# Current Affairs JEE Main & Advanced

#### Formation of Differential Equation

Formulating a differential equation from a given equation representing a family of curves means finding a differential equation whose solution is the given equation. The equation so obtained is the differential equation of order  for the family of given curves.     Algorithm for formation of differential equations     Step (i) : Write the given equation involving independent variable  (say), dependent variable  (say) and the arbitrary constants.     Step (ii) : Obtain the number of arbitrary constants in step (i). Let there be  arbitrary constants.     Step (iii) : Differentiate the relation in step (i)  times with respect to .     Step (iv) : Eliminate arbitrary constants with the help of n equations involving differential coefficients obtained in step (iii) and an equation in step (i). The equation so obtained is the desired differential equation.

#### Definition

An equation involving independent variable $x,$ dependent variable $y$ and the differential coefficients  $\frac{dy}{dx},\,\frac{{{d}^{2}}y}{d{{x}^{2}}},........$ is called differential equation.     Examples :     (i) $\frac{dy}{dx}=1+x+y$         (ii) $\frac{dy}{dx}+xy=\cot x$     (iii)${{\left( \frac{{{d}^{4}}y}{d{{x}^{4}}} \right)}^{3}}-4\frac{dy}{dx}+4y=5\cos 3x$     (iv) ${{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}=0$     (1) Order of a differential equation : The order of a differential equation is the order of the highest derivative occurring in the differential equation. For example, the order of above differential equations are 1, 1, 4 and 2 respectively.     The order of a differential equation is a positive integer. To determine the order of a differential equation, it is not needed to make the equation free from radicals.     (2) Degree of a differential equation : The degree of a differential equation is the degree of the highest order derivative, when differential coefficients are made free from radicals and fractions. The degree of above differential equations are 1, 1, 3 and 2 respectively.

#### Application of Vectors in 3-dimensional Geometry

(1) Direction cosines of $\mathbf{r}=a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$ are $\frac{a}{|\mathbf{r}|},\,\frac{b}{|\mathbf{r}|},\,\frac{c}{|\mathbf{r}|}$.   (2) Incentre formula : The position vector of the incentre of $\Delta ABC$ is $\frac{a\,\mathbf{a}+b\,\mathbf{b}+c\,\mathbf{c}}{a+b+c}$.   (3) Orthocentre formula : The position vector of the orthocentre of $\left| \,\begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} & 1 \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} & 1 \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} & 1 \\ {{d}_{1}} & {{d}_{2}} & {{d}_{3}} & 1 \\ \end{matrix}\, \right|\,=0$ is $\frac{\mathbf{a}\tan A+\mathbf{b}\tan B+\mathbf{c}\tan C}{\tan A+\tan B+\tan C}$.     (4) Vector equation of a straight line passing through a fixed point with position vector $\mathbf{a}$ and parallel to a given vector $\mathbf{b}$ is $\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$.     (5) The vector equation of a line passing through two points with position vectors $\mathbf{a}$ and $\mathbf{b}$ is $\mathbf{r}=\mathbf{a}+\lambda (\mathbf{b}-\mathbf{a})$.     (6) If the lines $\mathbf{r}={{\mathbf{a}}_{1}}+\lambda {{\mathbf{b}}_{1}}$ and $\mathbf{r}={{\mathbf{a}}_{2}}+\lambda {{\mathbf{b}}_{2}}$ are coplanar, then $[{{\mathbf{a}}_{1}}\,{{\mathbf{b}}_{1}}\,{{\mathbf{b}}_{2}}]=[{{\mathbf{a}}_{2}}\,{{\mathbf{b}}_{1}}\,{{\mathbf{b}}_{2}}]$ and the equation of the plane containing them is $[\mathbf{r}\,{{\mathbf{b}}_{1}}\,{{\mathbf{b}}_{2}}]=[{{\mathbf{a}}_{1}}\,{{\mathbf{b}}_{1}}\,{{\mathbf{b}}_{2}}]$ or $[\mathbf{r}\,{{\mathbf{b}}_{1}}\,{{\mathbf{b}}_{2}}]=[{{\mathbf{a}}_{2}}\,{{\mathbf{b}}_{1}}\,{{\mathbf{b}}_{2}}]$.       (7) Perpendicular distance of a point from a line : Let L is the foot of perpendicular drawn from $P(\vec{\alpha })$ on the line $\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$. Since $\mathbf{r}$ denotes the position vector of any point on the line $\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$. So, let the position vector of $\vec{L}$ be $\mathbf{a}+\lambda \mathbf{b}$.     Then $\overrightarrow{PL}=\mathbf{a}-\overrightarrow{\alpha }+\lambda \mathbf{b}=(\mathbf{a}-\overrightarrow{\alpha })-\left( \frac{(\mathbf{a}-\overrightarrow{\alpha })\mathbf{b}}{|\mathbf{b}{{|}^{2}}} \right)\mathbf{b}$     The length PL, is the magnitude of$\overrightarrow{PL}$, and required length of perpendicular.               (8) Image of a point in a straight line : Let $Q(\overrightarrow{\beta })$ is the image of P in $\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$, then, $\overrightarrow{\beta }=2\mathbf{a}-\left( \frac{2(\mathbf{a}-\overrightarrow{\alpha }).\mathbf{b}}{|\mathbf{b}{{|}^{2}}} \right)\mathbf{b}-\overrightarrow{\alpha }$     (9) Shortest distance between two parallel lines : Let ${{l}_{1}}$ and ${{l}_{2}}$ be two lines whose equations are ${{l}_{1}}:\mathbf{r}={{\mathbf{a}}_{1}}+\lambda {{\mathbf{b}}_{1}}$ and ${{l}_{2}}:\mathbf{r}={{\mathbf{a}}_{2}}+\mu {{\mathbf{b}}_{2}}$ respectively.     Then, shortest distance     $PQ=\left| \frac{({{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{\mathbf{2}}})\,.\,({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}})}{|{{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}}|} \right|=\left| \frac{[{{\mathbf{b}}_{1}}\text{ }{{\mathbf{b}}_{\mathbf{2}}}\,({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}})]}{|{{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}}|} \right|$     Shortest distance between two parallel lines : The shortest distance between the parallel lines $\mathbf{r}={{\mathbf{a}}_{1}}+\lambda \mathbf{b}$ and $\mathbf{r}={{\mathbf{a}}_{2}}+\mu \mathbf{b}$ is given by $d=\frac{|({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}})\times \mathbf{b}|}{|\mathbf{b}|}$.     If the lines $\mathbf{r}={{\mathbf{a}}_{1}}+\lambda {{\mathbf{b}}_{1}}$ and $\mathbf{r}={{\mathbf{a}}_{2}}+\mu {{\mathbf{b}}_{2}}$ intersect, then the shortest distance between them is zero.     Therefore, $[{{\mathbf{b}}_{1}}\,{{\mathbf{b}}_{2}}\,({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}})]=0$     $\Rightarrow$$[({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}})\,\,{{\mathbf{b}}_{1}}{{\mathbf{b}}_{2}}]=0$$\Rightarrow$$({{\mathbf{a}}_{2}}-{{\mathbf{a}}_{1}}).({{\mathbf{b}}_{1}}\times {{\mathbf{b}}_{2}})=0$.     (10) If the lines $\mathbf{r}={{\mathbf{a}}_{1}}+\lambda \,{{\mathbf{b}}_{1}}$ and $\mathbf{r}={{\mathbf{a}}_{2}}+\lambda \,{{\mathbf{b}}_{2}}$ are coplanar, then $[{{\mathbf{a}}_{1}}{{\mathbf{b}}_{1}}{{\mathbf{b}}_{2}}]=[{{\mathbf{a}}_{2}}{{\mathbf{b}}_{1}}{{\mathbf{b}}_{2}}]$ and the equation of the plane containing them is $[\mathbf{r}\,{{\mathbf{b}}_{1}}\,{{\mathbf{b}}_{2}}]=[{{\mathbf{a}}_{1}}\,{{\mathbf{b}}_{1}}\,{{\mathbf{b}}_{2}}]$ or $[\mathbf{r}\,{{\mathbf{b}}_{1}}\,{{\mathbf{b}}_{2}}]=[{{\mathbf{a}}_{2}}\,{{\mathbf{b}}_{\mathbf{1}}}\,{{\mathbf{b}}_{\mathbf{2}}}]$.     (11) Vector equation of a plane through the point $A(\mathbf{a})$ and perpendicular to the vector $\mathbf{n}$ is $(\mathbf{r}-\mathbf{a}).\mathbf{n}=0$ or $\mathbf{r}.\mathbf{n}=\mathbf{a}.\mathbf{n}$ or $\mathbf{r}\,.\,\mathbf{n}=d$, where $d=\mathbf{a}\,.\,\mathbf{n}$.  This is known as the scalar product form of a plane.     (12) Vector equation of a plane normal to unit vector $\mathbf{\hat{n}}$ and at a distance d from the origin is $\mathbf{r}\mathbf{.\hat{n}}=d$.     If $\mathbf{n}$ is not a unit vector, then to reduce the equation $\mathbf{r}.\mathbf{n}=d$ to normal form we divide both sides by $|\mathbf{n}|$ to obtain $\mathbf{r}\cdot \frac{\mathbf{n}}{|\mathbf{n}|}=\frac{d}{|\mathbf{n}|}$ more...

#### Rotation of a Vector About an Axis

Let $\mathbf{a}=({{a}_{1}},\,{{a}_{2}},\,{{a}_{3}})$. If system is rotated about   (i) x-axis through an angle $\alpha$, then the new components of $\mathbf{a}$ are $({{a}_{1}},\,{{a}_{2}}\cos \alpha +{{a}_{3}}\sin \alpha ,\,-{{a}_{2}}\sin \alpha +{{a}_{3}}\cos \alpha )$.   (ii) y-axis through an angle $\alpha$, then the new components of  $\mathbf{a}$ are $(-{{a}_{3}}\sin \alpha +{{a}_{1}}\cos \alpha ,\,{{a}_{2}},\,{{a}_{3}}\cos \alpha +{{a}_{1}}\sin \alpha )$.   (iii) z-axis through an angle $\alpha$, then the new components of $\mathbf{a}$ are $({{a}_{1}}\cos \alpha +{{a}_{2}}\sin \alpha ,\,-{{a}_{1}}\sin \alpha +{{a}_{2}}\cos \alpha ,{{a}_{3}})$.

#### Vector Product of Four Vectors

(1) $(\mathbf{a}\times \mathbf{b})\,\times (\mathbf{c}\times \mathbf{d})$ is a vector product of four vectors.     It is the cross product of the vectors $\mathbf{a}\times \mathbf{b}$ and $\mathbf{c}\times \mathbf{d}$.     (2) $\mathbf{a}\times \{\mathbf{b}\times (\mathbf{c}\times \mathbf{d})\},\,\{(\mathbf{a}\times \mathbf{b})\times \mathbf{c}\}\times \mathbf{d}$ are also different vector products of four vectors $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ and $\mathbf{d}$.

#### Scalar Product of Four Vectors

$\frac{1}{2}|\overrightarrow{CB}\times \overrightarrow{CA}|$ is a scalar product of four vectors. It is the dot product of the vectors $\mathbf{a}$ and $\mathbf{c}\times \mathbf{d}$.     It is a scalar triple product of the vectors $\mathbf{a},\,\mathbf{b}$ and $\mathbf{c}\,\times \mathbf{d}$ as well as scalar triple product of the vectors $\mathbf{a}\times \mathbf{b},\,\mathbf{c}$ and d.     $(\mathbf{a}\times \mathbf{b})\,.\,(\mathbf{c}\times \mathbf{d})\,=\left| \,\begin{matrix} \mathbf{a}.\mathbf{c} & \mathbf{a}\,.\,\mathbf{d} \\ \mathbf{b}.\mathbf{c} & \mathbf{b}\,.\,\mathbf{d} \\ \end{matrix}\, \right|$

#### Vector Triple Product

Let $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ be any three vectors, then the vectors $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})$ and $(\mathbf{a}\times \mathbf{b})\,\,\times \mathbf{c}$ are called vector triple product of $\mathbf{a,b,}\,\mathbf{c}$ .     Thus, $\mathbf{a}\,\times \,(\mathbf{b}\times \mathbf{c})=(\mathbf{a}\,.\,\mathbf{c})\,\mathbf{b}-(\mathbf{a}\,.\,\mathbf{b})\mathbf{c}$     Properties of vector triple product     (i) The vector triple product $\mathbf{a}\times (\mathbf{b}\,\times \,\mathbf{c})$ is a linear combination of those two vectors which are within brackets.     (ii) The vector $\mathbf{r}=\mathbf{a}\times (\mathbf{b}\times \mathbf{c})$ is perpendicular to $\mathbf{a}$ and lies in the plane of $\mathbf{b}$ and $\mathbf{c}$.     (iii) The formula $\mathbf{r}={{\mathbf{a}}_{1}}+\lambda \mathbf{b}$ is true only when the vector outside the bracket is on the left most side. If it is not, we first shift on left by using the properties of cross product and then apply the same formula.     Thus, $\mathbf{(b\times c)\times a=}-\mathbf{\{a\times (b\times c)\}=}-\mathbf{\{(a}\mathbf{.c)b}-\mathbf{(a}\mathbf{.b)c\}=(a}\mathbf{.b)c}-\mathbf{(a}\mathbf{.c)b}$     (iv) Vector triple product is a vector quantity.     (v) $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})\ne (\mathbf{a}\times \mathbf{b})\times \mathbf{c}$

#### Scalar Triple Product

(1) Scalar triple product of three vectors : If $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ are three vectors, then their scalar triple product is defined as the dot product of two vectors $\mathbf{a}$ and $\mathbf{b\times c}$. It is generally denoted by $\mathbf{a}$. $\mathbf{(b\times c)}$ or $\mathbf{[a}\,\,\mathbf{b}\,\,\,\mathbf{c]}$.     (2) Properties of scalar triple product     (i) If $\mathbf{a},\,\mathbf{b},\mathbf{c}$ are cyclically permuted, the value of scalar triple product remains the same. i.e., $(\mathbf{a}\times \mathbf{b})\,.\,\mathbf{c}=(\mathbf{b}\times \mathbf{c})\,.\,\mathbf{a}=(\mathbf{c}\times \mathbf{a})\,.\,\mathbf{b}$ or $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=[\mathbf{b}\,\mathbf{c}\,\mathbf{a}]=[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]$     (ii) The change of cyclic order of vectors in scalar triple product changes the sign of the scalar triple product but not the magnitude i.e., $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=-[\mathbf{b}\,\mathbf{a}\,\mathbf{c}]=-[\mathbf{c}\,\mathbf{b}\,\mathbf{a}]=-[\mathbf{a}\,\mathbf{c}\,\mathbf{b}]$     (iii) In scalar triple product the positions of dot and cross can be interchanged provided that the cyclic order of the vectors remains same i.e., $(\mathbf{a}\times \mathbf{b})\,.\,\mathbf{c}=\mathbf{a}\,.\,(\mathbf{b}\times \mathbf{c})$     (iv) The scalar triple product of three vectors is zero if any two of them are equal.     (v) For any three vectors $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ and scalar $\lambda$, $[\lambda \,\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\,$ $=\,\lambda [\mathbf{a}\,\mathbf{b}\,\mathbf{c}]$     (vi) The scalar triple product of three vectors is zero if any two of them are parallel or collinear.     (vii) If $\mathbf{a},\,\mathbf{b},\,\mathbf{c},\,\mathbf{d}$ are four vectors, then $[(\mathbf{a}+\mathbf{b})\,\,\mathbf{c}\,\,\mathbf{d}]=$ $[\mathbf{a}\,\mathbf{c}\,\mathbf{d}]+\mathbf{[b}\text{ }\mathbf{c}\text{ }\mathbf{d}]$     (viii) The necessary and sufficient condition for three non-zero non-collinear vectors $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ to be coplanar is that $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=0$.     (ix) Four points with position vectors $\mathbf{a,}\,\,\mathbf{b,}\,\,\mathbf{c}$ and $\mathbf{d}$ will be coplanar, if $[\mathbf{a}\,\,\mathbf{b}\ \,\mathbf{c}]+[\mathbf{d}\,\,\mathbf{c}\,\,\mathbf{a}]+[\mathbf{d}\ \,\mathbf{a}\,\,\mathbf{b}]$$=[\mathbf{a}\,\ \mathbf{b}\,\ \mathbf{c}]$.     (x) Volume of parallelopiped whose coterminous edges are $\mathbf{a,}\,\,\mathbf{b,}\,\,\mathbf{c}$ is $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]$or  $\mathbf{a}\,\mathbf{(b\times c)}$.     (3) Scalar triple product in terms of components     (i) If $\mathbf{a}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k},\,\,\mathbf{b}={{b}_{1}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{b}_{3}}\mathbf{k}$ and $\mathbf{c}={{c}_{1}}\mathbf{i}+{{c}_{2}}\mathbf{j}+{{c}_{3}}\mathbf{k}$ be three vectors then, $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=\left| \,\begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix}\, \right|$     (ii) If $\mathbf{a}={{a}_{1}}\mathbf{l}+{{a}_{2}}\mathbf{m}+{{a}_{3}}\mathbf{n},\,\mathbf{b}={{b}_{1}}\mathbf{l}+{{b}_{2}}\mathbf{m}+{{b}_{3}}\mathbf{n}$  and $\mathbf{c}={{c}_{1}}\mathbf{l}+{{c}_{2}}\mathbf{m}+{{c}_{3}}\mathbf{n}$, then $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix}\, \right|\,[\mathbf{l}\,\mathbf{m}\,\mathbf{n}]$     (iii) For any three vectors $\mathbf{a},\,\mathbf{b}$ and $\mathbf{c}$     (a) $[\mathbf{a}+\mathbf{b}\,\,\,\mathbf{b}+\mathbf{c}\,\,\,\mathbf{c}+\mathbf{a}]=2[\mathbf{a}\,\,\mathbf{b}\,\,\mathbf{c}]$       (b) $[\mathbf{a}-\mathbf{b}\,\,\,\mathbf{b}-\mathbf{c}\,\,\,\mathbf{c}-\mathbf{a}]=0$       (c) $[\mathbf{a}\times \mathbf{b}\,\,\,\mathbf{b}\times \mathbf{c}\,\,\,\mathbf{c}\times \mathbf{a}]={{[\mathbf{a}\,\,\mathbf{b}\,\,\mathbf{c}]}^{2}}$     (4) Tetrahedron : A tetrahedron is a three-dimensional figure formed by four triangle $OABC$ is a tetrahedron with $\Delta ABC$ as the base. $OA,\,OB,OC,AB,\,BC$ and $CA$ are known as edges of the tetrahedron. $OA,\,BC;\,OB,\,CA$ and $OC,\,AB$ are known as the pairs of opposite edges. A tetrahedron in which all edges are equal, is called a regular tetrahedron. Any two edges of regular tetrahedron are perpendicular to each other.         Volume of tetrahedron     (i) The volume of a tetrahedron  $=\frac{1}{3}(\text{area of the base) (corresponding altitude)}$$=\frac{1}{6}[\overrightarrow{AB}\,\text{ }\overrightarrow{BC}\,\text{ }\overrightarrow{AD}]$     (ii) If $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ are position vectors of vertices $A,\,\,B$ and $C$ with respect to $O,$ then volume of tetrahedron $OABC=\frac{1}{6}\,[\mathbf{a}\,\,\mathbf{b}\ \,\mathbf{c}]$.     (iii) If $\mathbf{a},\,\mathbf{b},\,\mathbf{c},\,\mathbf{d}$ are position vectors of vertices $A,B,C,D$ of a tetrahedron $ABCD,$ then its volume $=\frac{1}{6}[\mathbf{b}-\mathbf{a}\,\,\mathbf{c}-\mathbf{a}\,\,\mathbf{d}-\mathbf{a}]$.     (5) Reciprocal more...

#### Vector or Cross Product

(1) Vector product of two vectors : Let $\mathbf{a},\,\mathbf{b}$ be two non-zero, non-parallel vectors.     Then $\mathbf{a}\times \mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|$$\sin \theta \,\hat{\eta }$, and $|\mathbf{a}\times \mathbf{b}|\text{ }=\text{ }|\mathbf{a}||\mathbf{b}|\sin \theta$, where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$, $\hat{\eta }$ is a unit vector perpendicular to the plane of $\mathbf{a}$ and $\mathbf{b}$ such that $\mathbf{a},\,\mathbf{b},\,\hat{\eta }$ form a right-handed system.     (2) Properties of vector product     (i) Vector product is not commutative i.e., if $\mathbf{a}$ and $\mathbf{b}$ are any two vectors, then $\mathbf{a}\times \mathbf{b}\ne \mathbf{b}\times \mathbf{a}$, however, $\mathbf{a}\times \mathbf{b}=-(\mathbf{b}\times \mathbf{a})$     (ii) If $\mathbf{a},\,\mathbf{b}$ are two vectors and m, n are scalars, then $m\mathbf{a}\times n\mathbf{b}=mn(\mathbf{a}\times \mathbf{b})=m\,(\mathbf{a}\times n\mathbf{b})=n(m\,\mathbf{a}\times \mathbf{b})$.     (iii) Distributivity of vector product over vector addition.     Let $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ be any three vectors. Then     (a) $\mathbf{a}\times (\mathbf{b}+\mathbf{c})=\mathbf{a}\times \mathbf{b}+\mathbf{a}\times \mathbf{c}$        (Left distributivity)     (b) $(\mathbf{b}+\mathbf{c})\times \mathbf{a}=\mathbf{b}\times \mathbf{a}+\mathbf{c}\times \mathbf{a}$        (Right distributivity)     (iv) For any three vectors $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ we have $\mathbf{a}\times (\mathbf{b}-\mathbf{c})=$ $\mathbf{a}\times \mathbf{b}-\mathbf{a}\times \mathbf{c}$.     (v) The vector product of two non-zero vectors is zero vector iff they are parallel (Collinear) i.e., $\mathbf{a}\times \mathbf{b}=0\Leftrightarrow \mathbf{a}||\mathbf{b},\,\mathbf{a},\,\mathbf{b}$ are non-zero vectors.     It follows from the above property that $\mathbf{a}\times \mathbf{a}=0$ for every non-zero vector $\mathbf{a}$, which in turn implies that $\mathbf{i}\times \mathbf{i}=$ $\mathbf{j}\times \mathbf{j}=\mathbf{k}\times \mathbf{k}=0$.     (vi) Vector product of orthonormal triad of unit vectors $\mathbf{i},\text{ }\mathbf{j},\text{ }\mathbf{k}$ using the definition of the vector product, we obtain $\mathbf{i}\times \mathbf{j}=\mathbf{k},\,\mathbf{j}\times \mathbf{k}=\mathbf{i},\,\mathbf{k}\times \mathbf{i}=\mathbf{j}$, $\mathbf{j}\times \mathbf{i}=-\mathbf{k},\,\mathbf{k}\times \mathbf{j}=-\mathbf{i},\,\mathbf{i}\times \mathbf{k}=-\mathbf{j}$.     (3) Vector product in terms of components: If $\mathbf{a}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k}$ and $\mathbf{b}={{b}_{1}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{b}_{3}}\mathbf{k}$.     Then, $\mathbf{a}\times \mathbf{b}=\left| \,\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ \end{matrix}\, \right|$     (4) Angle between two vectors : If $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$, then $\sin \theta =\frac{|\mathbf{a}\times \mathbf{b}|}{|\mathbf{a}|\,|\mathbf{b}|}$.     (5) (i) Right handed system of vectors : Three mutually perpendicular vectors $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ form a right handed system of vector iff $\mathbf{a}\times \mathbf{b}=\mathbf{c},\,\,\mathbf{b}\times \mathbf{c}=\mathbf{a},\,\,\mathbf{c}\times \mathbf{a}=\mathbf{b}$     Examples: The unit vectors $\mathbf{i}\,\mathbf{,}\,\mathbf{j,}\,\,\mathbf{k}$ form a right-handed system, $\mathbf{i}\times \mathbf{j}=\mathbf{k},\,\mathbf{j}\times \mathbf{k}=\mathbf{i},\,\mathbf{k}\times \mathbf{i}=\mathbf{j}$       (ii) Left handed system of vectors : The vectors $\mathbf{a},\,\mathbf{b},\,\mathbf{c}$ mutually perpendicular to one another form a left handed system of vector iff $\mathbf{c}\times \mathbf{b}=\mathbf{a},\,\mathbf{a}\times \mathbf{c}=\mathbf{b},\mathbf{b}\times \mathbf{a}=\mathbf{c}$     (6) Vector normal to the plane of two given vectors : If $\mathbf{a},\,\mathbf{b}$ be two non-zero, nonparallel  vectors and let $\theta$ be the angle between them. $\mathbf{a}\times \mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|\sin \theta \,\hat{\eta }$ where $\hat{\eta }$ is a unit vector perpendicular to the plane of $\mathbf{a}$ and $\mathbf{b}$ such that $\mathbf{a},\,\mathbf{b},\,\eta$ form a right-handed system.     $\Rightarrow$   $(\mathbf{a}\times \mathbf{b})=|\mathbf{a}\times \mathbf{b}|\,\hat{\eta }$$\Rightarrow$ $\hat{\eta }=\frac{\mathbf{a}\times \mathbf{b}}{|\mathbf{a}\times \mathbf{b}|}$     Thus, $\frac{\mathbf{a}\times \mathbf{b}}{|\mathbf{a}\times \mathbf{b}|}$ is a unit vector perpendicular to the plane of $-\frac{\sqrt{3}}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}$ and $\mathbf{b}$. Note that $-\frac{\mathbf{a}\times \mathbf{b}}{|\mathbf{a}\times \mathbf{b}|}$ is also more...

#### Scalar or Dot Product

(1) Scalar or Dot product of two vectors : If $\mathbf{a}$ and $\mathbf{b}$ are two non-zero vectors and $\theta$ be the angle between them, then their scalar product (or dot product) is denoted by $\mathbf{a}\,.\,\mathbf{b}$ and is defined as the scalar $|\mathbf{a}|\,|\mathbf{b}|\cos \theta$, where $|\mathbf{a}|\,\text{ and }|\mathbf{b}|$are modulii of $\mathbf{a}$ and $\mathbf{b}$ respectively and $0\le \theta \le \pi$. Dot product of two vectors is a scalar quantity.     Angle between two vectors : If $\mathbf{a},\,\mathbf{b}$ be two vectors inclined at an angle $\theta$, then $\mathbf{a}\,.\,\mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|\,\cos \theta$     $\Rightarrow$ $\cos \theta =\frac{\mathbf{a}\,.\,\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}$$\Rightarrow$ $\theta ={{\cos }^{-1}}\left( \frac{\mathbf{a}\,.\,\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|} \right)$     If $\mathbf{a}\,={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k}$ and $\mathbf{b}={{b}_{1}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{b}_{3}}\mathbf{k}$; then     $\theta ={{\cos }^{-1}}\left( \frac{{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}\sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}} \right)$.     (2) Properties of scalar product     (i) Commutativity : The scalar product of two vector is commutative i.e., $\mathbf{a}\,.\,\mathbf{b}\,=\mathbf{b}\,.\,\mathbf{a}$.     (ii) Distributivity of scalar product over vector addition  The scalar product of vectors is distributive over vector addition i.e., (a) $\mathbf{a}\,.\,(\mathbf{b}\,+\mathbf{c})\,=\,\mathbf{a}\,.\,\mathbf{b}+\mathbf{a}\,.\,\mathbf{c}$,     (Left distributivity)      (b) $(\mathbf{b}+\mathbf{c})\,.\,\mathbf{a}=\mathbf{b}\,.\,\mathbf{a}+\mathbf{c}\,.\,\mathbf{a}$,   (Right distributivity)     (iii) Let $\mathbf{a}$ and $\mathbf{b}$ be two non-zero vectors $\mathbf{a}\,.\,\mathbf{b}=0\Leftrightarrow \mathbf{a}\bot \mathbf{b}$. As $\mathbf{i},\,\mathbf{j},\,\mathbf{k}$ are mutually perpendicular unit vectors along the co-ordinate axes, therefore, $\mathbf{i}\,.\,\mathbf{j}=\mathbf{j}\,.\,\mathbf{i}=0$; $\mathbf{j}\,.\,\mathbf{k}=\mathbf{k}\,.\,\mathbf{j}=0;$ $\mathbf{k}\,.\,\mathbf{i}=\mathbf{i}\,.\,\mathbf{k}\,=0$.     (iv) For any vector $\mathbf{a},\,\text{ }\mathbf{a}\,.\,\mathbf{a}=|\mathbf{a}{{|}^{2}}$.     As $\mathbf{i},\,\mathbf{j},\,\mathbf{k}$ are unit vectors along the co-ordinate axes, therefore $\mathbf{i}\,.\,\mathbf{i}=|\mathbf{i}{{|}^{2}}=1$, $\mathbf{j}\,.\,\mathbf{j}=|\mathbf{j}{{|}^{2}}=1$ and $\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}+\mu \mathbf{c}$     (v) If m, n are scalars and $\mathbf{a},\,\mathbf{b}$ be two vectors, then $m\mathbf{a}\,.\,n\mathbf{b}=mn(\mathbf{a}\,.\,\mathbf{b})=(mn\,\mathbf{a})\,.\,\mathbf{b}=\mathbf{a}\,.\,(mn\,\mathbf{b})$     (vi) For any vectors $\mathbf{a}$ and $\mathbf{b}$, we have     (a) $\mathbf{a}\,.\ (-\mathbf{b})=-(\mathbf{a}\,.\,\mathbf{b})=(-\mathbf{a})\,.\,\mathbf{b}$     (b) $(-\mathbf{a})\,.\,(-\mathbf{b})=\mathbf{a}\,.\,\mathbf{b}$     (vii) For any two vectors $\mathbf{a}$ and $\mathbf{b}$, we have      (a) $|\mathbf{a}+\mathbf{b}{{|}^{2}}=\ |\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}+\ 2\mathbf{a}\,.\,\mathbf{b}$                     (b) $|\mathbf{a}-\mathbf{b}{{|}^{2}}=\ |\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}-\ 2\mathbf{a}\,.\,\mathbf{b}$     (c) $(\mathbf{a}+\mathbf{b})\,.\,(\mathbf{a}-\mathbf{b})=\ |\mathbf{a}{{|}^{2}}-|\mathbf{b}{{|}^{2}}$                                      (d) $|\mathbf{a}+\mathbf{b}|\ =\ |\mathbf{a}|+|\mathbf{b}|$$\Rightarrow$ $\mathbf{a}||\mathbf{b}$     (e) $|\mathbf{a}+\mathbf{b}{{|}^{2}}=\ |\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}\Rightarrow \mathbf{a}\bot \mathbf{b}$     (f) $|\mathbf{a}+\mathbf{b}|\ =\ |\mathbf{a}-\mathbf{b}|\ \Rightarrow \mathbf{a}\bot \mathbf{b}$     (3) Scalar product in terms of components: If $\mathbf{a}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k}$ and $\mathbf{b}={{b}_{1}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{b}_{3}}\mathbf{k}$, then, $\mathbf{a}\,.\,\mathbf{b}={{a}_{1}}{{b}_{1}}$ $+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}$.     The components of $\mathbf{b}$ along and perpendicular to $\mathbf{a}$ are $\left( \frac{\mathbf{a}\,.\ \mathbf{b}}{|\mathbf{a}{{|}^{2}}} \right)\,\mathbf{a}$ and $\mathbf{b}-\left( \frac{\mathbf{a}\,.\,\mathbf{b}}{|\mathbf{a}{{|}^{2}}} \right)\,\mathbf{a}$ respectively.     (4) Work done by a force :      Work done $=|\mathbf{F}|\,|\overrightarrow{OA}|\,\cos \theta =\mathbf{F}\,.\,\overrightarrow{OA}\,=\mathbf{F}\,.\mathbf{d}$, where $\mathbf{d}=\overrightarrow{OA}$     Work done = (Force). (Displacement)     If a number of forces are acting on a particle, then the sum of the works done by the separate forces is equal to the work done by the resultant force.

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