NCERT Solutions for Exercise 10.2 Class 12 Maths Chapter 10

NCERT Solutions for Exercise 10.2 Class 12 Maths Chapter 10 - Vector Algebra

Edited By Ramraj Saini | Updated on Dec 04, 2023 08:51 AM IST | #CBSE Class 12th

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NCERT Solutions For Class 12 Maths Chapter 10 Exercise 10.2

NCERT Solutions for Exercise 10.2 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 10.2 Class 12 Maths chapter 10 introduces a few more concepts of vectors. The questions of exercise 10.2 Class 12 Maths deals with the concepts like addition of vectors, magnitude of vectors, unit vectors etc. Also in NCERT solutions for Class 12 Maths chapter 10 exercise 10.2 questions related to collinear vectors and components of vectors. As the Class 12 Maths syllabus is concerned, it is important to practice Class 12 Maths chapter 10 exercise 10.2. Questions related to the Class 12th Maths chapter 10 exercise 10.2 can be expected for CBSE Class 12 Board Exams as well as competitive exams like JEE Mains.

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The chapter Vector Algebra exercises will be helpful for both Mathematics and Physics. To solve more problems of NCERT Class 12 Maths, students can also make use of NCERT Exemplar Solutions for Class 12 Maths. 12th class Maths exercise 10.2 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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Access NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2

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Vector Algebra-Exercise: 10.2

Question:1 Compute the magnitude of the following vectors:

(1) $\vec a = \hat i + \hat j + \hat k$

Answer:

Here

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

Magnitude of $\vec a$

$\vec a=\sqrt{1^2+1^2+1^2}=\sqrt{3}$

Question:1 Compute the magnitude of the following vectors:

(2) $\vec b = 2 \hat i - 7 \hat j - 3 \hat k$

Answer:

Here,

$\vec b = 2 \hat i - 7 \hat j - 3 \hat k$

Magnitude of $\vec b$

$\left | \vec b \right |=\sqrt{2^2+(-7)^2+(-3)^2}=\sqrt{62}$

Question:1 Compute the magnitude of the following vectors:

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

Answer:

Here,

$\vec c = \frac{1}{\sqrt 3 }\hat i + \frac{1}{\sqrt 3 }\hat j -\frac{1}{\sqrt 3 }\hat k$

Magnitude of $\vec c$

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

Question:2 Write two different vectors having same magnitude

Answer:

Two different Vectors having the same magnitude are

$\vec a= 3\hat i+6\hat j+9\hat k$

$\vec b= 9\hat i+6\hat j+3\hat k$

The magnitude of both vector

$\left | \vec a \right |=\left | \vec b \right | = \sqrt{9^2+6^2+3^2}=\sqrt{126}$

Question:3 Write two different vectors having same direction.

Answer:

Two different vectors having the same direction are:

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

$\vec b=2\hat i+4\hat j+6\hat k$

Question:4 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.

Answer:

$2 \hat i + 3 \hat j$ will be equal to $x \hat i + y \hat j$ when their corresponding components are equal.

Hence when,

$x=2$ and

$y=3$

Question:5 Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).

Answer:

Let point P = (2, 1) and Q = (– 5, 7).

Now,

$\vec {PQ}=(-5-2)\hat i+(7-1)\hat j=-7\hat i +6\hat j$

Hence scalar components are (-7,6) and the vector is $-7\hat i +6\hat j$

Question:6 Find the sum of the vectors $\vec a = \hat i - 2 \hat j + \hat k , \vec b = -2 \hat i + 4 \hat j + 5 \hat k \: \: and\: \: \: \vec c = \hat i - 6 \hat j - 7 \hat k$

Answer:

Given,

$\\ \vec a = \hat i - 2 \hat j + \hat k ,\\ \vec b = -2 \hat i + 4 \hat j + 5 \hat k \: \: and\: \: \: \\\vec c = \hat i - 6 \hat j - 7 \hat k$

Now, The sum of the vectors:

$\vec a +\vec b+\vec c = \hat i - 2 \hat j + \hat k + -2 \hat i + 4 \hat j + 5 \hat k + \hat i - 6 \hat j - 7 \hat k$

$\vec a +\vec b+\vec c = (1-2+1)\hat i +(-2+4-6) \hat j + (1+5-7)\hat k$

$\vec a +\vec b+\vec c =-4\hat j-\hat k$

Question:7 Find the unit vector in the direction of the vector $\vec a = \hat i + \hat j + 2 \hat k$

Answer:

Given

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

Magnitude of $\vec a$

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

A unit vector in the direction of $\vec a$

$\vec u = \frac{\hat i}{\left | a \right |} + \frac{\hat j}{\left | a \right |} +\frac{2\hat k}{\left | a \right |} =\frac{\hat i}{\sqrt{6}}+\frac{\hat j}{\sqrt{6}}+\frac{2\hat k}{\sqrt{6}}$

Question:8 Find the unit vector in the direction of vector $\vec { PQ}$ , where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.

Answer:

Given P = (1, 2, 3) and Q = (4, 5, 6)

A vector in direction of PQ

$\vec {PQ}=(4-1)\hat i+(5-2)\hat j +(6-3)\hat k$

$\vec {PQ}=3\hat i+3\hat j +3\hat k$

Magnitude of PQ

$\left | \vec {PQ} \right |=\sqrt{3^2+3^2+3^2}=3\sqrt{3}$

Now, unit vector in direction of PQ

$\hat u=\frac{\vec {PQ}}{\left | \vec {PQ} \right |}=\frac{3\hat i+3\hat j+3\hat k}{3\sqrt{3}}$

$\hat u=\frac{\hat i}{\sqrt{3}}+\frac{\hat j}{\sqrt{3}}+\frac{\hat k}{\sqrt{3}}$

Question:9 For given vectors, $\vec a = 2 \hat i - \hat j + 2 \hat k$ and $\vec b = - \hat i + \hat j - \hat k$ , find the unit vector in the direction of the vector $\vec a + \vec b$ .

Answer:

Given

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

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

Now,

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

$\vec a + \vec b=\hat i+\hat k$

Now a unit vector in the direction of $\vec a + \vec b$

$\vec u= \frac{\vec a + \vec b}{\left |\vec a + \vec b \right |}=\frac{\hat i+\hat j}{\sqrt{1^2+1^2}}$

$\vec u= \frac{\hat i}{\sqrt{2}}+\frac{\hat j}{\sqrt{2}}$

Question:10 Find a vector in the direction of vector $5 \hat i - \hat j + 2 \hat k$ which has magnitude 8 units.

Answer:

Given a vector

$\vec a=5 \hat i - \hat j + 2 \hat k$

the unit vector in the direction of $5 \hat i - \hat j + 2 \hat k$

$\vec u=\frac{5\hat i - \hat j + 2 \hat k}{\sqrt{5^2+(-1)^2+2^2}}=\frac{5\hat i}{\sqrt{30}}-\frac{\hat j}{\sqrt{30}}+\frac{2\hat k}{\sqrt{30}}$

A vector in direction of $5 \hat i - \hat j + 2 \hat k$ and whose magnitude is 8 =

$8\vec u=\frac{40\hat i}{\sqrt{30}}-\frac{8\hat j}{\sqrt{30}}+\frac{16\hat k}{\sqrt{30}}$

Question:11 Show that the vectors $2 \hat i -3 \hat j + 4 \hat k$ and $- 4 \hat i + 6 \hat j - 8 \hat k$ are collinear.

Answer:

Let

$\vec a =2 \hat i -3 \hat j + 4 \hat k$

$\vec b=- 4 \hat i + 6 \hat j - 8 \hat k$

It can be seen that

$\vec b=- 4 \hat i + 6 \hat j - 8 \hat k=-2(2 \hat i -3 \hat j + 4 \hat k)=-2\vec a$

Hence here $\vec b=-2\vec a$

As we know

Whenever we have $\vec b=\lambda \vec a$ , the vector $\vec a$ and $\vec b$ will be colinear.

Here $\lambda =-2$

Hence vectors $2 \hat i -3 \hat j + 4 \hat k$ and $- 4 \hat i + 6 \hat j - 8 \hat k$ are collinear.

Question:12 Find the direction cosines of the vector $\hat i + 2 \hat j + 3 \hat k$

Answer:

Let

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

$\left |\vec a \right |=\sqrt{1^2+2^2+3^2}=\sqrt{14}$

Hence direction cosine of $\vec a$ are

$\left ( \frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}} ,\frac{3}{\sqrt{14}}\right )$

Question:13 Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, –2, 1), directed from A to B.

Answer:

Given

point A=(1, 2, –3)

point B=(–1, –2, 1)

Vector joining A and B Directed from A to B

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

$\vec {AB}=-2\hat i +-4\hat j+4\hat k$

$\left | \vec {AB} \right |=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{36}=6$

Hence Direction cosines of vector AB are

$\left ( \frac{-2}{6},\frac{-4}{6},\frac{4}{6} \right )=\left ( \frac{-1}{3},\frac{-2}{3},\frac{2}{3} \right )$

Question:14 Show that the vector $\hat i + \hat j + \hat k$ is equally inclined to the axes OX, OY and OZ.

Answer:

Let

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

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

Hence direction cosines of this vectors is

$\left ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right )$

Let $\alpha$ , $\beta$ and $\gamma$ be the angle made by x-axis, y-axis and z- axis respectively

Now as we know,

$cos\alpha=\frac{1}{\sqrt{3}}$ , $cos\beta=\frac{1}{\sqrt{3}}$ $and\:cos\gamma=\frac{1}{\sqrt{3}}$

Hence Given vector is equally inclined to axis OX,OY and OZ.

Question:15 (1) Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $i + 2 j - k$ and $- i + j + k$ respectively, in the ratio 2 : 1 internally

Answer:

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n internally:

$\vec r=\frac{m\vec b+n\vec a}{m+n}$

Here

position vector os P = $\vec a$ = $i + 2 j - k$

the position vector of Q = $\vec b=- i + j + k$

m:n = 2:1

And Hence

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

$\vec r = \frac{-2\hat i+2\hat j +2\hat k+\hat i+2\hat j-\hat k}{3}=\frac{-\hat i+4\hat j+\hat k}{3}$

$\vec r = \frac{-\hat i}{3}+\frac{4\hat j}{3}+\frac{\hat k}{3}$

Question:15 (2) Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat i + 2 \hat j - \hat k$ and $- \hat i + \hat j + \hat k$ respectively, in the ratio 2 : 1 externally

Answer:

As we know

The position vector of the point R which divides the line segment PQ in ratio m:n externally:

$\vec r=\frac{m\vec b-n\vec a}{m-n}$

Here

position vector os P = $\vec a$ = $i + 2 j - k$

the position vector of Q = $\vec b=- i + j + k$

m:n = 2:1

And Hence

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

$\vec r = -3\hat i +3\hat k$

Question:16 Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).

Answer:

Given

The position vector of point P = $2\hat i+3\hat j +4\hat k$

Position Vector of point Q = $4\hat i+\hat j -2\hat k$

The position vector of R which divides PQ in half is given by:

$\vec r =\frac{2\hat i+3\hat j +4\hat k+4\hat i+\hat j -2\hat k}{2}$

$\vec r =\frac{6\hat i+4\hat j +2\hat k}{2}=3\hat i+2\hat j +\hat k$

Question:17 Show that the points A, B and C with position vectors, $\vec a = 3 \hat i - 4 \hat j - 4 \hat k , \vec b = 2 \hat i - \hat j + \hat k \: \: and \: \: \: \vec c = \hat i - 3 \hat j - 5 \hat k$ , respectively form the vertices of a right angled triangle.

Answer:

Given

the position vector of A, B, and C are

$\\\vec a = 3 \hat i - 4 \hat j - 4 \hat k ,\\\vec b = 2 \hat i - \hat j + \hat k \: \: and \\\vec c = \hat i - 3 \hat j - 5 \hat k$

Now,

$\vec {AB}=\vec b-\vec a=-\hat i+3\hat j+5\hat k$

$\vec {BC}=\vec c-\vec b=-\hat i-2\hat j-6\hat k$

$\vec {CA}=\vec a-\vec c=2\hat i-\hat j+\hat k$

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

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

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

AS we can see

$\left | \vec {BC} \right |^2=\left | \vec {CA} \right |^2+\left | \vec {AB} \right |^2$

Hence ABC is a right angle triangle.

Question:18 In triangle ABC (Fig 10.18), which of the following is not true:

$A ) \overline{AB}+ \overline{BC}+ \overline{CA} = \vec 0 \\\\ B ) \overline{AB}+ \overline{BC}- \overline{AC} = \vec 0 \\\\ C ) \overline{AB}+ \overline{BC}- \overline{CA} = \vec 0 \\\\ D ) \overline{AB}- \overline{CB}+ \overline{CA} = \vec 0$

Answer:

From triangles law of addition we have,

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

From here

$\vec {AB}+\vec {BC}-\vec {AC}=0$

also

$\vec {AB}+\vec {BC}+\vec {CA}=0$

Also

$\vec {AB}-\vec {CB}+\vec {CA}=0$

Hence options A,B and D are true SO,

Option C is False.

Question:19 If are two collinear vectors, then which of the following are incorrect:
(A) $\vec b = \lambda \vec a$ for some saclar $\lambda$
(B) $\vec a = \pm \vec b$
(C) the respective components of $\vec a \: \:and \: \: \vec b$ are not proportional
(D) both the vectors $\vec a \: \:and \: \: \vec b$ have same direction, but different magnitudes.

Answer:

If two vectors are collinear then, they have same direction or are parallel or anti-parallel.
Therefore,
They can be expressed in the form $\vec{b}= \lambda \vec{a}$ where a and b are vectors and $\lambda$ is some scalar quantity.

Therefore, (a) is true.
Now,
(b) $\lambda$ is a scalar quantity so its value may be equal to $\pm 1$

Therefore,
(b) is also true.

C) The vectors and are proportional,
Therefore, (c) is not true.

D) The vectors and can have different magnitude as well as different directions.

Therefore, (d) is not true.

Therefore, the correct options are (C) and (D).

Importance of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2

The Class 12 Maths NCERT exercises are solved by math experts and can be used for preparations of various examinations.
By practising the NCERT Solutions for Class 12 Maths chapter 10 exercise 10.2 students will be able to revise the concepts studied before the exercise and also will be able to clarify doubts.

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Key Features Of NCERT Solutions for Exercise 10.2 Class 12 Maths Chapter 10

Comprehensive Coverage: The solutions encompass all the topics covered in ex 10.2 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 10.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 10.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 10.2 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 10.2 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 10.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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Frequently Asked Questions (FAQs)

1. Can two different vectors have the same magnitude?

Yes, two different vectors can have the same magnitude. These vectors may be pointing in a different direction. An example is shown in question number 2 of NCERT solutions for Class 12 Maths chapter 10 exercise 10.2.

2. Can two different vectors have the same direction?

Yes, two different vectors can have the same direction. It can be vectors with different magnitudes. The third question of exercise 10.2 Class 12 Maths gives an example of this.

3. What type of questions are expected from Class 12th Maths chapter 10 exercise 10.2 for the CBSE board exam?

The question may be to find a unit vector in the direction of a given vector, or it may be related to collinear vectors etc.

4. How many questions are covered in NCERT Class 12 Maths exercise 10.2?

There are a total of 19 questions. Practising these may help in the CBSE Class 12 Maths Paper

5. What is the use of Vector Algebra in NEET exam?

The concepts in Vector Algebra help to solve NEET physics problems. A good part of Class 11 and C lass 12 NCERT Physics Syllabus uses the concepts of vectors. So vectors will be useful for the NEET exam as well as for engineering entrance exams like JEE Main.

6. If 7i+6j=2xi+3yj, then the value of x and y is.

The given vectors are equal. Which implies that 2x=7. Therefore x=3.5. And 3y=6, therefore y=2

7. How to verify whether the given vector forms a right angled triangle?

One method to solve such kinds of problems is to find the magnitudes of given vectors and verify Pythagoras Theorem.

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Can I change my board from CBSE to CHSE Odisha in Class 12th?

Hello,

Yes, you can switch from CBSE to CHSE Odisha in Class 12th, but there are a few conditions:

Eligibility : Check if CHSE Odisha allows lateral entry into Class 12th. Generally, students can transfer from CBSE to state boards with prior approval.
Document Requirements : You may need to provide transcripts, mark sheets, and a transfer certificate.
Subject Alignment : Ensure that the subjects you studied in CBSE align with the CHSE curriculum.
Application Process : Contact the respective CHSE Odisha authority or school for specific transfer procedures and deadlines.

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Prachi Kumari 30 December,2024

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

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Central Board of Secondary Education Class 12th Examination

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Exercise 10.2 Class 12 Maths Chapter 10 - Vector Algebra

NCERT Solutions For Class 12 Maths Chapter 10 Exercise 10.2

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