# Vector Solutions, Shortcuts, Formulas

Some Basic Concept of Vector: Let l be any straight line in plane or three dimensional space. This line can be given two directions by means of arrowheads. A line with one of these directions prescribed is called a directed line {Fig 10.1 (i), (ii)}. Now observe that if we restrict the line l to the line segment AB, then a magnitude is prescribed on the line l with one of the two directions, so that we obtain a directed line segment {Fig 10.1(iii)}. Thus, a directed line segment has magnitude as well as direction. Definition: A quantity that has magnitude as well as direction is called a vector .

Notice: That a directed line segment is a vector fig {10.1(iii)}, denoted by as $\vec{AB}$ or simply as $\vec{a}$ , and read a ‘ vector $\vec{AB}$’ or ‘vector $\vec{a}$’.

The point A from where the vector $\vec{AB}$’ starts is called its initial point , and the point B where it ends is called its terminal point. The distance between initial and terminal points of a vector is called the magnitude (or length) of the vector, denoted as $|\vec{AB}|\: or\: |\vec{a}|$ , or a .The arrow indicates the direction of the vector.

## Position Vector

A vector $\vec{AB}$ simply means the displacement from a point A to the point B. Now consider a situation that a girl moves from A to B and then from B to C (Fig 10.7). The net displacement made by the girl from point A to the point C, is given by the vector $\vec{AC}$ and express $\vec{AC}=\vec{AB}+\vec{AC}$
This is known as the triangle law of vector addition.

If we have two vectors $\vec{a}\: and\: \vec{b}$ by the two adjacent sides of a parallelogram in magnitude and direction (Fig 10.9), then their sum is represented in magnitude and direction by the diagonal of the parallelogram through their common point. This is known as the parallelogram law of vector addition.
Using triangle law , $\vec{OA}+\vec{AC}=\vec{OC}$
$\vec{OA}+\vec{OB}=\vec{OC}$ (since $\vec{AC}=\vec{OB}$ Which is parallelogram law. Thus, we may say that the two laws of vector addition are equivalent to each other

1) For any two vector $\vec{a}\: and\: \vec{b}$
$\vec{a}+\vec{b}=\vec{b}+\vec{a}$ (commutative properties)

2) For any three properties $\vec{a},\vec{b}\:and\: \vec{c}$
$(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$ (Associative properties )

Note: The associative property of vector addition enables us to write the sum of three vectors $\vec{a},\vec{b,\vec{c}}\: as\:\vec{a}+\vec{b}+\vec{c}$ as without using brackets. Note that for any vector $\vec{a}$, we have $\vec{a}+\vec{0}=\vec{0}+\vec{a}=\vec{a}$
Here , the zero vector is called additive identity for the vector addition .

## Multiplication of a Vector by a Scalar

Let be a given vector and λ a scalar. Then the product of the vector a by the scalar λ , denoted as $\lambda \vec{a}$, is called the multiplication of vector $\vec{a}$, by the scalar λ. Note that , $\lambda \vec{a}$, is also a vector, collinear to the vector $\vec{a}$, . The vector $\lambda \vec{a}$, has the direction same (or opposite) to that of vector $\vec{a}$, according as the value of λ is positive (or negative).Also, the magnitude of vector $\lambda \vec{a}$, Is |λ| times the magnitude of the vector $\vec{a}$ , i.e , $\left | \lambda \vec{a} \right |=\left | \lambda \right |\left | \vec{a} \right |$
A geometric visualisation of multiplication of a vector by a scalar is given in( Fig 10.12) When $\lambda =-1,\: then\: \lambda \vec{a}=-\vec{a}$ which is a vector having magnitude equal to the magnitude of $\vec{a}$ the vector $-\vec{a}$ is called negative vector .(or additive inverse ) of vector a $\vec{a}$ and we always have .
$\vec{a}+(-\vec{a})=(-\vec{a})+(\vec{a})$ . Also , if $\lambda =\frac{1}{|\vec{a}|}$ , provide 0 , i.e $\vec{a}$ is not a null vector , then $\lambda \vec{a}$ represents the unit vector in the direction of. We write it as $\hat{a}=\frac{1}{|\vec{a}|}$

Note:

• One may observe that whatever be the value of λ , the vector $\lambda \vec{a}$ is always collinear to the vector $\vec{a}$. In fact, two vectors $\vec{a}\: and\: \vec{b}$ are collinear if and only if there exists a nonzero scalar λ such that $\vec{b}=\lambda \vec{a}$ . If the vectors $\vec{a}\: and\: \vec{b}$ are given in component form , i.e , $\vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}\: and\: \vec{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}$
• If two vectors are collinear if and only if $\lambda (a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k})\:=\:b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}\Rightarrow \lambda a_{1},b_{2}=\lambda a_{2},b_{3}=\lambda a_{3}$
• $\frac{b_{1}}{a_{1}}=\frac{b_{2}}{a_{2}}=\frac{b_{3}}{a_{3}}$
• If $(a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k})$ ,this is called the direction of ratio $\vec{a}$
• In case if it is given that l ,m ,n are direction cosine of a vector , then $l\hat{i}+m\hat{j}+n\hat{k}=(\cos\alpha)\hat{i}+(\cos\beta)\hat{j}+(\cos \gamma )\hat{k}$ is the unit vector in the direction of that vector, where α , β and γ are the angles which the vector makes with x , y and z axes respectively .

Example: find the unit vector in the direction of vector $\vec{a}=2\hat{i}+3\hat{j}+\hat{k}$
Solution: The unit vector in the direction of a vector $\vec{a}$ is given by $\hat{a}=\frac{1}{|\vec{a}|}\vec{a}$
Now $|\vec{a}|=\sqrt{2^{2}+3^{2}+1^{2}}=\sqrt{14}$

Therefore $\hat{a}=\frac{1}{\sqrt{14}}(2\hat{i}+3\hat{j}+\hat{k})=\frac{2}{\sqrt{14}}\hat{i}+\frac{3}{\sqrt{14}}\hat{j}+\frac{1}{\sqrt{14}}\hat{k}$

Example: Find the unit vector in the direction of the sum of the vector $\vec{a}=2\hat{i}+2\hat{j}-5\hat{k}\: and\: \vec{b}=2\hat{i}+\hat{j}+3\hat{k}$
Solution: The sum of the given vector $\vec{a}+\vec{b}=4\hat{i}+3\hat{j}+2\hat{k}$, Let $\vec{a}+\vec{b}=\vec{c}$
$|\vec{c}|=\sqrt{4^{2}+3^{^{3}}+(-2)^{2}}=\sqrt{29}$ , Thus the require unit vector is

$\hat{c}=\frac{1}{|\vec{c}|}\vec{c}=\frac{1}{\sqrt{29}}(4\hat{i}+3\hat{j}-2\hat{k})=\frac{4}{\sqrt{29}}\hat{i}+\frac{3}{\sqrt{29}\hat{j}}-\frac{2}{\sqrt{29}}\hat{k}$

## Components of a Vector

Let us take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x -axis, y -axis and z -axis, respectively. Then, clearly .
$|\vec{OA}|=1,|\vec{OB}|=1\:and\:\:\vec{OC}=1$ The vector $|\vec{OA}|,|\vec{OB}|,\vec{OC}$ each having magnitude 1,are called unit vectors along the axes OX, OY and OZ respectively and denoted by $\hat{i},\hat{j}\: and\: \hat{k}$ respectively (Fig 10.13).
Now, consider the position vector OP of a point P (x , y, z) as in Fig 10.14. Let P1 be the foot of the perpendicular from P on the plane XOY. We, thus, see that P1 P is parallel to z – axis.
Therefore it follows that $\vec{OP_{1}}=\vec{OQ}+\vec{QP_{1}}=x\hat{i}+y\hat{j}$
$\vec{OP_{1}}(\vec{a})=x\hat{i}+y\hat{j}+z\hat{k}$
The length of any vector = , is readily determined by applying the Pythagoras theorem twice. We note that in the right angle triangle OQP1. (Fig 10.14) $\vec{op} = \ \sqrt{X^{2}+Y^{2}}$

$| \vec{r} |= \sqrt{x^{2}+y^{2}+z^{2}}$
Vector joining two point : $p_{1}(x_{1},y_{1},z_{1})\: and\: p_{2}(x_{2},y_{2},z_{1})$ are any two points $p_{1} and p_{2}$ with the origin O, and applying triangle law , from the triangle $OP_{1}P_{2}$ ,we have $p_{1}p_{2}=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

## Section Formula

Let P and Q be two points represented by the position vectors OP and OQ , respectively, with respect to the origin O. Then the line segment joining the points P and Q may be divided by a third point, say R, in two ways – internally (Fig 10.16) and externally (Fig 10.17). Here, we intend to find the position vector OR for the point R with respect to the origin O .we take the two cases ;
Case I) where R divided PQ internally . where m and n point R divides PQ internally in the ratio of m : n . Now from triangles ORQ and OPR, we have,
$\vec{r}=\frac{m\vec{b}+n\vec{a}}{m+n}$
Hence , the position vector of the point R which divides P and Q internally in the ratio of m: n is given b $\vec{OR}=\frac{m\vec{b}+n\vec{a}}{m+n}$ Case II) When R divides PQ externally (Fig 10.17). We leave it to the reader as an exercise to verify
that the position vector of the point R which divides the line segment PQ externally in the ratio m : n , i.e
$\frac{PR}{QR}=\frac{m}{n},\vec{or}=\frac{m\vec{b}-n\vec{a}}{m+n}$
Product of two vector : other algebraic operation which we intend to discuss regarding vectors is their product. We may recall that product of two numbers is a number, product of two matrices is again a matrix. But in case of functions, we may multiply them in two ways, namely, multiplication of two functions point wise and composition of two functions. Similarly, multiplication of two vectors is also defined in two ways, namely, scalar (or dot) product where the result is a scalar, and vector (or cross) product where the result is a vector. Based upon these two types of products for vectors, they have found various applications in geometry, mechanics and engineering. In this section, we will discuss these two types of products.
Scalar (or dot ) product of two vector :The scalar product of of two non zero vector a and vector b , denoted by vector $\vec{a}\cdot \vec{b}$ is define as $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \Theta$
where, θ is the angle between and $\vec{a}\:and\:\vec{b}$, 0 is not defined, and in this case

## Observation

• $\vec{a}\cdot \vec{b}$ is real number
• Let $\vec{a}\:and\:\vec{b},$ be two nonzero vector , then $\vec{a}\cdot \vec{b}=0$ if and only if $\vec{a}\:and\:\vec{b},$ are perpendicular to each other i.e $\vec{a}\cdot \vec{b}=0$,
• If θ = 0, then , $\vec{a}.\vec{b}=|\vec{a}||\vec{b}|$ in particular $\vec{a}\cdot \vec{a}=|\vec{a}^{2}|$ as θ in this case is 0.
• If θ = π , then $\vec{a}\cdot \vec{b}=-|\vec{a}||\vec{b}|$ in particular $\vec{a}\cdot \vec{a}=|\vec{a}^{2}|$ as θ in this case is $\Pi$
• In view of the Observations 2 and 3, for mutually perpendicular unit vector$\hat{i},\hat{j},\hat{k}$ we have , $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1\: or\: \hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0$
• The angle between two non zero vector $\vec{a}\:and\:\vec{b}$ is given by $\cos\:\:\frac{\vec{a}\:\:\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\:\:or\:\:\theta$
$=\cos^{-1}\:\left ( \frac{\vec{a}\:\:\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |} \right )$
• The scalar product is commutative i.e $\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$
• Cross product of two vector : he three dimensional right handed rectangular coordinate system. In this system, when the positive x -axis is rotated counter clockwise nto the positive y -axis, a right handed (standard) screw would advance in the direction of the positive z -axis (Fig 10.22(i)). In a right handed coordinate system, the thumb of the right hand points in the direction of the positive z -axis when the fingers are curled in the direction away from the positive x-axis toward the positive y -axis (Fig 10.22(ii). ## Definition

The vector product of two nonzero vectors $\vec{a}\:and\:\vec{b}$ , is denoted by $\vec{a}\times \vec{b}=|\vec{b}||\vec{a}|\sin\theta \hat{n}$ where, θ is the angle between $\vec{a}\:and\:\vec{b}$ and ≤θ≤π and $\hat{n}$ is a unit vector perpendicular to both $\vec{a}\:and\:\vec{b}$ , such that , $\vec{a}\:,\:\vec{b}\:and\: \hat{n}$ and form a right handed system (Fig 10.23). i.e., the right handed system rotated from $\vec{a}to\vec{b}$ moves in the direction of . If either , then θ is not defined and in this case, we define of $\hat{n}$

## Observation

• $\vec{a}\times \vec{b}$ is a vector
• Let $\vec{a}\:and\:\vec{b}$ be two nonzero vector ,Then $\vec{a}\times \vec{b}=\vec{0}$ if and only if$\vec{a}\:and\:\vec{b}$are parallel (or collinear) to each other . in particular $\vec{a}\times \vec{b}=\vec{0}\: and\: \vec{a}\times (-\vec{a})$ , since in the first situation, θ = 0 and in the second one, θ = π , making the value of sinθ to be 0
• In view of the Observations 2 and 3, for mutually perpendicular $\hat{i},\hat{k},\hat{k}$ (fig 10.24)
• we have , $\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=\hat{0}$ unit vector $\hat{i},\hat{k},\hat{k}$
$\hat{i}\times \hat{j}=\hat{k},\hat{j}\times \hat{k}=\hat{k},\hat{k}\times \hat{i}=\hat{j}$
• In terms of vector product, the angle between two vectors and may be given as $\vec{a}\:and\:\vec{b}$
• It is always true that the vector product is not commutative, as $\vec{a}\times \vec{b}=-\vec{b}\times \vec{a}$ Indeed, , where $\vec{a}\times \vec{b}=|\vec{a}||\vec{b}|\sin \theta\hat{n}$ form a right handed system, i.e., θ is traversed from . $\vec{a}\:to\: \vec{b}$ is a vector Fig 10.25 (i). While, $\vec{b}\times \vec{a}=|\vec{a}||\vec{b}|\sin \theta \vec{n_{1}}$ ,where, $\vec{b},\vec{a}$ ,and $n_{1}$form a right handed system i.e. θ is traversed from . ## Points to Remember

• position vector of a point p(x,y,z) is given as $\vec{OP}(\vec{r})=x\hat{i}+y\hat{j}+z\hat{k}$ , and it’s magnitude by $\sqrt{x^{2}+y^{2}+z^{2}}$
• The scalar components of a vector are its direction ratios, and represent its projections along the respective axes.
• The magnitude (r ), direction ratios (a , b, c) and direction cosines (l, m, n) of any vector are related as$l=\frac{a}{r},m\frac{b}{r},n=\frac{c}{r}$
• Th e vector sum of the three sides of a triangle taken in order is 0
• The vector sum of two coinitial vectors is given by the diagonal of the parallelogram whose adjacent sides are the given vectors.
• The multiplication of a given vector by a scalar λ , changes the magnitude of the vector by the multiple |λ|, and keeps the direction same (or makes it opposite) according as the value of λ is positive (or negative).
• For a given a , the vector $\hat{a}=\frac{\vec{a}}{|\vec{a}|}$ gives the unit vector in the direction of vector a.
• The position vector of a point R dividing a line segment joining the points P and Q whose position vectors are $\vec{a}\:and\:\vec{b}$respectively in the ratio m: nInternally is given by : $\frac{n\vec{a}+m\vec{b}}{m+n}$Externally , is given by : $\frac{n\vec{b}-m\vec{a}}{m+n}$
• The scalar product of two given vector $\vec{a}\:and\:\vec{b}$ having angle θ between them is define as \vec{a}\cdot $\vec{b}=|\vec{a}||\vec{b}|\cos \theta$Also when $\vec{a}\cdot \vec{b}$ is given the angle , θ between the vector $\vec{a}\:and\:\vec{b}$ may be determine by $\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$
• If θ is the angle between two vectors $\vec{a}\:and\:\vec{b}$ , then their cross product is given as$\vec{a}\times \vec{b}=|\vec{a}||\vec{b}|\sin \theta \hat{n}$, where unit vector $\hat{n}$, perpendicular to the plane containing $\vec{a}\:and\:\vec{b}$. Such that $\vec{a},\vec{b}, hat{n}$. form right handed system of coordinate axes.

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