# Vectors Class 12 Notes Maths Chapter 10 - CBSE

## What are Vectors ?

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Vectors come in handy when we study those physical objects that have direction as well as magnitude. Let’s look at some of the terms associated with vectors :

## Scalar

It is a quantity which has magnitude but no direction. Examples : Mass, length, distance, speed etc.

## Vector

It is a quantity which has a magnitude as well as a specific direction. Examples : Weight, displacement, velocity etc. Denoted as

$$\overrightarrow{AB}$$

## Magnitude

The distance between initial and terminal points of a vector is called the magnitude of the vector.

$$\text{Denoted as}\space|\overrightarrow{AB}| \space \text{which is read as modulus of}\space \overrightarrow{AB}.$$

## Direction Cosines

Direction cosines of a vector are the cosines of the angle between the vector and the three coordinate axes.

• As you see in the figure the angle α, β,γ made by the vector A with the positive directions of x,y and z-axis respectively, are called its direction angles. The cosine values of these angles ,i.e., cos α, cos β and cos γ are called direction cosines.
• Denoted by l,m and n.

## Direction Ratios

Numbers that are proportional to the direction cosines of the line are called direction ratios of the line.

• Ax = A cos α, Ay = A cos β and Az = A cos γ
• Denoted as a, b and c.
• Let (l, m, n) be the direction cosines of a line and direction ratios of the line be (a, b, c). Then

$$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=\frac{\sqrt{l^2+m^2+n^2}}{\sqrt{a^2+b^2+c^2}}=\pm \space \frac{1}{\sqrt{a^2+b^2+c^2}}\qquad\\l=\pm\frac{a}{\sqrt{a^2+b^2+c^2}}(\because l^2+m^2+n^2=1)\\m=\pm\frac{b}{\sqrt{a^2+b^2+c^2}}\space\space\space\space n=\pm\frac{c}{\sqrt{a^2+b^2+c^2}}$$

## Vectors

Equal Vectors

Two vectors are said to be equal vectors if they have same magnitude and direction.

$$\overrightarrow{AB}=\overrightarrow{CD}$$

## Unit Vectors

If the magnitude of the vector is one , then it is called unit vector.

$$•\space\hat {a} \text{ is a unit vector in the direction of}\space \vec{a} \space\text{and}\space \hat a =\frac{\vec{a}}{|\vec{a}|}\\ \text{• Unit vectors in the directions of the positive x-axis, y-axis and z-axis are i, j, k.}$$

## Zero Vectors

If the magnitude of the vector is zero , then it is called null vector or zero vector. It can have any arbitrary direction.

## Parallel Vectors

Vectors which have same direction or exactly the opposite direction are called parallel vectors . The angle between them is 0° or 180°. Parallel vectors
are of two types :

## Like Parallel Vectors

If the angle between the vectors is 0°, then they are called like parallel vectors.

## Unlike Parallel Vectors

If the angle between the vectors is 180°, then they are called unlike parallel vectors.

## Collinear Vectors

Two or more vectors are said to be collinear vectors if they are parallel in same line irrespective of their magnitudes.

## Position Vector Of A Point

The vector OA is said to be a position vector of A with respect to Origin O, if A be any terminal point and O is the origin which is fixed.

## Negative Of A Vector

Two vectors are called negative vectors of each other if they have same magnitude but opposite direction. Denoted as

$$-\vec{a}$$

## Components Of A Vector

$$\overrightarrow{\text{OA}},\overrightarrow{\text{OB}}\space\text{AND}\space \overrightarrow{\text{OC}}\text{are the unit vectors along the axes OX, OY, and OZ respectively,}\text{and denoted by}\space \hat{i}, \hat{j}, \hat{k}.$$

$$\text{Now, consider the position vector}\space\overrightarrow{\text{OP}}\text{of a point P (x,y,z) as in fig}.\\\text{Let}\space \text{P}_1\space \text{be the foot of the perpendicular from P on the plane XOY. We, thus, See that}\\\text{P}_1 \text{P is parallel to z-}\hat{axis}.\hat{}\space \hat{\text{As}}\text{i, j, k are the unit vectors along the x,y and z-axes respectively, and by the definition of the coordinates of P,}\text{we have}\overrightarrow{P_1P}=\overrightarrow{OR}=Z\hat{k}. \space\text{Similarly,}\\\overrightarrow{QP}_1=\overrightarrow{OS}=y\hat{j}\space\text{and}\space\overrightarrow{OQ}=x\hat{i}.$$

$$\text{Therefore, it follows that}\space\overrightarrow{\text{OP}_1}={\overrightarrow{\text{OQ}}}+{\overrightarrow{\text{OP}}_1}=x\hat{i}+y\hat{j}\\\text{And}\overrightarrow{\text\space{OP}}=\overrightarrow{\text{OP}}_1+\overrightarrow{\text{P}_1P}=x\hat{i}+y\hat{j}+z\hat{k}\\\text{Hence , the position vector of P with reference to O is given by}\\\overrightarrow{\text{OP}}=r=x\hat{i}+y\hat{j}+z\hat{k}\\\text{This form of any vector is called its component form. Here x,y,z are called as the scalar components of }\text{are called the vector components of r along the respective axes.}$$

Triangle Law

$$\overrightarrow{\text{AC}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}\implies \vec{c}=\vec{a}+\vec{b}\\•\space \vec{\text{c}}\space \text{is called the resultant of\space }\vec{\text{a}}\space\text{and}\space\vec{\text{b}}$$

The addition of vectors is achieved by ‘tail to nose’ placing of the directed line segments in a triangle.

Parallelogram Law

The result of adding two co-initial vectors is the vector represented by the diagonal of the parallelogram formed with the component vectors as adjacent sides. eg: in the fig. below,

## Multiplication Of A Vector By A Scalar

$$\text{\text{The product of}}\space \text{by a scalar ‘m’ is denoted as m.}\vec{a}\space \text{such that :}\\\text{• The support of ma is same or parallel to that of }\vec{a},\\ • |\text{m}\vec{\text{a}}|=|\text{m}||\vec{\text{a}}|\\ \text{• m}\vec{\text{a}}\space \text{has the direction same or opposite to that of a, according as m is +ve or -ve.}$$

## Position Vector Of A Point Dividing A Line Segment In A Given Ratio

For Internal Division

$$\text{\text{Let}}\space\vec{\text{a}}\space \text{and}\space \vec{\text{b}}\space \text{be the position vectors of two points A and B respectively and let C be a point with position vector}\space\vec{\text{c}}\space \text{dividing AB internally in the ratio m : n. Then the position vector of C is}\\\vec{\text{c}}=\overrightarrow{\text{OC}}=\frac{\text{m}\vec{\text{b}}+\text{n}\vec{\text{a}}}{\text{m+n}}$$

For External Division

$$\text{\text{Let}}\space\vec{\text{a}}\space \text{and}\space \vec{\text{b}}\space \text{be the position vectors of two points A and B respectively and let C be the point with position vector}\space \vec{\text{c}} \text{dividing AB externally in the ratio m : n. Then the position vector of C is}\\ \vec{\text{c}}=\overrightarrow{\text{OC}}=\frac{\text{m}\vec{\text{b}}-\text{n}\vec{\text{a}}}{\text{m+n}}$$

## Properties And Application Of Scalar(Dot) Product Of Vectors

$$\text{(i)}\space \vec{a}.\vec{a}=|\vec{a}|^2=a^2\\\qquad\text{(ii)}\space \vec{a}.\vec{b}=\vec{b}.\vec{a} \space \text{(commutativity)}\\\qquad\text{(iii)} \space \vec{a}.\vec{0}=0\\\qquad\text{(iv)}\space \vec{a}.(\vec{b}+\vec{c})=\vec{a}.\vec{b}+\vec{a}.\vec{c}\space\text{(Distributivity)}\\\qquad\text{(v)} \space\vec{a}.(\vec{b}-\vec{c})=\vec{a}.\vec{b}-\vec{a}.\vec{c}\\\qquad\text{(vi)}\space \vec{a}.\vec{b}=|\vec{a}|.|\vec{b}|.\text{cos} \space \theta\\\qquad \text{(vii) If x is any scalar, then}\space (x\vec{a}).\vec{b}=x(\vec{a}.\vec{b})=\vec{a}.(x\vec{b})\\\qquad\text{(viii)}\space(x\vec{a}).(y\vec{b})=\text{xy}(\vec{a}.\vec{b})\text{where x, y are scalars.}\\\qquad\text{(ix)}\space (\vec{a}+\vec{b}).(\vec{a}-\vec{b})=|\vec{a}|^2-|\vec{b}|^2\\\qquad\text{(x)}\space \text{(a + b)}^2=|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}\\\qquad\text{(xi)}\space (\vec{\text{a}}-\vec{\text{b}})^2=|\vec{a}|^2+|\vec{b}|^2-2\vec{a}.\vec{b}\\\qquad\text{(xii) If}\space \hat i, \hat j, \hat k\space\text{are three-unit vectors along three mutually perpendicular lines, then}\\ \qquad •\space\space \hat i .\hat i=\hat j.\hat j=\hat k.\hat k=1\\•\space \hat{i} .\hat{j} = \hat{j} . \hat{k} = \hat{k} . \hat{i} = \hat{j} . \hat{i} = \hat{k} . \hat{j} = \hat{i} . \hat{k} = 0.\\\qquad\text{(xiii) If OX, OY, OZ are three mutually perpendicular axes and}\space\hat{i},\hat{j},\hat{k}\space \text{are unit vectors along these axes. If co-ordinates of any point P, are (x, y, z) then its position vector is}\\\text{x}\hat{i}+\text{y}\hat{j}+\text{z}\hat{k}.$$

## Properties Of Vector (Cross) Product Of Vectors

$$\text{(i)}\space \vec{a}×\vec{b}=-(\vec{b}×\vec{a})\\\qquad\text{(ii)}\space\vec{a}×\vec{a}=\vec{0}\\\qquad\text{(iii)}\space x(\vec{a}×\vec{b})=(x\vec{a})×\vec{b}=\vec{a}×(x\vec{b})\\\qquad\text{(iv)}\space(x\vec{a})×(y\vec{b})=\text{xy}(\vec{a}×\vec{b}).\\\qquad\text{(v)}\space|\vec{a}×\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}.\vec{b})^2.\\\qquad\text{(vi)}\space \hat{i}×\hat{i}=\hat{j}×\hat{j}=\hat{k}×\hat{k}=0.\\\qquad\text{(vii)}\space \hat{i}×\hat{j}=\hat{k},\hat{j}×\hat{k}=\hat{i},\hat{k}×\hat{i}=\hat{j}\\\qquad\text{(viii) Two non-zero vectors are collinear if their cross product is equal to zero vector.}\\\qquad\text{(ix)} \space\text{If} \space \vec{a}_1\hat{i}+\vec{a_2}\hat{j}+\vec{a_3}\hat{k}\\\text{And} \space\vec{b}=\vec{b}_1\hat{i}+\vec{b_2}\hat{j}+\vec{b_3}\hat{k}\\\text{Then}\space\vec{a}×\vec{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}=(a_2b_3-a_3b_2)\hat{i}-(a_1b_3-a_3b_1)\hat{j}+(a_1b_2-a_2b_1)\hat{k}\\\qquad\text{(x) If a, b, c are three vectors, then}\\\vec{a}×(\vec{b}+\vec{c})=\vec{a}×\vec{b}+\vec{a}×\vec{c}\\(\vec{b}+\vec{c})×\vec{a}=\vec{b}×\vec{a}+\vec{c}×\vec{a}$$