Future Study Point

NCERT Solutions for Class 12 Maths Exercise 10.4 of Chapter 10 Vector Algebra

NCERT Solutions for Class 12 Maths Exercise 10.4 of Chapter 10 Vector Algebra is compulsory to study for clearing the concept of vector algebra of class 12 maths NCERT.NCERT Solutions for Class 12 Maths Exercise 10.4 of Chapter 10 Vector Algebra will help students for the preparation of the maths paper for the CBSE Board exam term 2. All the unsolved questions of exercise 10.4 of chapter 10 Vector Algebra are created by a CBSE Maths expert as per the CBSE norms.Here you can study all the study material, NCERT Solutions, Science and Maths notes, preparation for the competitive entrance exams, and e-books free of cost.

What is differentiation?

NCERT Solutions Exercise 10.1 :Vector Algebra

NCERT Solutions Exercise 10.2: Vector Algebra

NCERT Solutions Exercise 10.3 Vector Algebra

NCERT Solutions Exercise Miscellaneous: Vector Algebra

NCERT Solutions for Class 12 Maths Exercise 10.4 of Chapter 10 Vector Algebra

Q1. Find  $\fn_cm \left | \overrightarrow{a}\times \overrightarrow{b} \right |$   ,if  $\fn_cm \overrightarrow{a}=\hat{i}-7\hat{j}+7\hat{k}$ and $\fn_cm \overrightarrow{b}=3\hat{i}-2\hat{j}+2\hat{k}$.

Solutions: We are given that

$\fn_cm \overrightarrow{a}=\hat{i}-7\hat{j}+7\hat{k}$$\fn_cm \overrightarrow{b}=3\hat{i}-2\hat{j}+2\hat{k}$

$\fn_cm \left | \overrightarrow{a}\times \overrightarrow{b} \right |=\begin{vmatrix} i & j & k\\ 1 & -7 &7 \\ 3& -2 & 2 \end{vmatrix}$

$\fn_cm \left | \overrightarrow{a}\times \overrightarrow{b} \right |= \left ( -14+14 \right )\hat{i}-\left ( 2-21 \right )\hat{j}+\left ( -2+21 \right )\hat{k}$

$\fn_cm \left | \overrightarrow{a}\times \overrightarrow{b} \right |= 19\hat{j}+19\hat{k}$

Q2.Find a unit vector perpendicular to each of the vector    $\fn_cm \overrightarrow{a}+\overrightarrow{b}$ and$\fn_cm \overrightarrow{a}-\overrightarrow{b}$, where   $\fn_cm \overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and

$\fn_cm \overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$

Solutions: We are given  vectors   $\fn_cm \overrightarrow{a}+\overrightarrow{b}$ and$\fn_cm \overrightarrow{a}-\overrightarrow{b}$, where

$\fn_cm \overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}$, $\fn_cm \overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$

$\fn_cm \overrightarrow{a}-\overrightarrow{b}=\left ( 3-1 \right )\hat{i}+\left ( 2+2 \right )\hat{k}=2\hat{i}+4\hat{k}$

$\fn_cm \overrightarrow{a}+\overrightarrow{b}=\left ( 3+1 \right )\hat{i}+\left ( 2+2 \right )\hat{k}=4\hat{i}+4\hat{k}$

The unit vector that is perpendicular to both vectors = cross product of both vectors/Magnitude of the cross product of both vectors

$\fn_cm (\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})=\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 4& 4 &0 \\ 2&0 & 4 \end{vmatrix}$

$\fn_cm =\left ( 16-0 \right )\hat{i}-\left ( 16-0 \right )\hat{j}+\left ( 0-8 \right )\hat{k}=16\hat{i}-16\hat{j}-8\hat{k}$

$\fn_cm \left | 16\hat{i}-16\hat{j}-8\hat{k} \right |$

$\fn_cm =\sqrt{16^{2}+\left ( -16 \right )^{2}+\left ( -8 \right )^{2}}=\sqrt{256+256+64}=\sqrt{576}=24$

The unit vector that is perpendicular to both vectors

$\fn_cm =\pm \frac{2\hat{i}-2\hat{j}-\hat{k}}{3}$$\fn_cm =\pm \frac{2\hat{i}}{3}\mp \frac{2\hat{j}}{3}\mp \frac{\hat{k}}{3}$

Q3.If a unit vector  $\fn_cm \overrightarrow{a}$ makes an angle π/3 with  $\fn_cm \hat{i}$, π/4,  with  $\fn_cm \hat{j}$ and an acute angle θ with $\fn_cm \hat{k}$, then find θ and hence ,the components of $\fn_cm \overrightarrow{a}$

Solution: Let the unit vector$\fn_cm \overrightarrow{a}$  has the components a1,a2 and a3

Therefore

$\fn_cm \overrightarrow{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$

Since $\fn_cm \overrightarrow{a}$ is a unit vector

$\fn_cm \left | \overrightarrow{a} \right |=1$

It is also given that   $\fn_cm \overrightarrow{a}$ makes an angle π/3 with  $\fn_cm \hat{i}$, π/4,  with  $\fn_cm \hat{j}$ and an acute angle θ with $\fn_cm \hat{k}$

We know that

$\fn_cm \overrightarrow{a}$ and $\fn_cm \overrightarrow{b}$ ,two vectors that makes an angle θ with each other then we have

$\fn_cm \overrightarrow{a}.\overrightarrow{b}=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |cos\theta$

$\fn_cm \left | \overrightarrow{a} \right |=1$

$\fn_cm \overrightarrow{b} =\hat{i}=1.\hat{i}+0.\hat{j}+0.\hat{k}$

$\fn_cm \left | \hat{i} \right |=1$

$\fn_cm (a_{1} \hat{i}+ a_{2} \hat{j}+ a_{3} \hat{k}).\left ( 1.\hat{i} +0.\hat{j}+0.1.\hat{k}\right )=1.1cos\pi /3$

$\fn_cm a_{1} =cos\frac{\pi }{3}=\frac{1}{2}$

Similarly, we can get

$\fn_cm a_{2} =cos\frac{\pi }{4}=\frac{1}{\sqrt{2}},a_{3}=cos\theta$

Now, we have

$\fn_cm \left | \overrightarrow{a} \right |=1$

$\fn_cm \sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}=1$

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

$\fn_cm \sqrt{\frac{1}{4}+\frac{1}{2}+cos^{2}\theta }=1$

Squaring both sides

cos²θ + 3/4 =1

cos²θ =1/4

cos θ = 1/2⇒a3=1/2

Hence components of $\fn_cm \overrightarrow{a}$ are 1/2,1/√2 and 1/2

Q4. Show that

$\fn_cm \left ( \overrightarrow{a} - \overrightarrow{b}\right )\times \left ( \overrightarrow{a} + \overrightarrow{b}\right )=2\left ( \overrightarrow{a}\times \overrightarrow{b} \right )$

Solution: Taking LHS

$\fn_cm \left ( \overrightarrow{a} - \overrightarrow{b}\right )\times \left ( \overrightarrow{a} + \overrightarrow{b}\right )$

By the distributive property of the vector product

$\fn_cm =\overrightarrow{a}\times \left ( \overrightarrow{a} +\overrightarrow{b}\right )-\overrightarrow{b}\times \left ( \overrightarrow{a} +\overrightarrow{b}\right )$

$\fn_cm =\overrightarrow{a}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{b}\times \overrightarrow{a}-\overrightarrow{b}\times \overrightarrow{b}$

$\fn_cm =0+\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{b}\times \overrightarrow{a}-0$

$\fn_cm \because -\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{a}$

$\fn_cm \therefore 2(\overrightarrow{a}\times \overrightarrow{b})=RHS$

Q5.Find λ and μ if

$\fn_cm \left ( 2\hat{i} +6\hat{j}+27\hat{k}\right )\times \left ( \hat{i}+\lambda \hat{j}+\mu \hat{k} \right )=0$

Solutions:We are given that

$\fn_cm \left ( 2\hat{i} +6\hat{j}+27\hat{k}\right )\times \left ( \hat{i}+\lambda \hat{j}+\mu \hat{k} \right )=0$

Simplifying it

$\fn_cm =\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 6 & 27\\ 1& \lambda & \mu \end{vmatrix}$

$\fn_cm \Rightarrow \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 6 & 27\\ 1& \lambda & \mu \end{vmatrix}=0.\hat{i}+0.\hat{j}+0.\hat{k}$

$\fn_cm \left ( 6\mu -27\lambda \right )\hat{i}-\left ( 2\mu -27 \right )\hat{27}+\left ( 2\lambda -6 \right )\hat{k}=0.\hat{i}+0.\hat{j}+0.\hat{k}$

Comparing both sides

6μ – 27λ = 0….(i) ,2μ  – 27=0…..(ii), 2λ – 6 = 0…….(iii)

From equation (ii) , (ii)

μ = 27/2, λ = 3 and

Q6.Given that   $\fn_cm \overrightarrow{a}.\overrightarrow{b}=0$  and $\fn_cm \overrightarrow{a}\times \overrightarrow{b}=0$ ,what can you conclude about the vectors $\fn_cm \overrightarrow{a}$ and $\fn_cm \overrightarrow{b}$ ?

Solution: It is given that

$\fn_cm \overrightarrow{a}.\overrightarrow{b}=0$

$\fn_cm \overrightarrow{a}.\overrightarrow{b}=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |cos\theta$

$\fn_cm \left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |cos\theta=0 ...\left ( i \right )$

Equation (i) implies that either $\fn_cm \left | \overrightarrow{a} \right |=0,\left | \overrightarrow{b} \right |=0$  or cos θ=0(i.e both vectors are perpendicular to each other)

It is also given that

$\fn_cm \overrightarrow{a}\times \overrightarrow{b}=0$

$\fn_cm \left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |sin\theta =0 ...\left ( ii \right )$

Equation (ii) implies that either $\fn_cm \left | \overrightarrow{a} \right |=0,\left | \overrightarrow{b} \right |=0$  or sin θ=0(i.e both vectors are parallel to each other)

From Equation (i) and Equation (ii), it is to be concluded that both vectors can’t be parallel  and perpendicular simultaneously.

Therefore

$\fn_cm \left | \overrightarrow{a} \right |=0\: or\: \left | \overrightarrow{b} \right |=0$

Q7. Let the vectors $\fn_cm \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ given as $\fn_cm \overrightarrow{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$, $\fn_cm \overrightarrow{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}$, $\fn_cm \overrightarrow{c}=c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}$, then show that

$\fn_cm \overrightarrow{a}\times \left ( \overrightarrow{b} +\overrightarrow{c}\right )=\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}$

Solutions:

The given vectors are $\fn_cm \overrightarrow{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$, $\fn_cm \overrightarrow{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}$,$\fn_cm \overrightarrow{c}=c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}$

Evaluating LHS

$\fn_cm \overrightarrow{a}\times \left ( \overrightarrow{b} +\overrightarrow{c}\right )$

First of all evaluating $\fn_cm \overrightarrow{b} +\overrightarrow{c}$

$\fn_cm \overrightarrow{b} +\overrightarrow{c}=\left ( b_{1}+c_{1} \right )\hat{i}+\left ( b_{2}+c_{2} \right )\hat{j}+\left ( b_{3}+c_{3} \right )\hat{k}$

Now, simplifying LHS

$\fn_cm \overrightarrow{a}\times (\overrightarrow{b} +\overrightarrow{c})=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_{1}& a_{2} &a_{3} \\ b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3} \end{vmatrix}$

$\fn_cm =\left [ a_{2}\left ( b_{3} +c_{3}\right ) -a_{3}\left ( b_{2} +c_{2} \right )\right ]\hat{i}-\left [ a_{1}\left ( b_{3} +c_{3}\right ) -a_{3}\left ( b_{1} +c_{1} \right )\right ]\hat{j}+\left [ a_{1}\left ( b_{2} +c_{2}\right ) -a_{2}\left ( b_{1} +c_{1} \right )\right ]\hat{k}$……..(i)

Simplifying RHS

$\fn_cm \overrightarrow{a}\times \overrightarrow{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_{1}&a_{2} &a_{3} \\ b_{1} & b_{2} & b_{3} \end{vmatrix}$

$\fn_cm =\left ( a_{2}b_{3}-a_{3}b_{2} \right )\hat{i}-\left ( a_{1}b_{3}-a_{3}b_{1} \right )\hat{j}+\left ( a_{1}b_{2}-a_{2}b_{1} \right )\hat{k}$……(ii)

And

$\fn_cm \overrightarrow{a}\times \overrightarrow{c}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_{1}&a_{2} &a_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix}$

$\fn_cm =\left ( a_{2}c_{3}-a_{3}c_{2} \right )\hat{i}-\left ( a_{1}c_{3}-a_{3}c_{1} \right )\hat{j}+\left ( a_{1}c_{2}-a_{2}c_{1} \right )\hat{k}$…….(iii)

$\fn_cm \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}$

$\fn_cm =\left [ a_{2}\left ( b_{3} +c_{3}\right ) -a_{3}\left ( b_{2} +c_{2} \right )\right ]\hat{i}-\left [ a_{1}\left ( b_{3} +c_{3}\right ) -a_{3}\left ( b_{1} +c_{1} \right )\right ]\hat{j}+\left [ a_{1}\left ( b_{2} +c_{2}\right ) -a_{2}\left ( b_{1} +c_{1} \right )\right ]\hat{k}$

Hence

$\fn_cm \overrightarrow{a}\times \left ( \overrightarrow{b}+\overrightarrow{c} \right )=\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}$

Q8. If either $\fn_cm \overrightarrow{a}=0$ or $\fn_cm \overrightarrow{b}=0$ then $\fn_cm \overrightarrow{a}\times \overrightarrow{b}=0$.Is the converse true?Justify your answer with an example.

Solution:

We are given that if either $\fn_cm \overrightarrow{a}=0$ or $\fn_cm \overrightarrow{b}=0$ then $\fn_cm \overrightarrow{a}\times \overrightarrow{b}=0$.

Let’s prove if $\fn_cm \overrightarrow{a}\times \overrightarrow{b}=0$ then $\fn_cm \overrightarrow{a}=0$ or $\fn_cm \overrightarrow{b}=0$

Taking two vector$\fn_cm \overrightarrow{a}$ and$\fn_cm \overrightarrow{b}$ parallel to each other

Let $\fn_cm \overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k}\: and\: \overrightarrow{b}=2\hat{i}+4\hat{j}+2\hat{k}$

$\fn_cm \overrightarrow{a}\times \overrightarrow{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1& 2 & 1\\ 2 & 4& 2 \end{vmatrix}$

$\fn_cm =\left ( 4-4 \right )\hat{i}-\left ( 2-2 \right )\hat{j}+\left ( 4-4 \right )\hat{k}=0$

$\fn_cm \left | \overrightarrow{a} \right |=\sqrt{1^{2}+2^{2}+1^{2}}=\sqrt{6},\left | \overrightarrow{b} \right |=\sqrt{2^{2}+4^{2}+2^{2}}=2\sqrt{6}$

$\fn_cm \therefore \left | \overrightarrow{a} \right |\neq 0,\left | \overrightarrow{b} \right |\neq 0$

Hence converse of the given situation is not true.

NCERT Solutions and CBSE notes for classes 9,10,11 and 12 of Maths and Science

NCERT Solutions and CBSE notes for Class 9,10,11 and 12 of Maths and Science

You can also study

Active Voice to Passive Voice Rules

Learn Tenses in English and translate Hindi sentences into English language

Download PDF-Learn Tenses in English and translate Hindi sentences into the English language

You can also study

Percentage questions for competitive entrance exams with Solutions

NCERT Questions on mensuration helpful for competive entrance exams

Tips to get success in competitive exams

The best books for cracking the competitive entrance exams of CPO,NDA, IAS, CDS and Bank PO

Government jobs after 10 or 12 th pass : Qualify Entrance exams of SSC MTS and SSC CHSL

Science  Notes

What is the difference between virtual and real images?

Image formation by Convex and Concave Lenses

Image formation by Convex and Concave Mirrors

Difference between Convex and Concave lenses

What are the factors affecting evaporation?

How does the water kept in an earthen pot become cold during summer ?

Functioning of Soda-Acid Fire Extinguisher

What is DDT?

Structure and Function of Cell : Cell Biology

Three laws of motion

Archimedes Principle: Complete detail

Average Speed and Average velocity

Three equation of Motions

Recoil velocity of the gun

Mole concept

The second  law of motion

The universal  law of gravitational force

Thrust and Pressure : Difference

Evoporation,Vapourization and Latent heat

Important salts class 10 CBSE sceience notes

Reflection, Refraction, Dispersion, and Scattering

Atom, Molecule, and Atomicity

Determining Valency, Net Charge and Molecular Formula

Ozone Layer and How it is Getting depleted.

Human Eye – Structure and functions

Myopia, Hypermetropia, and Presbyopia

Why do the star twinkle?

Electric Current and Heating effect of Electric Current

Complete detail of electrical resistance and conductance

Type of Chemical Reactions with Complete detail

Class 10 chemistry Viva Voce Questions and Answers for CBSE Board 2020-21

What are the physical and chemical properties of metals?

Important maths notes

Tricks – How to write linear equations

Tricks- How to solve question from algebraic equations

Three ways of solving quadratic equation

Mean, Mode and Median

Solutions- Specific questions of mensuration

Finding the roots of the polynomial by Complete square method

Technics – Achieving 100% marks in Maths

You can compensate us

Paytm number 9891436286

The money collected by us will be used for the education of poor students who leaves their study because of a lack of money.

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 science

CBSE Class 9-Question paper of science 2020 with solutions

CBSE Class 9-Sample paper of science

CBSE Class 9-Unsolved question paper of science 2019

NCERT Solutions for class 10 maths

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

Solutions of Class 10 Science Sample Paper and Question Papers for Term-1 and Term 2 2021-22 CBSE Board

Solution of Class 10 Science Question Paper Preboard 2021-22:Term 2 CBSE Board Exam

Solutions of Class 10 Science Sample Paper Term-1 2021-22 CBSE Board

Class 10 Science Sample Paper for Term 2 CBSE Board Exam 2021-22 with Solution

Solutions of Class 10 Science Question Paper Preboard Examination (First) 2021 -22 Class 10 Science

CBSE Class 10 – Question paper of science 2020 with solutions

CBSE class 10 -Sample paper of Science 2020

NCERT Solutions for class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution