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NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 12 - Vector Algebra

NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 12 - Vector Algebra

Edited By Ramraj Saini | Updated on Dec 04, 2023 01:46 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Chapter 10 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 10 class 12 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 Class 12 Maths chapter 10 miscellaneous exercise gives explanations to 19 questions given in the Class 12 Mathematics NCERT book. Class 12 Maths chapter 10 miscellaneous exercise solutions cover questions from all the main topics of the chapter. Compared to the exercise questions NCERT solutions for Class 12 Maths chapter 10 miscellaneous exercises are a bit more tricky. A total of five exercises including miscellaneous are present in the NCERT Class 12 Book. On this page, the NCERT solutions for Class 12 Maths chapter 10 miscellaneous exercises are given. Along with the NCERT questions students can practice NCERT exemplar and also Class 12 CBSE Previous Year Questions.

Try to solve all these questions yourself before going to the Class 12 Maths chapter 10 miscellaneous solutions. So that students can understand their depth of understanding of the concepts and revisit or revise the concepts required. Miscellaneous exercise class 12 chapter 5 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 enumerated in NCERT Book together using the link provided below.

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Vector Algebra Class 12 Chapter 10 -Miscellaneous Exercise

Question:1 Write down a unit vector in XY-plane, making an angle of 30 \degree with the positive direction of x-axis.

Answer:

As we know

a unit vector in XY-Plane making an angle \theta with x-axis :

\vec r=cos\theta \hat i+sin\theta \hat j

Hence for \theta = 30^0

\vec r=cos(30^0) \hat i+sin(30^0) \hat j

\vec r=\frac{\sqrt{3}}{2} \hat i+\frac{1}{2} \hat j

Answer- the unit vector in XY-plane, making an angle of 30 \degree with the positive direction of x-axis is

\vec r=\frac{\sqrt{3}}{2} \hat i+\frac{1}{2} \hat j

Question:2 Find the scalar components and magnitude of the vector joining the points
P(x_1, y_1, z_1) \: \: and \: \: Q(x_2, y_2, z_2).

Answer:

Given in the question

P(x_1, y_1, z_1) \: \: and \: \: Q(x_2, y_2, z_2).

And we need to finrd the scalar components and magnitude of the vector joining the points P and Q

\vec {PQ}=(x_2-x_1)\hat i +(y_2-y_1)\hat j+(z_2-z_1)\hat k

Magnitiude of vector PQ

|\vec {PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

Scalar components are

(x_2-x_1),(y_2-y_1),(z_2-z_1)

Question:3 A girl walks 4 km towards west, then she walks 3 km in a direction 30 \degree east of north and stops. Determine the girl’s displacement from her initial point of departure.

Answer:

As the girl walks 4km towards west

Position vector = -4\hat i

Now as she moves 3km in direction 30 degree east of north.

-4\hat i+3sin30^0\hat i+3cos30^0\hat j

-4\hat i+\frac{3}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j

\frac{-5}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j

hence final position vector is;

\frac{-5}{2}\hat i+3\frac{\sqrt{3}}{2}\hat j

Question:4 If \vec a = \vec b + \vec c , then is it true that |\vec a| =| \vec b |+| \vec c | ? Justify your answer.

Answer:

No, if \vec a = \vec b + \vec c then we can not conclude that |\vec a| =| \vec b |+| \vec c | .

the condition \vec a = \vec b + \vec c satisfies in the triangle.

also, in a triangle, |\vec a| <| \vec b |+| \vec c |

Since, the condition |\vec a| =| \vec b |+| \vec c | is contradicting with the triangle inequality, if \vec a = \vec b + \vec c then we can not conclude that |\vec a| =| \vec b |+| \vec c |

Question:5 Find the value of x for which x ( \hat i+ \hat j + \hat k ) is a unit vector.

Answer:

Given in the question,

a unit vector, \vec u=x ( \hat i+ \hat j + \hat k )

We need to find the value of x

|\vec u|=1

|x ( \hat i+ \hat j + \hat k )|=1

x\sqrt{1^2+1^2+1^2}=1

x\sqrt{3}=1

x=\frac{1}{\sqrt{3}}

The value of x is \frac{1}{\sqrt{3}}

Question:6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors \vec a = 2 \hat i + 3 \hat j - \hat k \: \: and \: \: \vec b = \hat i - 2 \hat j + \hat k

Answer:

Given two vectors

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

Resultant of \vec a and \vec b :

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

Now, a unit vector in the direction of \vec R

\vec u =\frac{3\hat i+\hat j}{\sqrt{3^2+1^2}}=\frac{3}{\sqrt{10}}\hat i+\frac{1}{\sqrt{10}}\hat j

Now, a unit vector of magnitude in direction of \vec R

\vec v=5\vec u =5*\frac{3}{\sqrt{10}}\hat i+5*\frac{1}{\sqrt{10}}\hat j=\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j

Hence the required vector is \frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j

Question:7 If \vec a = \hat i + \hat j + \hat k , \vec b = 2 \hat i - \hat j + 3 \hat k \: \: and\: \: \vec c = \hat i - 2 \hat j + \hat k , find a unit vector parallel to the vector 2\vec a - \vec b + 3 \vec c .

Answer:

Given in the question,

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

Now,

let vector \vec V=2\vec a - \vec b + 3 \vec c

\vec V=2(\hat i +\hat j +\hat k) - (2\hat i-\hat j+3\hat k)+ 3 (\hat i-2\hat j+\hat k)

\vec V=3\hat i-3\hat j+2\hat k

Now, a unit vector in direction of \vec V

\vec u =\frac{3\hat i-3\hat j+2\hat k}{\sqrt{3^2+(-3)^2+2^2}}=\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}} \hat j+\frac{2}{\sqrt{22}}\hat k

Now,

A unit vector parallel to \vec V

\vec u =\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}} \hat j+\frac{2}{\sqrt{22}}\hat k

OR

-\vec u =-\frac{3}{\sqrt{22}}\hat i+\frac{3}{\sqrt{22}} \hat j-\frac{2}{\sqrt{22}}\hat k

Question:8 Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Answer:

Given in the question,

points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7)

\vec {AB }=(5-1)\hat i+(0-(-2))\hat j+(-2-(-8))\hat k=4\hat i+2\hat j+6\hat k

\vec {BC }=(11-5)\hat i+(3-0)\hat j+(7-(-2))\hat k=6\hat i+3\hat j+9\hat k

\vec {CA }=(11-1)\hat i+(3-(-2))\hat j+(7-(-8))\hat k=10\hat i+5\hat j+15\hat k

now let's calculate the magnitude of the vectors

|\vec {AB }|=\sqrt{4^2+2^2+6^2}=\sqrt{56}=2\sqrt{14}

|\vec {BC }|=\sqrt{6^2+3^2+9^2}=\sqrt{126}=3\sqrt{14}

|\vec {CA }|=\sqrt{10^2+5^2+15^2}=\sqrt{350}=5\sqrt{14}

As we see that AB = BC + AC, we conclude that three points are colinear.

we can also see from here,

Point B divides AC in the ratio 2 : 3.

Question:9 Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are ( 2 \vec a + \vec b ) \: \:and \: \: ( \vec a - 3 \vec b ) externally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

Answer:

Given, two vectors \vec P=( 2 \vec a + \vec b ) \: \:and \: \:\vec Q= ( \vec a - 3 \vec b )

the point R which divides line segment PQ in ratio 1:2 is given by

=\frac{2(2\vec a +\vec b)-(\vec a-3\vec b)}{2-1}=4\vec a +2\vec b -\vec a+3\vec b=3\vec a+5\vec b

Hence position vector of R is 3\vec a+5\vec b .

Now, Position vector of the midpoint of RQ

=\frac{( 3\vec a + 5\vec b + \vec a - 3 \vec b )}{2}=2\vec a+\vec b

which is the position vector of Point P . Hence, P is the mid-point of RQ

Question:10 The two adjacent sides of a parallelogram are 2 \hat i - 4 \hat j + 5 \hat k \: \:and \: \: \hat i - 2 \hat j - 3 \hat k . Find the unit vector parallel to its diagonal. Also, find its area.

Answer:

Given, two adjacent sides of the parallelogram

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

The diagonal will be the resultant of these two vectors. so

resultant R:

\vec R=2 \hat i - 4 \hat j + 5 \hat k \: +\: \hat i - 2 \hat j - 3 \hat k=3\hat i-6\hat j+2\hat k

Now unit vector in direction of R

\vec u=\frac{3\hat i-6\hat j+2\hat k}{\sqrt{3^2+(-6)^2+2^2}}=\frac{3\hat i-6\hat j+2\hat k}{\sqrt{49}}=\frac{3\hat i-6\hat j+2\hat k}{7}

Hence unit vector along the diagonal of the parallelogram

\vec u={\frac{3}{7}\hat i-\frac{6}{7}\hat j+\frac{2}{7}\hat k}

Now,

Area of parallelogram

A=(2 \hat i - 4 \hat j + 5 \hat k )\: \times \: \: (\hat i - 2 \hat j - 3 \hat k)

A=\begin{vmatrix} \hat i &\hat j &\hat k \\ 2& -4 &5 \\ 1&-2 &-3 \end{vmatrix}=|\hat i(12+10)-\hat j(-6-5)+\hat k(-4+4)|=|22\hat i +11\hat j|

A=\sqrt{22^2+11^2}=11\sqrt{5}

Hence the area of the parallelogram is 11\sqrt{5} .

Question:11 Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are \pm \left ( \frac{1}{\sqrt 3 } , \frac{1}{\sqrt 3 } , \frac{1}{\sqrt 3 } \right )

Answer:

Let a vector \vec a is equally inclined to axis OX, OY and OZ.

let direction cosines of this vector be

cos\alpha,cos\alpha \:and \:cos\alpha

Now

cos^2\alpha+cos^2\alpha +cos^2\alpha=1

cos^2\alpha=\frac{1}{3}

cos\alpha=\frac{1}{\sqrt{3}}

Hence direction cosines are:

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

Question:12 Let \vec a = \hat i + 4 \hat j + 2 \hat k , \vec b = 3 \hat i - 2 \hat j + 7 \hat k \: \:and \: \: \vec c = 2 \hat i - \hat j + 4 \hat k . Find a vector \vec d which is perpendicular to both \vec a \: \: and \: \: \vec b \: \: and \: \: \vec c . \vec d = 15

Answer:

Given,

\vec a = \hat i + 4 \hat j + 2 \hat k , \vec b = 3 \hat i - 2 \hat j + 7 \hat k \: \:and \: \: \vec c = 2 \hat i - \hat j + 4 \hat k

Let \vec d=d_1\hat i+d_2\hat j +d_3\hat k

now, since it is given that d is perpendicular to \vec a and \vec b , we got the condition,

\vec b.\vec d=0 and \vec a.\vec d=0

(\hat i+4\hat j +2\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0 And (3\hat i-2\hat j +7\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=0

d_1+4d_2+2d_3=0 And 3d_1-2d_2+7d_3=0

here we got 2 equation and 3 variable. one more equation will come from the condition:

\vec c . \vec d = 15

(2\hat i-\hat j +4\hat k)\cdot(d_1\hat i +d_2\hat j+d_3\hat k)=15

2d_1-d_2+4d_3=15

so now we have three equation and three variable,

d_1+4d_2+2d_3=0

3d_1-2d_2+7d_3=0

2d_1-d_2+4d_3=15

On solving this three equation we get,

d_1=\frac{160}{3},d_2=-\frac{5}{3}\:and\:d_3=-\frac{70}{3} ,

Hence Required vector :

\vec d=\frac{160}{3}\hat i-\frac{5}{3}\hat j-\frac{70}{3}\hat k

Question:13 The scalar product of the vector \hat i + \hat j + \hat k with a unit vector along the sum of vectors 2\hat i + 4 \hat j -5 \hat k and \lambda \hat i + 2 \hat j +3 \hat k is equal to one. Find the value of \lambda .

Answer:

Let, the sum of vectors 2\hat i + 4 \hat j -5 \hat k and \lambda \hat i + 2 \hat j +3 \hat k be

\vec a=(\lambda +2)\hat i + 6 \hat j -2 \hat k

unit vector along \vec a

\vec u=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{(\lambda+2)^2+6^2+(-2)^2}}=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{\lambda^2+4\lambda+44}}

Now, the scalar product of this with \hat i + \hat j + \hat k

\vec u.(\hat i+\hat j +\hat k)=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{\lambda^2+4\lambda+44}}.(\hat i+\hat j +\hat k)

\vec u.(\hat i+\hat j +\hat k)=\frac{(\lambda +2) + 6 -2 }{\sqrt{\lambda^2+4\lambda+44}}=1

\frac{(\lambda +2) + 6 -2 }{\sqrt{\lambda^2+4\lambda+44}}=1

\frac{(\lambda +6) }{\sqrt{\lambda^2+4\lambda+44}}=1

\lambda =1

Question:14 If \vec a , \vec b , \vec c are mutually perpendicular vectors of equal magnitudes, show that the vector \vec a+\vec b +\vec c is equally inclined to \vec a , \vec b \: \: and \: \: \vec c .

Answer:

Given

|\vec a|=|\vec b|=|\vec c| and

\vec a.\vec b=\vec b.\vec c=\vec c.\vec a=0

Now, let vector \vec a+\vec b +\vec c is inclined to \vec a , \vec b \: \: and \: \: \vec c at \theta_1,\theta_2\:and\:\theta_3 respectively.

Now,

cos\theta_1=\frac{(\vec a+\vec b+\vec c).\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{\vec a.\vec a +\vec a.\vec b +\vec c.\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{\vec a.\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{|\vec a|}{|\vec a+\vec b+\vec c|}

cos\theta_2=\frac{(\vec a+\vec b+\vec c).\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{\vec a.\vec b +\vec b.\vec b +\vec c.\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{\vec b.\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{|\vec b|}{|\vec a+\vec b+\vec c|}

cos\theta_3=\frac{(\vec a+\vec b+\vec c).\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{\vec a.\vec c +\vec b.\vec c +\vec c.\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{\vec c.\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{|\vec c|}{|\vec a+\vec b+\vec c|}

Now, Since, |\vec a|=|\vec b|=|\vec c|

cos\theta_1=cos\theta_2=cos\theta_3

\theta_1=\theta_2=\theta_3

Hence vector \vec a+\vec b +\vec c is equally inclined to \vec a , \vec b \: \: and \: \: \vec c .

Question:15 Prove that ( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a ^2 | + |\vec b |^2 , if and only if \vec a , \vec b are perpendicular, given \vec a \neq 0 , \vec b \neq 0

Answer:

Given in the question,

\vec a , \vec b are perpendicular and we need to prove that ( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a ^2 | + |\vec b |^2

LHS= ( \vec a + \vec b ) . (\vec a + \vec b ) = \vec a .\vec a+\vec a.\vec b+\vec b.\vec a+\vec b.\vec b

= \vec a .\vec a+2\vec a.\vec b+\+\vec b.\vec b

= |\vec a |^2+2\vec a.\vec b+\+|\vec b|^2

if \vec a , \vec b are perpendicular, \vec a.\vec b=0

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

= |\vec a |^2+|\vec b|^2

= RHS

LHS ie equal to RHS

Hence proved.

Question:16 Choose the correct answer If \theta is the angle between two vectors \vec a \: \: and \: \: \vec b , then \vec a \cdot \vec b \geq 0 only when
\\A ) 0 < \theta < \frac{\pi }{2} \\\\ \: \: \: \: B ) 0 \leq \theta \leq \frac{\pi }{2} \\\\ \: \: \: C ) 0 < \theta < \pi \\\\ \: \: \: D) 0 \leq \theta \leq\pi

Answer:

Given in the question

\theta is the angle between two vectors \vec a \: \: and \: \: \vec b

\vec a \cdot \vec b \geq 0

|\vec a| | \vec b |cos\theta\geq 0

this will satisfy when

cos\theta\geq 0

0\leq\theta\leq \frac{\pi}{2}

Hence option B is the correct answer.

\\A ) \theta = \frac{\pi }{4} \\\\ B ) \theta = \frac{\pi }{3} \\\\ C ) \theta = \frac{\pi }{2} \\\\ D ) \theta = \frac{2\pi }{3}

Answer:

Gicen in the question

\vec a \: \: and \: \: \vec b be two unit vectors and \theta is the angle between them

|\vec a|=1,\:and\:\:|\vec b|=1

also

|\vec a + \vec b|=1

|\vec a + \vec b|^2=1

|\vec a|^2 + |\vec b|^2+2\vec a.\vec b=1

1 + 1+2\vec a.\vec b=1

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

|\vec a||\vec b|cos\theta =-\frac{1}{2}

cos\theta =-\frac{1}{2}

\theta =\frac{2\pi}{3}

Then \vec a + \vec b is a unit vector if \theta =\frac{2\pi}{3}

Hence option D is correct.

Question:18 The value of \hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j ) is

(A) 0

(B) –1

(C) 1

(D) 3

Answer:

To find the value of \hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j )

\\\hat i ( \hat j \times \hat k ) + \hat j ( \hat i \times \hat k ) + \hat k ( \hat i \times \hat j ) \\=\hat i.\hat i+\hat j(-\hat j)+\hat k.\hat k\\=1-1+1\\=1

Hence option C is correct.

Question:19 Choose the correct. If \theta is the angle between any two vectors \vec a \: \:and \: \: \vec b , then |\vec a \cdot \vec b |=|\vec a \times \vec b | when \theta
is equal to

\\A ) 0 \\\\ B ) \pi /4 \\\\ C ) \pi / 2 \\\\ D ) \pi

Answer:

Given in the question

\theta is the angle between any two vectors \vec a \: \:and \: \: \vec b and |\vec a \cdot \vec b |=|\vec a \times \vec b |

To find the value of \theta

Hence option D is correct.

More About NCERT Solutions for Class 12 Maths Chapter 10 Miscellaneous Exercises

Class 12 Maths miscellaneous exercises are designed by the in-house expert faculties and are according to the CBSE pattern. Many state boards also follow the NCERT Syllabus, so for these boards definitely, the NCERT solutions for Class 12 Maths chapter 10 miscellaneous exercises will be useful. Class 12 Maths chapter 10 miscellaneous gives an insight into the chapter vector algebra.

Also Read | Vector Algebra class 12 Chapter 10 Notes

Benefits of ncert solutions for Class 12 Maths chapter 10 miscellaneous exercises

  • By using miscellaneous exercise chapter 10 Class 12 students will be able to get an idea of the complete chapter.
  • Each question explained in the Class 12 Maths chapter 10 miscellaneous solutions are useful for the CBSE Class 12 board exam and also for exams like JEE main, VITEEE, BITSAT, etc.
  • The students can revise the chapter vector algebra by practising the ncert solutions for Class 12 Maths chapter 10 miscellaneous exercises
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Key Features Of NCERT Solutions For Class 12 Chapter 10 Miscellaneous Exercise

  • Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 10, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 chapter 10 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 10 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this class 12 maths ch 10 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for class 12 chapter 10 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 10 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
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Frequently Asked Question (FAQs)

1. Why should I study Class 12 Maths chapter Vectors?

Vectors is an important chapter of Class 12 Maths NCERT syllabus. The concepts studied in this chapter are used not only on mathematics but also to solve problems in Class 11 and 12 physics problems also. Also Vectors is an important chapter for CBSE Board exam. Solving the NCERT questions and referring to the NCERT exemplar will be beneficial for board exam preparation.

2. What are the topics coverd under Vectors Class 12 Maths NCERT syllabus?

The topics covered are vectors basic concepts, algebra of vectors, direction cosine and ratios, dot product and cross product of vectors.

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Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

  • Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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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

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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)

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

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

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

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

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