting in \( \mathbf{b} = \langle 2, 3, -6 \rangle \).

Direction vectors are essential for calculating angles between lines or vectors, as they provide the needed orientation in space.

The **dot product** is a fundamental operation in vector calculus, allowing us to find the cosine of the angle between two vectors. Think of it as a way to measure how much one vector extends in the direction of another.Given two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as below:\[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]For the vectors in our exercise: \( \mathbf{a} = \langle -1, 2, -2 \rangle \) and \( \mathbf{b} = \langle 2, 3, -6 \rangle \), we find their dot product:\[ \mathbf{a} \cdot \mathbf{b} = (-1)(2) + (2)(3) + (-2)(-6) = -2 + 6 + 12 = 16 \]This scalar (16 in our case) helps us in the next steps to find the angle between the vectors.

The **magnitude of a vector** is similar to measuring its length. It's like asking, "How far does this vector stretch?" For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), the magnitude \( \|\mathbf{a}\| \) is found using the Pythagorean theorem in three dimensions.The formula is:\[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]For example, to find the magnitudes of our direction vectors:

• \( \|\mathbf{a}\| = \sqrt{(-1)^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \)
• \( \|\mathbf{b}\| = \sqrt{(2)^2 + (3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \)

Magnitudes are vital for normalizing vectors or finding distances in vector spaces.

After finding the dot product and magnitudes, we use the **inverse cosine** to discover the angle between vectors. The formula connecting these concepts is:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|} \]From this, the angle \( \theta \) itself can be determined by applying the inverse cosine (often written as \( \cos^{-1} \)).In our solved example, first calculate:\[ \cos \theta = \frac{16}{3 \times 7} = \frac{16}{21} \]Then using inverse cosine:\[ \theta = \cos^{-1}\left(\frac{16}{21}\right) \approx 39.81^\circ \]This concludes how we find the specific angle between the two direction vectors, allowing us to understand their spatial relationship comprehensively.