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Vector addition is a foundational concept in vector calculus and physics. It involves adding two vectors together to form a resultant vector, which is simply a combination of the two vectors involved. This is achieved by adding their respective components.

• The components of a vector are usually represented with unit vectors such as \( \hat{\mathrm{i}} \), \( \hat{\mathrm{j}} \), and \( \hat{\mathrm{k}} \), which correspond to the x, y, and z axes, respectively.
• To perform vector addition, add the corresponding unit vector components from each vector.
• For example, if you are given vectors \( \vec{a} = a_1 \hat{\mathrm{i}} + a_2 \hat{\mathrm{j}} + a_3 \hat{\mathrm{k}} \) and \( \vec{b} = b_1 \hat{\mathrm{i}} + b_2 \hat{\mathrm{j}} + b_3 \hat{\mathrm{k}} \), then their sum \( \vec{a} + \vec{b} \) is \( (a_1+b_1) \hat{\mathrm{i}} + (a_2+b_2) \hat{\mathrm{j}} + (a_3+b_3) \hat{\mathrm{k}} \).

In the exercise, adding \( \vec{d}_1 = 3 \hat{\mathrm{i}} - 2 \hat{\mathrm{j}} + 4 \hat{\mathrm{k}} \) and \( \vec{d}_2 = -5 \hat{\mathrm{i}} + 2 \hat{\mathrm{j}} - \hat{\mathrm{k}} \) results in:\[ \vec{d}_1 + \vec{d}_2 = (-2) \hat{\mathrm{i}} + 0 \hat{\mathrm{j}} + 3 \hat{\mathrm{k}} \]This shows the simplicity of vector addition, as you just align and sum up similar components.

The cross product, also known as the vector product, is an operation on two vectors in three-dimensional space. The result is another vector that is perpendicular to the plane containing the two vectors.
To calculate the cross product, you can use a determinant approach that involves the unit vectors \( \hat{\mathrm{i}} \), \( \hat{\mathrm{j}} \), and \( \hat{\mathrm{k}} \).

• Arrange the two vectors in a matrix with the unit vectors. For vectors \( \vec{a} = a_1 \hat{\mathrm{i}} + a_2 \hat{\mathrm{j}} + a_3 \hat{\mathrm{k}} \) and \( \vec{b} = b_1 \hat{\mathrm{i}} + b_2 \hat{\mathrm{j}} + b_3 \hat{\mathrm{k}} \), their cross product is given by the determinant:

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

• This yields a new vector, calculated by taking each unit vector and subtracting products of other rows and columns in what is known as the "Sarrus rule." This results in a vector with each component being a determinant of its corresponding 2x2 sub-matrix.

The cross product of \( \vec{d}_1 \) and \( 4\vec{d}_2 \) is calculated similarly by finding:\[-24 \hat{\mathrm{i}} + 68 \hat{\mathrm{j}} - 16 \hat{\mathrm{k}}\]This product describes a vector that is orthogonal to both \( \vec{d}_1 \) and \( 4\vec{d}_2 \).

The dot product, also known as the scalar product, combines two vectors to yield a scalar quantity. It measures how much one vector extends in the direction of another.

• Given vectors \( \vec{a} = a_1 \hat{\mathrm{i}} + a_2 \hat{\mathrm{j}} + a_3 \hat{\mathrm{k}} \) and \( \vec{b} = b_1 \hat{\mathrm{i}} + b_2 \hat{\mathrm{j}} + b_3 \hat{\mathrm{k}} \), the dot product is calculated as:

\[\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\]

• Observe it is a sum of the products of the corresponding components of the vectors.
• The dot product is useful for finding angles between vectors and determining vector projections.

In the given problem, the dot product of the result from vector addition and the resultant cross product allows you to mix their directions into a single scalar value. Using the components from the vectors:\[-2\hat{\mathrm{i}} + 0\hat{\mathrm{j}} + 3\hat{\mathrm{k}}\] with \[-24\hat{\mathrm{i}} + 68\hat{\mathrm{j}} - 16\hat{\mathrm{k}}\], the dot product is calculated as:\[ (-2)(-24) + (0)(68) + (3)(-16) = 0 \]This result, zero, signifies that the two vectors (originally from addition and cross product) are orthogonal, or that they are perpendicular to each other in 3D space.