Vector Operations

Vector Addition

• The sum of two vectors is obtained by adding their corresponding components.
• The resultant vector is also a vector of the same dimension.

Method:

Example:

---

Scalar Multiplication

• Multiplying a vector by a scalar means multiplying each component of the vector by the scalar.
• The resultant vector is in the same direction as the original vector but scaled by the scalar.

Method:

$$k\vec{a}=k\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)=\left(\begin{array}{c}k{a}_1\\ k{a}_2\\ k{a}_3\end{array}\right)$$

Example:

$$3\left(\begin{array}{c}2\\ 4\\ 6\end{array}\right)=\left(\begin{array}{c}3\cdot 2\\ 3\cdot 4\\ 3\cdot 6\end{array}\right)=\left(\begin{array}{c}6\\ 12\\ 18\end{array}\right)$$

---

Dot Product

• The dot product (also called scalar product) is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number.
• The dot product of two vectors is the sum of the products of their corresponding components.

Method:

$$\vec{a}\cdot \vec{b}=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)\cdot \left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right)=a_1{b}_1+a_2{b}_2+a_3{b}_3$$

Example:

$$\left(\begin{array}{c}1\\ 3\\ 5\end{array}\right)\cdot \left(\begin{array}{c}2\\ 4\\ 6\end{array}\right)=1\cdot 2+3\cdot 4+5\cdot 6=2+12+30=44$$

---

Cross Product

• The cross product of two vectors results in a third vector that is perpendicular to both of the original vectors.
• It is only defined for three-dimensional vectors.
• The magnitude of the cross product vector is equal to the area of the parallelogram that the vectors span.

Method:

$$\vec{a}\times \vec{b}=\left(\begin{array}{c}{a}_2{b}_3-a_3{b}_2\\ a_3{b}_1-a_1{b}_3\\ a_1{b}_2-a_2{b}_1\end{array}\right)$$

Example:

$$\vec{a}=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right),\vec{b}=\left(\begin{array}{c}4\\ 5\\ 6\end{array}\right)$$ $$\vec{a}\times \vec{b}=\left(\begin{array}{c}2\cdot 6-3\cdot 5\\ 3\cdot 4-1\cdot 6\\ 1\cdot 5-2\cdot 4\end{array}\right)=\left(\begin{array}{c}12-15\\ 12-6\\ 5-8\end{array}\right)=\left(\begin{array}{c}-3\\ 6\\ -3\end{array}\right)$$

---

Cross Product (Using Determinant Method)

• The cross product of two vectors $\vec{a}$ and $\vec{b}$ can be found using the determinant of a matrix composed of the unit vectors $(\mathbf{i},\mathbf{j},\mathbf{k})$ on the top row and the components of $\vec{a}$ and $\vec{b}$ on the second and third rows, respectively.
• This method simplifies remembering and computing the cross product by visualizing which components to multiply and subtract.

Method:

$$\vec{a}\times \vec{b}=\text{det}\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|=\mathbf{i}(a_2{b}_3-a_3{b}_2)-\mathbf{j}(a_1{b}_3-a_3{b}_1)+\mathbf{k}(a_1{b}_2-a_2{b}_1)$$

Example:

Given vectors $\vec{a}=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$ and $\vec{b}=\left(\begin{array}{c}4\\ 5\\ 6\end{array}\right)$,

$$\vec{a}\times \vec{b}=\text{det}\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 2& 3\\ 4& 5& 6\end{array}\right|=\mathbf{i}(2\cdot 6-3\cdot 5)-\mathbf{j}(1\cdot 6-3\cdot 4)+\mathbf{k}(1\cdot 5-2\cdot 4)$$ $$=\mathbf{i}(12-15)-\mathbf{j}(6-12)+\mathbf{k}(5-8)$$ $$=-3\mathbf{i}+6\mathbf{j}-3\mathbf{k}$$ $$=\left(\begin{array}{c}-3\\ 6\\ -3\end{array}\right)$$