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The physical quantities like distance and speed have magnitude but no direction. They are represented by scalars. The quantities like displacement and velocity have magnitude as well as direction. They are represented by vectors. They follow the rules of vector algebra like vector addition, vector multiplication etc. There are also physical quantities that are neither scalar not vector, they are represented by other mathematical constructs like matrices etc.

The magnitude of a vector $\vec{v}$ is denoted as $|\vec{v}|$. Two vectors, $\vec{v}_1$ and $\vec{v}_2$, are equal if and only if they have same magnitude (i.e., $|\vec{v}_1=\vec{v}_2|$) and both points in the same direction.

A null or zero vector is denoted as $\vec{0}$. Its magnitude is zero. A unit vector is a vector of magnitude 1.

A vector $\vec{v}$ is geometrically represented by a directed line segment $\vec{\mathrm{AB}}$. An arrow from A to B represents the direction of $\vec{v}$ and the length $\mathrm{AB}$ represents the magnitude of $\vec{v}$. For example, a vector of magnitude 1.5 and directed from left to right is represented by

A vector $\vec{v}_1$ can be multiplied by a real number $\lambda$ to get a new vector $\vec{v}_2=\lambda\vec{v}_1$. If $\lambda>0$ then $|\vec{v}_2|=\lambda|\vec{v}_1|$ and direction of $\vec{v}_2$ is same as that of $\vec{v}_1$. If $\lambda<0$ then $|\vec{v}_2|=-\lambda|\vec{v}_1|$ and direction of $\vec{v}_2$ is opposite to that of $\vec{v}_1$. For example, multiplying a vector $\vec{v}_1$ by real number $-1$ gives new vector $\vec{v}_2=-\vec{v}_1$. This vector $\vec{v}_2$ is called negative (or additive inverse) of $\vec{v}_1$. The direction of $\vec{v}_2$ is opposite to that of $\vec{v}_1$.

The addition of two vectors $\vec{v}_1=\vec{\mathrm{AB}}$ and $\vec{v}_2=\vec{\mathrm{BC}}$ gives a new vector $\vec{v}_3=\vec{\mathrm{AC}}$ i.e., \begin{align} \vec{v}_3=\vec{v}_1+\vec{v}_2. \end{align} The vector $\vec{v}_3$ is called resultant of $\vec{v}_1$ and $\vec{v}_2$. The vectors are added geometrically by triangle or parallelogram method. If $\theta$ is the angle between $\vec{v}_1$ and $\vec{v}_2$ then the magnitude of the resultant is given by \begin{align} |\vec{v}_3|=\sqrt{{|\vec{v}_1|}^2 + {|\vec{v}_2|}^2 + 2|\vec{v}_1|\, |\vec{v}_2| \,\cos\theta}.\nonumber \end{align} For $\theta=0$ deg (parallel), $|\vec{v}_3|=|\vec{v}_1|+|\vec{v}_2|$, for $\theta=180$ deg (anti-parallel), $|\vec{v}_3|=\left|\,|\vec{v}_1|-|\vec{v}_2|\,\right|$, and for $\theta=90$ deg (orthogonal), $|\vec{v}_3|=\sqrt{{|\vec{v}_1|}^2+{|\vec{v}_2|}^2}$. The resultant $\vec{v}_3$ makes an angle $\phi$ with $\vec{v}_1$ given by \begin{align} \phi=\tan^{-1}\left(\frac{|\vec{v}_2|\sin\theta}{|\vec{v}_1|+|\vec{v}_2|\,\cos\theta}\right).\nonumber \end{align}

The vector addition is commutative i.e., \begin{align} \vec{v}_1+\vec{v}_2=\vec{v}_2+\vec{v}_1. \end{align} It is also associative i.e., \begin{align} \vec{v}_1+(\vec{v}_2+\vec{v}_3)=(\vec{v}_1+\vec{v}_2)+\vec{v}_3. \end{align} The subtraction of $\vec{v}_2$ from $\vec{v}_1$ is same as addition of $\vec{v}_1$ with negative of $\vec{v}_2$ i.e., \begin{align} \vec{v}_1-\vec{v}_2=\vec{v}_1+(-\vec{v}_2). \end{align}

If $\theta$ is the angle between two vectors $\vec{v}_1$ and $\vec{v}_2$ then their scalar (or dot) product is defined as \begin{align} \vec{v}_1\cdot\vec{v}_2=|\vec{v}_1|\,|\vec{v}_2|\,\cos\theta. \end{align} The scalar product of two orthogonal vectors ($\theta=90$ deg) is zero. The angle between two vectors is given by \begin{align} \theta=\cos^{-1}\left(\frac{ \vec{v}_1\cdot\vec{v}_2}{|\vec{v}_1|\,|\vec{v}_2|}\right).\nonumber \end{align} The orthogonal projection of vector $\vec{v}_2$ along vector $\vec{v}_1$ is given by $\mathrm{AP}=|\vec{v}_2|\cos\theta=(\vec{v}_1\cdot\vec{v}_2)/|\vec{v}_1|$.

The cross product of two vectors $\vec{v}_1$ and $\vec{v}_2$ gives a new vector $\vec{v}_1\times\vec{v}_2$. The magnitude of cross product is $|\vec{v}_1\times\vec{v}_2|=|\vec{v}_1|\,|\vec{v}_2|\,\sin\theta$ and its direction is given by right hand screw rule. The cross product is not commutative ($\vec{v}_1\times\vec{v}_2=-\vec{v}_2\times\vec{v}_1$).

It is convenient to represent a vector in Cartesian coordinates. Consider a coordinate system with three orthogonal axes $x,y$ and $z$. The unit vectors in $x$, $y$ and $z$ directions are denoted as $\hat\imath$, $\hat\jmath$ and $\hat{k}$. A vector $\vec{v}$ can be written as \begin{align} \vec{v}=v_x\hat\imath+v_y\hat\jmath+v_z \hat{k}, \end{align} where $v_x$, $v_y$ and $v_z$ are components (coordinates) along $x$, $y$ and $z$ axes. Note that $v_x\hat\imath$ is a vector along $x$-axis, $v_y\hat\jmath$ is a vector along $y$-axis and $v_z\hat{k}$ is a vector along $z$-axis.

To multiply a vector $\vec{v}$ by a real number $\lambda$, multiply each component of $\vec{v}$ by $\lambda$ i.e., $\lambda\vec{v}=(\lambda v_x)\hat\imath+(\lambda v_y)\hat\jmath+(\lambda v_z)\hat{k}$.

The magnitude of a vector $\vec{v}={v}_{x}\hat\imath+{v}_{y}\hat\jmath+{v}_{z}\hat{k}$ is given by \begin{align} |\vec{v}|=\sqrt{v_x^2+v_y^2+v_z^2}. \end{align}

Two vectors, $\vec{v}_1={v}_{1x}\hat\imath+{v}_{1y}\hat\jmath+{v}_{1z}\hat{k}$ and $\vec{v}_2={v}_{2x}\hat\imath+{v}_{2y}\hat\jmath+{v}_{2z}\hat{k}$ are equal if and only if their components are equal i.e., $v_{1x}=v_{2x}$, $v_{1y}=v_{2y}$ and $v_{1z}=v_{2z}$.

Two vectors are added by adding their components i.e., \begin{align} \vec{v}_3 & =\vec{v}_1+\vec{v}_2 \\ &=(v_{1x}+v_{2x}) \hat\imath \\ &\quad+ (v_{1y}+v_{2y}) \hat\jmath \\ &\quad+(v_{1z}+v_{2z})\hat{k}. \end{align}

The scalar product of a unit vectors with itself is 1 and with another unit vector is zero. The scalar product of two vectors is given by \begin{align} \vec{v}_1\cdot\vec{v}_2=v_{1x}v_{2x}+v_{1y}v_{2y}+v_{1z}v_{2z}. \end{align}

The cross (or vector) product of a unit vectors with itself is zero. The cross product with another unit vectors is given by $\hat\imath\times\hat\jmath=\hat{k}$, $\hat\jmath\times\hat{k}=\hat\imath$ and $\hat{k}\times\hat\imath=\hat\jmath$. The cross product of two vectors is given by \begin{align} \vec{v}_1\times\vec{v}_2 &= (v_{1y}v_{2z}-v_{1z}v_{2y})\hat\imath \\ &\quad+(v_{1z}v_{2x}-v_{1x}v_{2z})\hat\jmath \\ &\quad+(v_{1x}v_{2y}-v_{1y}v_{2x})\hat{k}. \end{align}

The direction of the vector product is given by the right hand thumb rule.