NCERT Solutions for Exercise 10.3 Class 12 Maths Chapter 10 - Vector Algebra

# NCERT Solutions for Exercise 10.3 Class 12 Maths Chapter 10 - Vector Algebra

Edited By Ramraj Saini | Updated on Dec 04, 2023 08:54 AM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 10 Exercise 10.3

NCERT Solutions for Exercise 10.3 Class 12 Maths Chapter 10 Vector Algebra are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 10.3 Class 12 Maths chapter 10 explains the questions related to dot products of vectors. As far as the unit vector algebra is concerned dot product is an important topic. NCERT syllabus Exercise 10.3 Class 12 Math familiarise the students with the concept of dot products. NCERT solutions for Class 12 Maths chapter 10 exercise 10.3 are framed by maths expert and is in accordance with the CBSE syllabus. Students can use Class 12 Maths chapter 10 exercise 10.3 for the preparations of CBSE Class 12 Board Exams. Along with NCERT practice CBSE Previous Year Class 12 Maths Questions to get a good score in board exam.

12th class Maths exercise 10.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Vector Algebra Class 12 Chapter 10 Exercise: 10.3

Given

$\left | \vec a \right |=\sqrt{3}$

$\left | \vec b \right |=2$

$\vec a . \vec b = \sqrt 6$

As we know

$\vec a . \vec b = \left | \vec a \right |\left | \vec b \right |cos\theta$

where $\theta$ is the angle between two vectors

So,

$cos\theta =\frac{\vec a.\vec b}{\left | \vec a \right |\left | \vec b \right |}=\frac{\sqrt{6}}{\sqrt{3}*2}=\frac{1}{\sqrt{2}}$

$\theta=\frac{\pi}{4}$

Hence the angle between the vectors is $\frac{\pi}{4}$ .

Given two vectors

$\vec a=\hat i - 2 \hat j + 3 \hat k \: \:and \: \: \vec b=3 \hat i - 2 \hat j + \hat k$

Now As we know,

The angle between two vectors $\vec a$ and $\vec b$ is given by

$\theta=cos^{-1}\left ( \frac{\vec a.\vec b}{\left | \vec a \right |\left | \vec b \right |}\right )$

Hence the angle between $\vec a=\hat i - 2 \hat j + 3 \hat k \: \:and \: \: \vec b=3 \hat i - 2 \hat j + \hat k$

$\theta=cos^{-1}\left ( \frac{(\hat i-2\hat j+3\hat k).(3\hat i-2\hat j+\hat k)}{\left | \hat i-2\hat j+3\hat k \right |\left |3\hat i-2\hat j+\hat k \right |}\right )$

$\theta=cos^{-1}\left ( \frac{3+4+3}{\sqrt{1^2+(-2)^2+3^3}\sqrt{3^2+(-2)^2+1^2}} \right )$

$\theta=cos^{-1}\frac{10}{14}$

$\theta=cos^{-1}\frac{5}{7}$

Let

$\vec a=\hat i - \hat j$

$\vec b=\hat i + \hat j$

Projection of vector $\vec a$ on $\vec b$

$\frac{\vec a.\vec b}{\left | \vec b \right |}=\frac{(\hat i-\hat j)(\hat i+\hat j)}{\left |\hat i+\hat j \right |}=\frac{1-1}{\sqrt{2}}=0$

Hence, Projection of vector $\vec a$ on $\vec b$ is 0.

Let

$\vec a =\hat i + 3 \hat j + 7 \hat k$

$\vec b=7\hat i - \hat j + 8 \hat k$

The projection of $\vec a$ on $\vec b$ is

$\frac{\vec a.\vec b}{\left | \vec b \right |}=\frac{(\hat i+3\hat j+7\hat k)(7\hat i-\hat j+8\hat k)}{\left | 7\hat i-\hat j+8\hat k \right |}=\frac{7-3+56}{\sqrt{7^2+(-1)^2+8^2}}=\frac{60}{\sqrt{114}}$

Hence, projection of vector $\vec a$ on $\vec b$ is

$\frac{60}{\sqrt{114}}$

Given

$\\\vec a=\frac{1}{7}( 2 \hat i + 3 \hat j + 6 \hat k ), \\\ \vec b =\frac{1}{7}( 3 \hat i- 6 \hat j + 2 \hat k ),\\\vec c = \frac{1}{7}( 6\hat i + 2 \hat j -3\hat k )$

Now magnitude of $\vec a,\vec b \:and\: \vec c$

$\left | \vec a \right |=\frac{1}{7} \sqrt{2^2+3^2+6^2}=\frac{\sqrt{49}}{7}=1$

$\left | \vec b \right |=\frac{1}{7} \sqrt{3^2+(-6)^2+2^2}=\frac{\sqrt{49}}{7}=1$

$\left | \vec c \right |=\frac{1}{7} \sqrt{6^2+2^2+(-3)^2}=\frac{\sqrt{49}}{7}=1$

Hence, they all are unit vectors.

Now,

$\vec a.\vec b=\frac{1}{7}(2\hat i+3\hat j+6\hat k)\frac{1}{7}(3\hat i-6\hat j+2\hat k)=\frac{1}{49}(6-18+12)=0$

$\vec b.\vec c=\frac{1}{7}(3\hat i-6\hat j+2\hat k)\frac{1}{7}(6\hat i+2\hat j-3\hat k)=\frac{1}{49}(18-12-6)=0$

$\vec c.\vec a=\frac{1}{7}(6\hat i+2\hat j-3\hat k)\frac{1}{7}(2\hat i+3\hat j-6\hat k)=\frac{1}{49}(12+6-18)=0$

Hence all three are mutually perpendicular to each other.

Given in the question

$( \vec a + \vec b ). ( \vec a - \vec b )=8$

$\left | \vec a \right |^2-\left | \vec b \right |^2=8$

Since $|\vec a |\: \:= 8 \: \:|\vec b |$

$\left | \vec {8b} \right |^2-\left | \vec b \right |^2=8$

$\left | \vec {63b} \right |^2=8$

$\left | \vec {b} \right |^2=\frac{8}{63}$

$\left | \vec {b} \right |=\sqrt{\frac{8}{63}}$

So, answer of the question is

$\left | \vec {a} \right |=8\left | \vec {b} \right |=8\sqrt{\frac{8}{63}}$

To evaluate the product $( 3\vec a - 5 \vec b ). ( 2 \vec a + 7 \vec b )$

$( 3\vec a - 5 \vec b ). ( 2 \vec a + 7 \vec b )=6\vec a.\vec a+21\vec a.\vec b-10\vec b.\vec a-35\vec b.\vec b$

$=6\vec a.^2+11\vec a.\vec b-35\vec b^2$

$=6\left | \vec a \right |^2+11\vec a.\vec b-35\left | \vec b \right |^2$

Given two vectors $\vec a \: \: and \: \: \vec b$

$\left | \vec a \right |=\left | \vec b\right |$

$\vec a.\vec b=\frac{1}{2}$

Now Angle between $\vec a \: \: and \: \: \vec b$

$\theta=60^0$

Now As we know that

$\vec a.\vec b=\left | \vec a \right |\left | \vec b \right |cos\theta$

$\frac{1}{2}=\left | \vec a \right |\left | \vec a \right |cos60^0$

$\left | a \right |^2=1$

Hence, the magnitude of two vectors $\vec a \: \: and \: \: \vec b$

$\left | a \right |=\left | b \right |=1$

Given in the question that

$( \vec x -\vec a ) . ( \vec x + \vec a ) = 12$

And we need to find $\left | \vec x \right |$

$\left | \vec x \right |^2-\left | \vec a \right |^2 = 12$

$\left | \vec x \right |^2-1 = 12$

$\left | \vec x \right |^2 = 13$

$\left | \vec x \right | = \sqrt{13}$

So the value of $\left | \vec x \right |$ is $\dpi{100} \sqrt{13}$

Given in the question is

$\vec a = 2 \hat i + 2 \hat j + 3 \hat k , \vec b = - \hat i + 2 \hat j + \hat k \: \: and \: \: \vec c = 3 \hat i + \hat j$

and $\vec a + \lambda \vec b$ is perpendicular to $\vec c$

and we need to find the value of $\lambda$ ,

so the value of $\vec a + \lambda \vec b$ -

$\vec a + \lambda \vec b=2\hat i +2\hat j +3\hat k+\lambda (-\hat i+2\hat j+\hat k)$

$\vec a + \lambda \vec b=(2-\lambda)\hat i +(2+2\lambda)\hat j +(3+\lambda)\hat k$

As $\vec a + \lambda \vec b$ is perpendicular to $\vec c$

$(\vec a + \lambda \vec b).\vec c=0$

$((2-\lambda)\hat i +(2+2\lambda)\hat j +(3+\lambda)\hat k)(3\hat i+\hat j)=0$

$3(2-\lambda)+2+2\lambda=0$

$6-3\lambda+2+2\lambda=0$

$\lambda=8$

the value of $\lambda=8$ ,

Given in the question that -

$\vec a \: \: \: and \: \: \vec b$ are two non-zero vectors

According to the question

$\left ( |\vec a | \vec b + |\vec b | \vec a\right )\left (|\vec a | \vec b - |\vec b | \vec a \right )$

$=|\vec a |^2 |\vec b|^2 - |\vec b |^2 |\vec a|^2+|\vec b||\vec a|\vec a.\vec b-|\vec a||\vec b|\vec b.\vec a=0$

Hence $|\vec a | \vec b + |\vec b | \vec a$ is perpendicular to $|\vec a | \vec b - |\vec b | \vec a$ .

Given in the question

$\\\vec a . \vec a = 0 \\|\vec a|^2=0$

$\\|\vec a|=0$

Therefore $\vec a$ is a zero vector. Hence any vector $\vec b$ will satisfy $\vec a . \vec b = 0$

Given in the question

$\vec a , \vec b , \vec c$ are unit vectors $\Rightarrow |\vec a|=|\vec b|=|\vec c|=1$

and $\vec a + \vec b + \vec c = \vec 0$

and we need to find the value of $\vec a . \vec b + \vec b. \vec c + \vec c . \vec a$

$(\vec a + \vec b + \vec c)^2 = \vec 0$

$\vec a^2 + \vec b^2 + \vec c ^2+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$|\vec a|^2 + |\vec b|^2 + |\vec c |^2+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$1+1+1+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$\vec a . \vec b + \vec b. \vec c + \vec c . \vec a=\frac{-3}{2}$

Answer- the value of $\vec a . \vec b + \vec b. \vec c + \vec c . \vec a$ is $\dpi{80} \frac{-3}{2}$

Let

$\vec a=\hat i-2\hat j +3\hat k$

$\vec b=5\hat i+4\hat j +1\hat k$

we see that

$\vec a.\vec b=(\hat i-2\hat j +3\hat k)(5\hat i+4\hat j +1\hat k)=5-8+3=0$

we now observe that

$|\vec a|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}$

$|\vec b|=\sqrt{5^2+4^2+1^2}=\sqrt{42}$

Hence here converse of the given statement is not true.

Given points,

A=(1, 2, 3),

B=(–1, 0, 0),

C=(0, 1, 2),

As need to find Angle between $\overline{BA}\: \: and\: \: \overline{BC} ]$

$\vec {BA}=(1-(-1))\hat i+(2-0)\hat j+(3-0)\hat k=2\hat i+2\hat j+3\hat k$

$\vec {BC}=(0-(-1))\hat i+(1-0)\hat j+(2-0)\hat k=\hat i+\hat j+2\hat k$

Hence angle between them ;

$\theta=cos^{-1}(\frac{\vec {BA}.\vec {BC}}{\left | \vec {BA} \right |\left | \vec {BC} \right |})$

$\theta=cos^{-1}\frac{2+2+6}{\sqrt{17}\sqrt{6}}$

$\theta=cos^{-1}\frac{10}{\sqrt{102}}$

Answer - Angle between the vectors $\overline{BA}\: \: and\: \: \overline{BC}$ is $\dpi{80} \theta=cos^{-1}\frac{10}{\sqrt{102}}$

Given in the question

A=(1, 2, 7), B=(2, 6, 3) and C(3, 10, –1)

To show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear

$\vec {AB}=(2-1)\hat i+(6-2)\hat j+(3-7)\hat k$

$\vec {AB}=\hat i+4\hat j-4\hat k$

$\vec {BC}=(3-2)\hat i+(10-6)\hat j+(-1-3)\hat k$

$\vec {BC}=\hat i+4\hat j-4\hat k$

$\vec {AC}=(3-1)\hat i+(10-2)\hat j+(-1-7)\hat k$

$\vec {AC}=2\hat i+8\hat j-8\hat k$

$|\vec {AB}|=\sqrt{1^2+4^2+(-4)^2}=\sqrt{33}$

$|\vec {BC}|=\sqrt{1^2+4^2+(-4)^2}=\sqrt{33}$

$|\vec {AC}|=\sqrt{2^2+8^2+(-8)^2}=2\sqrt{33}$

As we see that

$|\vec {AC}|=|\vec {AB}|+|\vec {BC}|$

Hence point A, B , and C are colinear.

Given the position vector of A, B , and C are

$2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k$

To show that the vectors $2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k$ form the vertices of a right angled triangle

$\vec {AB}=(1-2)\hat i + (-3-(-1))\hat j+(-5-1)\hat k=-1\hat i -2\hat j-6\hat k$

$\vec {BC}=(3-1)\hat i + (-4-(-3))\hat j+(-4-(-5))\hat k=-2\hat i -\hat j+\hat k$

$\vec {AC}=(3-2)\hat i + (-4-(-1))\hat j+(-4-(1))\hat k=\hat i -3\hat j-5\hat k$

$|\vec {AB}|=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{41}$

$|\vec {BC}|=\sqrt{(-2)^2+(-1)^2+(1)^2}=\sqrt{6}$

$|\vec {AC}|=\sqrt{(1)^2+(-3)^2+(-5)^2}=\sqrt{35}$

Here we see that

$|\vec {AC}|^2+|\vec {BC}|^2=|\vec {AB}|^2$

Hence A,B, and C are the vertices of a right angle triangle.

Given $\vec a$ is a nonzero vector of magnitude ‘a’ and $\lambda$ a nonzero scalar

$\lambda \vec a$ is a unit vector when

$|\lambda \vec a|=1$

$|\lambda|| \vec a|=1$

$| \vec a|=\frac{1}{|\lambda|}$

Hence the correct option is D.

## More About NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.3

There are a total of 18 questions from exercise 10.3 Class 12 Maths. These questions mainly are based on the concepts of dot products. Dot products are used to identify orthogonal vectors and the projections of a vector. Question number 4 of Class 12 Maths chapter 10 exercise 10.3 gives an example of the projection of a vector on another.

Also Read | Vector Algebra Class 10 Chapter 10 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.3

• Similar questions as in Class 12th Maths chapter 10 exercise 10.3 can be expected for the CBSE Class 12 Maths exam. The questions may not be exactly the same.
• Problems based on dot products are also asked in various competitive exams like JEE Main, KEAM etc.
• Problems discussed in exercise 10.3 Class 12 Maths are useful not only for Class 12 Maths but also for Class 11 and 12 Physics also.
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## Key Features Of NCERT Solutions for Exercise 10.3 Class 12 Maths Chapter 10

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 10.3 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 10.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 10.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 10.3 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 10.3 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 10.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
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## Subject Wise NCERT Exemplar Solutions

1. Is it possible to have two vectors whose magnitude is non zero but their dot product is zero?

Yes. For two perpendicular vectors, the dot product is zero.

2. What are the main topics covered in the chapter vector algebra?

The main three topics are the addition of vectors, the dot product of vectors and the cross product of vectors.

3. Why work is a scalar quantity even though force and displacement are vectors?

The work done is the dot product of force and displacement. The dot product of two vectors is a scalar (real number).

4. What is the quantity obtained from the dot product of force and velocity?

The dot product of force and velocity gives power.

5. Why i.k=0?

i.k=0 as the angle between them is 90 degrees

6. Can dot product be a negative number?

Yes, the dot product of two vectors can be either positive, negative or zero based on the angle between them.

7. How many exercises are solved in NCERT Class 12 chapter vector algebra?

There are a total of 5 exercises including miscellaneous.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

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In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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