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The dot product is a way to multiply two vectors that results in a scalar, rather than another vector. It helps determine how much of one vector goes in the direction of another. The formula to find the dot product of vectors \( \vec{a} = \langle a_x, a_y, a_z \rangle \) and \( \vec{b} = \langle b_x, b_y, b_z \rangle \) is:

• \( \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z \)

This formula calculates a sum of products of their corresponding components. In this specific problem, since \( \vec{a} \) has only \( y \) and \( z \) components and \( \vec{b} \) has only \( x \) and \( z \) components, only the \( z \) components contribute:

• \( \vec{a} \cdot \vec{b} = 2.85 \times 1.04 = 2.964 \)

The result, a scalar value (2.964), implies that the vectors have some degree of alignment in the \( z \)-axis, illustrating the usefulness of the dot product in assessing vector similarity.

The cross product of two vectors results in a third vector that is perpendicular to the plane formed by the original vectors. For vectors \( \vec{a} = \langle a_x, a_y, a_z \rangle \) and \( \vec{b} = \langle b_x, b_y, b_z \rangle \), the cross product \( \vec{a} \times \vec{b} \) is determined by the determinant of a matrix involving the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \):

• \( \vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_x & a_y & a_z \ b_x & b_y & b_z \end{vmatrix} \)

The result is computed as:

• \( \mathbf{i}(a_y b_z - a_z b_y) - \mathbf{j}(a_x b_z - a_z b_x) + \mathbf{k}(a_x b_y - a_y b_x) \)

For this exercise, computing the cross product gives:

• \( \vec{a} \times \vec{b} = \langle 1.508, -2.679, -1.363 \rangle \)

The resulting vector is perpendicular to both \( \vec{a} \) and \( \vec{b} \), and its direction is determined by the right-hand rule. The magnitude of the cross product reflects the area of the parallelogram formed by \( \vec{a} \) and \( \vec{b} \).

The angle between two vectors is a valuable measure to understand how closely aligned they are within a plane. Typically, the cosine of the angle \( \theta \) between vectors \( \vec{a} \) and \( \vec{b} \) can be found using the dot product and the magnitudes (lengths) of the vectors:

• \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \)

To find the angle, follow these steps:

• Calculate the dot product \( \vec{a} \cdot \vec{b} \).
• Determine the magnitudes of each vector using \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \) and \( |\vec{b}| = \sqrt{b_x^2 + b_y^2 + b_z^2} \).
• Use the above formula to solve for \( \cos \theta \).

Applying these steps for this scenario:

• \( \cos \theta = \frac{2.964}{3.20 \times 1.40} \approx 0.664 \)

The angle is finally calculated using the inverse cosine:

• \( \theta = \cos^{-1}(0.664) \approx 48.69^\circ \)

This angle provides insight into the positioning of \( \vec{a} \) and \( \vec{b} \) in space, with a smaller angle indicating a more parallel alignment and a larger angle indicating more perpendicular vectors.