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The dot product, also known as the scalar product, is a way of multiplying two vectors, resulting in a scalar quantity. To find the dot product, we multiply the corresponding components of the vectors and then sum up these products.
The formula for the dot product of two vectors \( \vec{a} = a_1 \hat{\mathrm{i}} + a_2 \hat{\mathrm{j}} + a_3 \hat{\mathrm{k}} \) and \( \vec{b} = b_1 \hat{\mathrm{i}} + b_2 \hat{\mathrm{j}} + b_3 \hat{\mathrm{k}} \) is:

• \( \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \)

This means we take each component of the first vector, multiply it by the corresponding component in the second vector, and then add them all together. This will give us a single number, not a vector.
In the original exercise, we calculated the dot product of \( \vec{d}_1 \) with \( \vec{d}_2 + \vec{d}_3 \), resulting in a scalar value.

The cross product, also known as the vector product, creates a new vector that is perpendicular to the plane containing the original vectors. Unlike the dot product, the cross product results in a vector, not a scalar.
The formula for the cross product of vectors \( \vec{a} \) and \( \vec{b} \) involves computing the determinant of a matrix composed of the unit vectors \( \hat{\mathrm{i}} \), \( \hat{\mathrm{j}} \), and \( \hat{\mathrm{k}} \) and the components of \( \vec{a} \) and \( \vec{b} \):

• \( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \)

To solve it:

• You compute the determinant, which involves ensuring that each component (\( \hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}} \)) of the resulting vector comes from the subtraction of products of the other components.

For the exercise, the cross product of \( \vec{d}_2 \) and \( \vec{d}_3 \) was used in further calculations.

Vector addition combines two or more vectors to yield another vector, known typically for its direction and magnitude. It involves summing the components of the vectors in the same dimension.
For vectors \( \vec{a} = a_1 \hat{\mathrm{i}} + a_2 \hat{\mathrm{j}} + a_3 \hat{\mathrm{k}} \) and \( \vec{b} = b_1 \hat{\mathrm{i}} + b_2 \hat{\mathrm{j}} + b_3 \hat{\mathrm{k}} \):

• \( \vec{a} + \vec{b} = (a_1 + b_1) \hat{\mathrm{i}} + (a_2 + b_2) \hat{\mathrm{j}} + (a_3 + b_3) \hat{\mathrm{k}} \)

This operation is quite straightforward. You add each corresponding component together to form a new vector.
In our initial problem, this was done with \( \vec{d}_2 \) and \( \vec{d}_3 \), bringing together their components to create the resultant vector.

The determinant is a useful property of square matrices, utilized not only in solving systems of linear equations, but also in calculating cross products between vectors.
To find the determinant of a 3x3 matrix:

• \( \begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix} = a \cdot (ei - fh) - b \cdot (di - fg) + c \cdot (dh - eg) \)

This sign-alternating sum is calculated by taking the minor determinants of the matrix, multiplied by the corresponding elements, to achieve the result.
In the problem scenario, determinants were calculated to solve the cross product operations, essential in deriving vector normalities and magnitudes.