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

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

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

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

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

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;

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

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 $|\vec a| =| \vec b |+| \vec c |$

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 $\dpi{80} \frac{1}{\sqrt{3}}$

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 $\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j$

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

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

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.

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

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 $11\sqrt{5}$ .

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:

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

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 :

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

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 $\hat i + \hat j + \hat k$

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

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

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

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

$\lambda =1$

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 $\vec a , \vec b \: \: and \: \: \vec c$ .

Given in the question,

are perpendicular and we need to prove that

LHS=

if are perpendicular,

= RHS

LHS ie equal to RHS

Hence proved.

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.

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.

(A) 0

(B) –1

(C) 1

(D) 3

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.

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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## Subject Wise NCERT Exemplar Solutions

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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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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 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

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

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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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