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Vector Class 12 Notes CBSE Maths Chapter 10 [Free PDF Download]

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Revision Notes for CBSE Class 12 Maths Chapter 10 (Vector algebra) - Free PDF Download

The topics included in the syllabus of Class 12 Mathematics focus on the development of basic and advanced mathematical concepts. Chapter 10 focuses on the topic of vector algebra. In this chapter, students will learn what is a vector quantity and how it can be represented or calculated. To comprehend this chapter easily, students can refer to the Vector Algebra Class 12 Maths revision notes prepared by the highly experienced teachers at Vedantu. The simplest explanation of the concepts in these revision notes Class 12 Chapter 10 will help them understand the topics and prepare the chapter more efficiently.

Students will also learn how to calculate its magnitude and the direction of the vector quantities. It is an algebraic subject that a student should learn to calculate velocity, acceleration, displacement, force, momentum, weight, and other vector quantities in Physics. Hence, it is one of the most important chapters of Class 12 maths syllabus.

CBSE Class 12 Maths Revision Notes 2023-24 - Chapter Wise PDF Notes for Free

In the table below we have provided the PDF links of all the chapters of CBSE Class 12 Maths whereby the students are required to revise the chapters by downloading the PDF. 


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Vector algebra Class 12 Notes Maths - Basic Subjective Questions

Section–A (1 Mark Questions)

1. Find the values of x  and y so that the vectors $2 \hat{i}+3\hat{j}$ and $x \hat{i}+y\hat{j}$ are equal.

Ans. We know that

$$\begin{aligned}& a_1 \hat{i}+b_1 j=a_2 \hat{i}+b_2 j \\& \Leftrightarrow a_1=a_2 \text { and } b_1=b_2 \\& \therefore 2 \hat{i}+3\hat{j}=x \hat{i}+y\hat{j} \\& \Rightarrow x=2 \text { and } y=3 .\end{aligned}$$


2. Find a unit vector parallel to the vector $-3\hat{i}+4\hat{j}$. 

Ans. Let, $\vec{a}=-3 \hat{i}+4 j$.

Then,

$$|\vec{a}|=\sqrt{(-3)^2+(4)^2}=5$$

$\therefore$ Unit vector parallel to

$$\vec{a}=a=\frac{\vec{a}}{|\vec{a}|}=\frac{1}{5}(-3 \hat{i}+4 j)=-\frac{3}{5}\hat{i}+\frac{4}{5} j .$$


3. The magnitude of the vector $\bar{a}=3 \hat{i}-6 j+2 k$  is.

Ans. Let $\bar{a}=3 \hat{i}-6 j+2 k$. Then,

$$|a|=\sqrt{3^2+(-6)^2+2^2}=7 \text {. }$$


4. Find the projection of the vector $\bar{a}=2 \hat{i}+3\hat{j}+2\hat{k}$ on the vector $\bar{b}=\hat{i}+2\hat{j}+\hat{k}$

Ans. The projection of vector $\vec{a}$ on the vector $\vec{b}$ is given by -

$$\frac{1}{\mid \vec{b}}(\vec{a} \cdot \vec{b})=\frac{(2 \times 1+3 \times 2+2 \times1)}{\sqrt{(1)^2+(2)^2+(1)^2}}=\frac{10}{\sqrt{6}}=\frac{5}{3} \sqrt{6}$$


5.  If l,m  and n are direction cosines of a given vector, then l2+m2+n2= ________.

Ans. We know that,

If $l, m$ and $n$ represents direction cosines of a given vector.

Then,

$$l^2+m^2+n^2=1 \text {. }$$


Section–2 (2 Mark Questions)

6. If the position vector $\vec{a}$ of a point (12,n) is such that $\left | \vec{a} \right |=13$,  find the value of n. 

Ans. The position vector of the point $(12, n)$ is $12 \hat{i}+n j$.

$$\begin{aligned}& \therefore \vec{a}=12 \hat{i}+n j \\& \Rightarrow|\vec{a}|=\sqrt{12^2+n^2}\end{aligned}$$

$|\vec{a}|=13$

$\Rightarrow 13=\sqrt{12^2+n^2}$

$\Rightarrow 169=144+n^2$

$\Rightarrow n^2=25$

$\Rightarrow n= \pm 5$


7. Find the sum of vectors $\hat{a}=\hat{i}-2\hat{i}+\hat{k}, \hat{b}=-2\hat{i}+4\hat{j}+5\hat{k}, \hat{c}=\hat{i}-6\hat{j}-7\hat{k}$.

Ans. We have,

$$\begin{aligned}& \vec{a}+\vec{b}+\vec{c}=(\vec{a}+\vec{b})+\vec{c} \\& \Rightarrow \vec{a}+\vec{b}+\vec{c}=\{(\hat{i}-2 j+k)+(-2 \hat{i}+4 j+5 k)\}+(\hat{i}-6 j-7 k) \\& \Rightarrow \vec{a}+\vec{b}+\vec{c}=\{(1-2) \vec{i}+(-2+4) j+(1+5) k\}+(\hat{i}-6 j-7 k) \\& \Rightarrow \vec{a}+\vec{b}+\vec{c}=(-\hat{i}+2 j+6 k)+(\hat{i}-6 j-7 k) \\& \Rightarrow \vec{a}+\vec{b}+\vec{c}=(-1+1) \hat{i}+(2-6) j+(6-7) k \\& \Rightarrow \vec{a}+\vec{b}+\vec{c}=0 \hat{i}-4 j-k .\end{aligned}$$


8. Find a vector in the direction of vector \vec{a}=\hat{i}-2 \hat{j} that has magnitude 7 units.

Ans. Given: $\vec{a}=\hat{i}-2 \hat{j}$

$$\therefore|\vec{a}|=\sqrt{1^2+(-2)^2}=\sqrt{5}$$

The unit vector in the direction of the a given vector $\dot{a}$ is

$$\hat{a}=\frac{1}{|\vec{a}|}=\frac{1}{\sqrt{5}}(\hat{i}-2\hat{j})=\frac{1}{\sqrt{5}}\hat{i}-\frac{2}{\sqrt{5}} \hat{j}$$

Therefore, the vector having magnitude equal to 7 and in the direction of $\vec{a}$ is

$$7 \hat{a}=7\left(\frac{1}{\sqrt{5}} \hat{i}-\frac{2}{\sqrt{5}}\hat{j}\right)=\frac{7}{\sqrt{5}} \hat{i}-\frac{14}{\sqrt{5}} \hat{j}$$


9. If a vector makes angles $\alpha ,\beta ,\gamma$ with OX, OY, and OZ respectively, prove that $sin^{2}\alpha +sin^{2}\beta +sin^{2}\gamma =2$ . 

Ans. Let $l, m, n$ be the direction cosines of the given vector. Then,

$$l=\cos \alpha, m=\cos \beta, n=\cos \gamma .$$

Now, $l^2+m^2+n^2=1$

$$\begin{aligned}& \Rightarrow \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\& \Rightarrow\left(1-\sin ^2 \alpha\right)+\left(1-\sin ^2 \beta\right)+\left(1-\sin ^2\gamma\right)=1 \\& \Rightarrow \sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=2 .\end{aligned}$$


10. Find $(\vec{a} + 3\vec{b}).(2\vec{a} - \vec{b})$ if $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=3\hat{i}+2\hat{j}-\hat{k}$. 

Ans. Given:

$\vec{a}=\hat{i}+j+2 k$ and $\vec{b}=3 \hat{i}+2 j-k$.

$$\Rightarrow \vec{a}+3 \vec{b}=(\hat{i}+j+2 k)+3(3 \hat{i}+2 j-k)=10 \hat{i}+7 j-k$$

And,

$\therefore ( 2 \vec{a}-\vec{b}=2(\hat{i}+j+2 k)-(3 \hat{i}+2 j-k)=-\hat{i}+0 j+5 k )$

$\therefore ( 2 \vec{a}-\vec{b}=2(\hat{i}+j+2 k)-(3 \hat{i}+2 j-k)=-\hat{i}+0 j+5 k )$

=(10)(-1)+(7)(0)+(-1)(5)

=-10+0-5

=-15


11. For given vectors, $\vec{a}=2\hat{i}-\hat{}{j}+2\hat{k} a$ and $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$, find the unit vector in the direction of the vector $\vec{a}+\vec{b}$ . 

Ans. The given vectors are

$\vec{a}=2 \hat{i}-j+2 k \text { and } \vec{b}=-\hat{i}+j-k$

$\therefore \vec{a}+\vec{b}=(2-1) \hat{i}+(-1+1) j+(2-1) k$

$=\hat{i}+k$

$|\vec{a}+\vec{b}|=\sqrt{1^2+1^2}-\sqrt{2}$

Hence, the unit vector in the direction of $\vec{a}+\vec{b}$ is

$\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{\hat{i}+k}{\sqrt{2}}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}} k \text {. }$


12. Find the magnitude of $\vec{a}$  given by $\vec{a}=(\hat{i}+3\hat{j}-2\hat{k})\times(-1\hat{i}+3\hat{k})$

Ans. We have,

$$\begin{aligned}& \vec{a}=(\hat{i}+3 j-2 k) \times(-\hat{i}+0 j+3 k) \\& \Rightarrow \vec{a}=\left|\begin{array}{ccc}\hat{i} & j & k \\1 & 3 & -2 \\-1 & 0 & 3\end{array}\right|=(9-0) \hat{i}-(3-2) j+(0+3) k \\& =9 \hat{i}-j+3 k \\& \therefore|\vec{a}|=\sqrt{9^2+(-1)^2+3^2}=\sqrt{91} .\end{aligned}$$


13. Find a unit vector perpendicular to both the vectors $\hat{i}-2\hat{j}+3\hat{k}$ and $\hat{i}+2\hat{j}-\hat{k}$. 

Ans. Let $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$.

Then,

$$\begin{aligned}& \vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & -2 & 3 \\1 & 2 & -1\end{array}\right|=(2-6) \hat{i}-(-1-3) \hat{j}+(2+2) \hat{k} \\& =-4 \hat{i}+4 \hat{j}+4 \hat{k} . \\& \Rightarrow|\vec{a} \times \vec{b}|=\sqrt{(-4)^2+4^2+4^2}=4 \sqrt{3} .\end{aligned}$$

So, a unit vector perpendicular to both the vectors $\vec{a}$ and $\vec{b}$ is given by

$$\hat{n}=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{(-4 \hat{i}+4\hat{j}+4 \hat{k})}{4 \sqrt{3}}=\frac{1}{\sqrt{3}}(-\hat{i}+\hat{j}+\hat{k})$$


PDF Summary - Class 12 Maths Vector algebra Notes (Chapter 10)

Vector:

Vector quantities are those quantities that have magnitude and direction. It is generally represented by a directed line segment. We represent a vector as \[\overrightarrow{\text{AB}}\], where initial point of vector is denoted by \[\text{A}\] and the terminal point by \[\text{B}\]. The magnitude of vector is expressed as \[\left| \overrightarrow{\text{AB}} \right|\].


Position Vector

Let us denote the origin as \[\text{O}\] such that this is a fixed point. There is a point, say \[\text{P}\] at a distance from \[\text{O}\]. Now, the position vector of a point \[\text{P}\] is given by the vector \[\overrightarrow{\text{OP}}\]. 

The next case is when there are two vectors, \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] which represent the position vectors of two points \[\text{A}\] and \[\text{B}\]. Then we can write the vector \[\overrightarrow{\text{AB}}\text{=}\overrightarrow{\text{b}}-\overrightarrow{\text{a}}\] or the position vector of \[\text{B}-\] the position vector of \[\text{A}\].


Types of Vectors

1. Zero Vector - It has zero magnitude. This means that vector has the same initial and terminal point. It is denoted by \[\overrightarrow{\text{O}}\]. The direction of zero vector is indeterminate.

2. Unit Vector - It has unit magnitude. Unit vector in direction of a vector \[\overrightarrow{\text{a}}\] is denoted by \[\widehat{\text{a}}\] and symbolically as \[\widehat{\text{a}}\text{=}\dfrac{\overrightarrow{\text{a}}}{\left| \overrightarrow{\text{a}} \right|}\].

3. Co-initial Vectors - Two or more vectors are said to be co-initial if they have the same initial point.

4. Equal Vectors - Two vectors are said to be equal if they have the same magnitude and direction. They represent the same physical quantity.

5. Collinear Vectors - Two or more vectors are said to be collinear if they are parallel to the same line irrespective of their direction. For this reason, they are also called parallel vectors. We have two sub-categories – like vectors (same direction) and unlike vectors (different directions). We can represent it mathematically by taking two non-zero vectors \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\]. They are collinear if and only, if \[\overrightarrow{\text{a}}\text{=K}\overrightarrow{\text{b}}\], where \[\text{K}\in \text{R-}\left\{ \text{0} \right\}\].

6. Coplanar Vectors - Those vectors which lie on the same plane and they are all parallel to the same plane. We must remember that two vectors are always coplanar.

7. Negative Vector - A vector which has same magnitude but opposite direction to another vector is called negative of that vector.


Addition of Vectors

1. Triangle Law - Consider a triangle $ABC$. Let the sum of two vectors \[\overrightarrow{\text{a}}\text{ and }\overrightarrow{\text{b}}\] be represented by $\vec{c}$. The position vectors are represented by \[\overrightarrow{\text{AB}}\text{ , }\overrightarrow{\text{BC}}\text{ }and\text{ }\overrightarrow{\text{AC}}\].


Triangle Law


Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

So, we can write that $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$.

2. Parallelogram Law - Consider a parallelogram $ABCD$. Let the sum of two vectors \[\overrightarrow{\text{a}}\text{ and }\overrightarrow{\text{b}}\] be represented by $\vec{c}$. The position vectors are represented as 

\[\vec{a}\text{ =}\overrightarrow{\text{AB}}\text{ = }\overrightarrow{\text{DC}}\]

\[\vec{b}\text{ =}\overrightarrow{\text{AD}}\text{ = }\overrightarrow{\text{BC}}\]

\[\vec{a}+\vec{b}\text{=}\overrightarrow{\text{AC}}\]


Parallelogram Law


According to the parallelogram law of vector addition if two vectors act along two adjacent sides of a parallelogram (having magnitude equal to the length of the sides) both pointing away from the common vertex, then the resultant is represented by the diagonal of the parallelogram passing through the same common vertex and in the same sense as the two vectors.

The sum is

\[\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\]

$\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}$

3. Properties of Vector Addition

  1. Commutative property - \[\overrightarrow{\text{a}}\text{+}\overrightarrow{\text{b}}\text{=}\overrightarrow{\text{b}}\text{+}\overrightarrow{\text{a}}\]

  2. Associative property - \[\left( \overrightarrow{\text{a}}\text{+}\overrightarrow{\text{b}} \right)\text{+}\overrightarrow{\text{c}}\text{=}\overrightarrow{\text{a}}\text{+}\left( \overrightarrow{\text{b}}\text{+}\overrightarrow{\text{c}} \right)\]

  3. Zero is the additive identity - \[\overrightarrow{\text{a}}\text{+}\overrightarrow{\text{0}}\text{=}\overrightarrow{\text{a}}\text{=}\overrightarrow{\text{0}}\text{+}\overrightarrow{\text{a}}\] 

  4. \[\overrightarrow{\text{a}}\text{+}\left( \text{-}\overrightarrow{\text{a}} \right)\text{=}\overrightarrow{\text{0}}\text{=}\left( \text{-}\overrightarrow{\text{a}} \right)\text{+}\overrightarrow{\text{a}}\] 


Multiplication of a Vector by a Scalar

If \[\overrightarrow{\text{a}}\] is a vector and \[\text{m}\] is a scalar, then their product is \[\text{m }\overrightarrow{\text{a}}\]. The magnitude would be \[\left| \text{m} \right|\] times the magnitude of \[\overrightarrow{\text{a}}\]. This is called scalar multiplication. If \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] are vectors and $m$ and $n$ are scalars, then

  1. \[\text{m}\left( \overrightarrow{\text{a}} \right)\text{=}\left( \overrightarrow{\text{a}} \right)\text{m=m}\overrightarrow{\text{a}}\] 

  2. \[\text{m}\left( \text{n}\overrightarrow{\text{a}} \right)\text{=n}\left( \text{m}\overrightarrow{\text{a}} \right)\text{=}\left( \text{mn} \right)\overrightarrow{\text{a}}\] 

  3. \[\left( \text{m+n} \right)\overrightarrow{\text{a}}\text{=m}\overrightarrow{\text{a}}\text{+n}\overrightarrow{\text{a}}\]

  4. \[\text{m}\left( \overrightarrow{\text{a}}\text{+}\overrightarrow{\text{b}} \right)\text{=m}\overrightarrow{\text{a}}\text{+m}\overrightarrow{\text{b}}\] 


Component Form of Vectors

  • We have to consider three axis - $x,y,z$ and a point in the coordinate axis. So, the position vector for such a point would be written as $\overrightarrow{OP}=x\hat{i}+y\hat{j}+z\hat{k}$. This is the component form of vector.

  • The scalar components are $x,y,z$ and the vector components are \[x\hat{i},y\hat{j},z\hat{k}\].

  • Consider two vectors as \[\vec{A}=a\hat{i}+b\hat{j}+c\hat{k}\] and $\vec{B}=p\hat{i}+q\hat{j}+r\hat{k}$, then

  1. Sum is given by $\vec{A}+\vec{B}=(a+p)\hat{i}+(b+q)\hat{j}+(c+r)\hat{k}$.

  2. Difference is given by $\vec{A}-\vec{B}=(a-p)\hat{i}+(b-q)\hat{j}+(c-r)\hat{k}$.

  3. Multiplication by a scalar $m$ is given by $m\vec{A}=ma\hat{i}+mb\hat{j}+mc\hat{k}$.

  4. The vectors are equal if \[a=p,b=q,c=r\].


Test for Collinearity

Three points \[\text{A,B,C}\] with position vectors \[\overrightarrow{\text{a}}\text{,}\overrightarrow{\text{b}}\text{,}\overrightarrow{\text{c}}\] respectively are collinear, if and only if there exist scalar \[\text{x,y,z}\] not all zero simultaneously such that; \[\text{x}\overrightarrow{\text{a}}\text{+y}\overrightarrow{\text{b}}\text{+z}\overrightarrow{\text{c}}\text{=0}\], where \[\text{x+y+z=0}\].


Test for Coplanar Points

Four points \[\text{A,B,C,D}\] with position vectors \[\overrightarrow{\text{a}}\text{,}\overrightarrow{\text{b}}\text{,}\overrightarrow{\text{c}}\text{,}\overrightarrow{\text{d}}\] respectively are coplanar if and only if there exist scalars \[\text{x,y,z,w}\] not all zero simultaneously such that; \[\text{x}\overrightarrow{\text{a}}\text{+y}\overrightarrow{\text{b}}\text{+z}\overrightarrow{\text{c}}\text{+w}\overrightarrow{\text{d}}\text{=0}\], where \[\text{x+y+z+w=0}\].


Section Formula

  1. Let \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] be the position vectors of two points \[\text{A}\] and \[\text{B}\]. A point $R$ with position vector as $\vec{r}$ divides $\overrightarrow{AB}$ such that $m\overrightarrow{RB}=n\overrightarrow{AR}$ and this denotes that $\overrightarrow{AB}$ is divided internally in the ratio \[\text{m:n}\] is given by \[\overrightarrow{\text{r}}\text{=}\dfrac{\text{m}\overrightarrow{\text{b}}+n\overrightarrow{\text{a}}}{\text{m+n}}\].

  2. Let \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] be the position vectors of two points \[\text{A}\] and \[\text{B}\]. A point $R$ with position vector as $\vec{r}$ divides $\overrightarrow{AB}$ such that $m\overrightarrow{RB}=n\overrightarrow{AR}$ and this denotes that $\overrightarrow{AB}$ is divided externally in the ratio \[\text{m:n}\] is given by \[\overrightarrow{\text{r}}\text{=}\dfrac{\text{m}\overrightarrow{\text{b}}-n\overrightarrow{\text{a}}}{m-n}\].

  3. Now if the ratio is $1:1$, then we can obtain the position vector of the midpoint as \[\dfrac{\overrightarrow{\text{a}}\text{+}\overrightarrow{\text{b}}}{\text{2}}\].


Magnitude of Vector

  1. For a vector \[\vec{A}=a\hat{i}+b\hat{j}+c\hat{k}\], magnitude is $\left| A \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$.

  2. For vector $\overrightarrow{AB}$ with \[\vec{A}=a\hat{i}+b\hat{j}+c\hat{k}\] and $\vec{B}=p\hat{i}+q\hat{j}+r\hat{k}$, the magnitude is $\left| \overrightarrow{AB} \right|=\sqrt{{{(p-a)}^{2}}+{{(q-b)}^{2}}+{{(r-c)}^{2}}}$.


Product of Vectors

1. Scalar Product

  • It is also called dot product. For two vectors \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\], the dot product can be represented as \[\vec{a}.\vec{b}\] and it is defined as \[\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}\text{=}\left| \overrightarrow{\text{a}} \right|\left| \overrightarrow{\text{b}} \right|\text{cos }\!\!\theta\!\!\text{ ;(0}\le \text{ }\!\!\theta\!\!\text{ }\le \text{ }\!\!\pi\!\!\text{ )}\].

  • From this, we can find the angle between vectors as $\cos \theta =\dfrac{\vec{a}.\vec{b}}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|}$.

  • We have the below possibilities:

  1. If \[\text{ }\!\!\theta\!\!\text{ }\] is acute, then \[\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}\text{0}\].

  2. If \[\text{ }\!\!\theta\!\!\text{ }\] is obtuse, then \[\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}\text{0}\].

  3. If \[\text{ }\!\!\theta\!\!\text{ }\] is zero, then \[\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\].

  4. If \[\text{ }\!\!\theta\!\!\text{ }\] is $\pi $, then \[\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}=-\left| {\vec{a}} \right|\left| {\vec{b}} \right|\].

  • If vectors \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] are non-zero and \[\vec{a}.\vec{b}=0\], then it is the condition for them to be perpendicular vectors.

  • Considering component form and above point, we get results as

  1. \[\widehat{\text{i}}\cdot \widehat{\text{i}}\text{=}\widehat{\text{j}}\cdot \widehat{\text{j}}\text{=}\widehat{\text{k}}\cdot \widehat{\text{k}}\text{=1}\]

  2. \[\widehat{\text{i}}\cdot \widehat{\text{j}}\text{=}\widehat{\text{j}}\cdot \widehat{\text{k}}\text{=}\widehat{\text{k}}\cdot \widehat{\text{i}}\text{=0}\]

  • If \[\overrightarrow{\text{a}}\text{=}{{\text{a}}_{\text{1}}}\widehat{\text{i}}\text{+}{{\text{a}}_{\text{2}}}\widehat{\text{j}}\text{+}{{\text{a}}_{\text{3}}}\widehat{\text{k}}\] and \[\overrightarrow{\text{b}}\text{=}{{\text{b}}_{\text{1}}}\widehat{\text{i}}\text{+}{{\text{b}}_{\text{2}}}\widehat{\text{j}}\text{+}{{\text{b}}_{\text{3}}}\widehat{\text{k}}\] then \[\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}\text{=}{{\text{a}}_{1}}{{\text{b}}_{1}}+{{\text{a}}_{2}}{{\text{b}}_{2}}+{{\text{a}}_{3}}{{\text{b}}_{3}}\].

  • Properties of Scalar Product

  1. \[\overrightarrow{\text{a}}\cdot \overrightarrow{\text{a}}\text{=}{{\left| \overrightarrow{\text{a}} \right|}^{\text{2}}}\text{=}{{\overrightarrow{\text{a}}}^{\text{2}}}\text{,}\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}\text{=}\overrightarrow{\text{b}}\cdot \overrightarrow{\text{a}}\] (Commutative) 

  2. \[\overrightarrow{\text{a}}\cdot \left( \overrightarrow{\text{b}}\text{+}\overrightarrow{\text{c}} \right)\text{=}\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}\text{+}\overrightarrow{\text{a}}\cdot \overrightarrow{\text{c}}\] (Distributive)

  3. \[\left( \text{m}\overrightarrow{\text{a}} \right)\cdot \overrightarrow{\text{b}}\text{=}\overrightarrow{\text{a}}\cdot \left( \text{m}\overrightarrow{\text{b}} \right)\text{=m}\left( \overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}} \right)\] (Associative), where \[\text{m}\] is scalar.

  • Projection of vector \[\overrightarrow{\text{a}}\] on \[\overrightarrow{\text{b}}\text{=}\dfrac{\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}}{\left| \overrightarrow{\text{b}} \right|}\].

  • Maximum value of \[\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}\text{=}\left| \overrightarrow{\text{a}} \right|\left| \overrightarrow{\text{b}} \right|\]

  • Minimum value of  \[\overrightarrow{\text{a}}\cdot \overrightarrow{\text{b}}\text{=-}\left| \overrightarrow{\text{a}} \right|\left| \overrightarrow{\text{b}} \right|\]

  • A vector in the direction of the bisector of the angle between two vectors \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] is \[\dfrac{\overrightarrow{\text{a}}}{\left| \overrightarrow{\text{a}} \right|}+\dfrac{\overrightarrow{\text{b}}}{\left| \overrightarrow{\text{b}} \right|}\].

  • Hence bisector of the angle between the two vectors \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] is \[\text{ }\!\!\lambda\!\!\text{ }\left( \widehat{\text{a}}+\widehat{\text{b}} \right)\], where \[\text{ }\!\!\lambda\!\!\text{ }\in {{\text{R}}^{+}}\].

  • Bisector of the exterior angle between \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] is \[\text{ }\!\!\lambda\!\!\text{ }\left( \widehat{\text{a}}\text{-}\widehat{\text{b}} \right)\text{ }\!\!\lambda\!\!\text{ }\in \text{R-}\left\{ \text{0} \right\}\].


2. Vector Product

  • It is also called cross product. For two vectors \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\], the vector product is represented as $\vec{a}\times \vec{b}$ and is defined by \[\overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}}\text{=}\left| \overrightarrow{\text{a}} \right|\left| \overrightarrow{\text{b}} \right|\text{sin }\!\!\theta\!\!\text{ }\widehat{\text{n}}\], where $\theta $ is the angle between them and \[\widehat{\text{n}}\] is the unit vector perpendicular to both \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] such that \[\overrightarrow{\text{a}}\], \[\overrightarrow{\text{b}}\] and \[\widehat{\text{n}}\] form a right handed screw system.

  • From this, we can write the angle between vectors as $\sin \theta =\dfrac{\left| \vec{a}\times \vec{b} \right|}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|}$.

  • If vectors \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] are non-zero and \[\vec{a}\times \vec{b}=0\], then it is the condition for them to be parallel vectors.

  • Considering component form and above point, we get results as

  1. \[\widehat{\text{i}}\text{ }\!\!\times\!\!\text{ }\widehat{\text{i}}\text{=}\widehat{\text{j}}\text{ }\!\!\times\!\!\text{ }\widehat{\text{j}}\text{=}\widehat{\text{k}}\text{ }\!\!\times\!\!\text{ }\widehat{\text{k}}\text{=0}\]

  2. \[\widehat{\text{i}}\text{ }\!\!\times\!\!\text{ }\widehat{\text{j}}\text{=}\widehat{\text{k}}\text{,}\widehat{\text{j}}\text{ }\!\!\times\!\!\text{ }\widehat{\text{k}}\text{=}\widehat{\text{i}}\text{,}\widehat{\text{k}}\text{ }\!\!\times\!\!\text{ }\widehat{\text{i}}\text{=}\widehat{\text{j}}\]

  • If \[\overrightarrow{\text{a}}\text{=}{{\text{a}}_{\text{1}}}\widehat{\text{i}}\text{+}{{\text{a}}_{\text{2}}}\widehat{\text{j}}\text{+}{{\text{a}}_{\text{3}}}\widehat{\text{k}}\] and \[\overrightarrow{\text{b}}\text{=}{{\text{b}}_{\text{1}}}\widehat{\text{i}}\text{+}{{\text{b}}_{\text{2}}}\widehat{\text{j}}\text{+}{{\text{b}}_{\text{3}}}\widehat{\text{k}}\] then 

\[\overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}}\text{=}\left| \begin{matrix} \widehat{\text{i}} & \widehat{\text{j}} & \widehat{\text{k}}  \\   {{\text{a}}_{\text{1}}} & {{\text{a}}_{\text{2}}} & \text{a}  \\   {{\text{b}}_{\text{1}}} & {{\text{b}}_{\text{2}}} & \text{b}  \\ \end{matrix} \right|\].

  • Geometrically, we can define \[\left| \overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}} \right|\text{=}\] area of the parallelogram whose two adjacent sides are represented by \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\].

  • Properties of Vector Product

  1. \[\overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}}\ne \overrightarrow{\text{b}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{a}}\] (Not Commutative)

  2. \[\left( \text{m}\overrightarrow{\text{a}} \right)\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}}\text{=}\overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\left( \text{m}\overrightarrow{\text{b}} \right)\text{=m}\left( \overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}} \right)\] (Associative) where \[\text{m}\] is scalar.

  3. \[\overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\left( \overrightarrow{\text{b}}\text{+}\overrightarrow{\text{c}} \right)\text{=}\left( \overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}} \right)\text{+}\left( \overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{c}} \right)\] (Distributive)

  • Unit vector perpendicular to the plane of \[\overrightarrow{\text{a}}\] and \[\overrightarrow{\text{b}}\] is \[\widehat{\text{n}}\text{= }\!\!\pm\!\!\text{ }\dfrac{\overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}}}{\left| \overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}} \right|}\]

  • A vector of magnitude \[\text{ }\!\!'\!\!\text{ r }\!\!'\!\!\text{ }\] and perpendicular to the plane of \[\overrightarrow{\text{a}}\text{ and }\overrightarrow{\text{b}}\] is  \[\text{ }\!\!\pm\!\!\text{ }\dfrac{\text{r}\left( \overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}} \right)}{\left| \overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}} \right|}\]

  • If \[\overrightarrow{\text{a}}\text{,}\overrightarrow{\text{b}}\text{ and }\overrightarrow{\text{c}}\] are the position vectors of vertices \[\text{A,B and C}\] of a triangle, then the vector area of triangle is given by \[\text{ABC=}\dfrac{\text{1}}{\text{2}}\left[ \overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}}+\overrightarrow{\text{b}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{c}}+\overrightarrow{\text{c}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{a}} \right]\] The points \[\text{A,B and C}\] are collinear if \[\overrightarrow{\text{a}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{b}}\text{+}\overrightarrow{\text{b}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{c}}\text{+}\overrightarrow{\text{c}}\text{ }\!\!\times\!\!\text{ }\overrightarrow{\text{a}}\text{=0}\].

  • Area of quadrilateral whose diagonal vectors are \[\overrightarrow{{{\text{d}}_{1}}}\text{ and }\overrightarrow{{{\text{d}}_{2}}}\] is given by \[\dfrac{1}{2}\left| \overrightarrow{{{\text{d}}_{1}}}\times \overrightarrow{{{\text{d}}_{2}}} \right|\].


Scalar Triple Product

  • The scalar triple product of three vectors \[\overrightarrow{\text{a}}\text{,}\overrightarrow{\text{b}}\text{ and }\overrightarrow{\text{c}}\] is defined as \[\overrightarrow{\text{a}}.\left( \overrightarrow{\text{b}}\times \overrightarrow{\text{c}} \right)\] and can be represented as \[\left[ \overrightarrow{\text{a}}\text{ }\overrightarrow{\text{b}}\text{ }\overrightarrow{\text{c}} \right]\]. It is also referred to as box product.

  • Geometrically, it represents the volume of the parallelepiped whose three coterminous edges are represented by \[\overrightarrow{\text{a}}\text{,}\overrightarrow{\text{b}}\text{ and }\overrightarrow{\text{c}}\] . So \[\text{V=}\left[ \overrightarrow{\text{a}}\text{ }\overrightarrow{\text{b}}\text{ }\overrightarrow{\text{c}} \right]\].

  • Scalar triple product is cyclic, i.e. the order of vectors can be interchanged in a cyclic manner as shown below, \[\overrightarrow{\text{a}}\cdot \left( \overrightarrow{\text{b}}\times \overrightarrow{\text{c}} \right)=\left( \overrightarrow{\text{a}}\times \overrightarrow{\text{b}} \right)\cdot \overrightarrow{\text{c}}\] or \[\left[ \overrightarrow{\text{a}}\text{ }\overrightarrow{\text{b}}\text{ }\overrightarrow{\text{c}} \right]=\left[ \overrightarrow{\text{b}}\text{ }\overrightarrow{\text{c}}\text{ }\overrightarrow{\text{a}} \right]=\left[ \overrightarrow{\text{c}}\text{ }\overrightarrow{\text{a}}\text{ }\overrightarrow{\text{b}} \right]\] \[\overrightarrow{\text{a}}\cdot \left( \overrightarrow{\text{b}}\times \overrightarrow{\text{c}} \right)=\text{-}\overrightarrow{\text{a}}\cdot \left( \overrightarrow{\text{c}}\times \overrightarrow{\text{b}} \right)\] or \[\left[ \overrightarrow{\text{a}}\text{ }\overrightarrow{\text{b}}\text{ }\overrightarrow{\text{c}} \right]=\text{-}\left[ \overrightarrow{\text{a}}\text{ }\overrightarrow{\text{c}}\text{ }\overrightarrow{\text{b}} \right]\]

  • If \[\overrightarrow{\text{a}}\text{=}{{\text{a}}_{\text{1}}}\widehat{\text{i}}\text{+}{{\text{a}}_{\text{2}}}\widehat{\text{j}}\text{+}{{\text{a}}_{\text{3}}}\widehat{\text{k}}\]; \[\overrightarrow{\text{b}}\text{=}{{\text{b}}_{\text{1}}}\widehat{\text{i}}\text{+}{{\text{b}}_{\text{2}}}\widehat{\text{j}}\text{+}{{\text{b}}_{\text{3}}}\widehat{\text{k}}\] and \[\overrightarrow{\text{c}}\text{=}{{\text{c}}_{\text{1}}}\widehat{\text{i}}\text{+}{{\text{c}}_{\text{2}}}\widehat{\text{j}}\text{+}{{\text{c}}_{\text{3}}}\widehat{\text{k}}\] then $\overrightarrow{\text{a}} \times \overrightarrow{\text{b}} \times \overrightarrow{\text{c}}$ $= \left|\begin{array}{ccc} a_{1} & a_{2} & a \\ b_{1} & b_{2} & b \\ c_{1} & c_{2} & c \end{array}\right|$

  • Scalar product of three vectors, two of which are equal or parallel is \[\text{0}\].

  • Vectors \[\overrightarrow{\text{a}},\overrightarrow{\text{b}},\overrightarrow{\text{c}}\] are coplanar if \[\left[ \overrightarrow{\text{a}}\text{ }\overrightarrow{\text{b}}\text{ }\overrightarrow{\text{c}} \right]\text{=0}\].


Vector Algebra Revision notes Class 12 Chapter 10

Download Revision Notes Class 12 Maths Chapter 10 PDF

Revision Notes Class 12 Chapter 10 Vector algebra is designed and prepared by the best teachers at Vedantu. All the important concepts are covered with a detailed explanation to help students understand concepts better. These revision notes will play a crucial role in your preparation for all exams conducted by the CBSE, including the JEE.

Class 12 Maths Chapter 9 Vector Algebra notes are prepared by subject expert teachers at Vedantu keeping in mind the latest syllabus and guidelines issued by the CBSE board. You can download Vector Algebra Class 12 Notes Maths Chapter 10 free pdf through the link provided below.

 

Benefits of Class 12 Notes Maths Vector Algebra

  • Class 12 Maths Chapter 10 Revision Notes provides brief information along with the relevant formulas of all the topics covered in Chapter 10. This enables students to quickly and easily revise the important topics of the chapter before the exams.

  • Vedantu revision notes Class 12 Maths Chapter 10 can be easily downloaded in a PDF format free of cost.

  • Class 12 Revision Notes Maths Chapter 10 pdf will be very helpful for the students for the last-minute revision of the chapter.

Revision Notes Class 12 Maths Chapter 10 PDF covers the following topics with important definitions and formulas.

 

What are Vectors?

A vector quantity is represented by an arrow. This arrow is called the ‘vector’. The length of the arrow represents the magnitude and the head of the arrow represents direction.

(Image will be uploaded soon)

 

Types of Vectors

There are ten different types of vectors frequently used in maths. The 10 types of vectors are:

  1. Zero vector

  2. Unit Vector

  3. Position Vector

  4. Co-initial Vector

  5. Like and Unlike Vectors

  6. Coplanar Vector

  7. Collinear Vector

  8. Equal Vector

  9. Displacement Vector

  10. Negative of a Vector

Let us understand each type of Vector Algebra in brief.

 

Zero Vector or Null Vector

A vector having the magnitude zero and the starting point and the endpoint of the vector is the same as a zero Vector. The zero vector doesn’t have any specific direction.

 

Unit Vector

A vector is a unit vector when the magnitude of the vector is 1 unit in length. Suppose if a  is a vector having a magnitude a then the unit vector is denoted by a^ in the direction of the vector and it has the magnitude equal to 1.

 

Position Vector

Any point X has taken in the plane simply denotes the position is said to be a position vector. Let us consider O is taken as reference origin and A is an arbitrary point in the plane then the vector is called the position vector of the point.

 

Co-initial Vectors

A vector is co-initial vectors when two or more vectors share the same starting point. For example, Vectors PQ and PR are called co-initial vectors because they have the same starting point P.

 

Like and Unlike Vectors

The vectors with the same directions are said to be like vectors whereas vectors with opposite directions are said to be unlike vectors.

 

Coplanar Vectors

When three vectors lie in the same plane, they are known as coplanar vectors.

 

Collinear Vectors

Vectors which lie in the same line with respect to their magnitude and direction are known to be collinear vectors.

 

Equal Vectors

Two vectors having both direction and magnitude equal are said to be equal vectors even if they have different initial points.

 

Displacement Vector

If a point is displaced from the position A to B then AB represents a displacement vector.

 

Negative of a Vector

Two Vectors with the same magnitude but having opposite directions are said to be negative vectors of each other.

Consider two vectors x and y, such that they have the same magnitude but opposite in direction then these vectors can be written as:

x = – y


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FAQs on Vector Class 12 Notes CBSE Maths Chapter 10 [Free PDF Download]

1. Provide Some Mathematical Tips on How to Practice and Prepare for CBSE Class 12 Vector Algebra.

Any process that aims to attain knowledge needs initiation by students themself, and not by a teacher or textbook. Keeping this view in mind, Vedantu has highlighted some motivational tips to help students go deep in their learning process; these are:

  1. Get a holistic view.

  2. Imagine and wonder.

  3. Understand the concepts.

  4. Practice and practice.

  5. If you find it tough, then someone else would too.

  6. Dig deeper in passion.

2. Mention Some of the Best Study Techniques Listed Out by the Vedantu Experts for Students Appearing in Class 12 Maths.

Some of the best study tips for students appearing in Class 12 exams are:

  • Eliminate distractions by silencing your mobile, or other background noise like TV, radio, etc.

  • Get enough sleep before you take hours of study.

  • Switch your environment as a change in the scenario will improve both your concentration level and memory.

  • Stick with an environment that works.

  • Listen to calm music as it can help to pay attention to the task.

  • Snack on smart food.

3. Should I practice all the Questions in Chapter 10 of Maths of Class 12th?

Maths is a subject learned better through practice. The number of formulas in chapter 10 of Class 12th Mathematics makes it a little complex for the students. Hence, it is better to practice by solving all the questions given in the book to learn the formulae and their application. It is also a great method to learn the formulas because the regular practice would help you store them in your long-term memory.

4. How will revision notes for Chapter 10 of Maths of Class 12th help for boards?

Revision Notes for Chapter 10 of Maths of Class 12th contain brief information as well as essential formulas for all of the subjects discussed in the chapter. This allows students to swiftly and simply review the key fundamentals in the chapter before the examinations. Time is a luxury during the examinations so it is better to focus your energy on revising the precisely created notes than flick through all the chapters intensively.

5. What are vectors according to Chapter 10 of Maths of Class 12th?

Vector quantity is a physical quantity with magnitude and direction. It denotes the movement of an object from one point to another. It is represented by an arrow. The magnitude is represented by the length of the arrow, and the direction is shown by the head of the arrow. It is an easy and scoring chapter for the students. If well prepared the students can gain full marks in the questions from this particular chapter.

6. What are like and unlike vectors according to Chapter 10 of Maths of Class 12th?

Like and Unlike vectors are two categories of vectors. In like vectors the direction is the same whereas unlike vectors are where the direction (with respect to each other) is different. There are other types of vectors too as explained in Revision Notes for Chapter 10 of Maths of Class 12th: Vector Algebra. The students get knowledge about all these types of vectors and solve questions based on that knowledge. The notes and solutions provided by Vedantu are free of cost. They Are also available on Vedantu Mobile app.

7. What is the negative of a vector according to Chapter 10 of Maths of Class 12th?

Two vectors are recognized as negatives of each other when their magnitude is the same but they have opposite directions. It is important to take note of the magnitude and the direction of a vector to distinguish between various types of vectors. The arrow usually depicts the direction of a vector. The students can gain more insight into the chapter using Revision Notes for Chapter 10 of Maths of Class 12th that are available on Vedantu.