RD Sharma Class 12 Exercise 22.5 Algebra of Vectors Solutions Maths

RD Sharma Class 12 Exercise 22.5 Algebra of vectors Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 06:19 PM IST

RD Sharma's book has set a standard for other publications as well. Chapter 22 of this book is Algebra of Vectors which includes a detailed study of vector quantities. Vector quantities are those that have magnitude as well as direction as compared to scalar amounts which have the only magnitude. This book could be dicey for some students, which is why RD Sharma Class 12th Exercise 22.5 has arrived to help.

Latest: JEE Main: high scoring chapters | Past 10 year's papers

Don't Miss: Most scoring concepts for NEET | NEET papers with solutions

New: Aakash iACST Scholarship Test. Up to 90% Scholarship. Register Now

This Story also Contains

RD Sharma Class 12 Solutions Chapter22 Algebra of vectors - Other Exercise
Algebra of Vectors Excercise:22.5
RD Sharma Chapter wise Solutions

RD Sharma Class 12th Chapter 22 Exercise 22.5 has exercise-wise questions in this book. This particular exercise has 16 questions from the chapter 'Algebra of Vectors' along with the examples. The concepts that are discussed are position vectors and the components of vectors in three dimensions.

RD Sharma Class 12 Solutions Chapter22 Algebra of vectors - Other Exercise

Algebra of Vectors Excercise:22.5

Algebra of vector exercise 22.5 question 1

Answer:

5
Hints: $\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$
Given: Position vector of a point (-4,-3) be $\left | \vec{a} \right |$
Find: $\left | \vec{a} \right |$
We know position vector of a point(x,y) is given by $x\hat{i} +y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are units vectors in x and y directions.
$\vec{a}=(-4)\hat{i} +(-3)\hat{j}$
Now we need to find magnitude of $\vec{a}$ i.e $\left |\vec{a} \right |$
$\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}$
Here, x=-4 and y=-3
$\left |\vec{a} \right |=\sqrt{(-4)^{2} +(-3)^{2}}\\ \left |\vec{a} \right |=\sqrt{16+9}\\ \left |\vec{a} \right |=\sqrt{25}\\ \left |\vec{a} \right |=5$
thus $\left |\vec{a} \right |=5$

Algebra of vector exercise 22.5 question 2

Answer:

$n=\pm 5$
Hints: $\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$
Given: Position vector of $\vec{a}$ point (12,n) be $\left | \vec{a} \right |$
Find: $\left | \vec{a} \right |$
We know position vector of a point(x,y) is given by $x\hat{i} +y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are units vectors in x and y directions.
$\vec{a}=12\hat{i} +n\hat{j}$
Then:
$\left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}$
Here, x=12 and y=n
$\left |\vec{a} \right |=\sqrt{12^{2} +n^{2}}\\ \left |\vec{a} \right |=13$ [given]
$13=\sqrt{(12)^{2} +(n)^{2}}\\$
Squaring both sides
$13^{2}=(12)^{2} +(n)^{2}\\ 169=144+n^{2}\\ n^{2}=25\\ n=\pm 5$

Algebra of vector exercise 22.5 question 4 (i)

Answer:

$\vec{AB}=-3\hat{i}+4\hat{j}\\ \left |\vec{AB} \right |=5$
Hint: $\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$
Position vector $\left |\vec{AB} \right |$ is given x, y
$\left |\vec{AB} \right |$ = position vector of B - position vector of A
Given:
A=(4,-1) and B(1,3)
Solution:
We know position vector of a point (x, y) is given by $x\hat{i} +y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are unit vector in
x and y direction
$A=4\hat{i}+(-1)\hat{j}\\ B=i+3\hat{j}$
Then the position vector $\vec{AB}$ is given by
$\vec{AB}$ = position vector of B - position vector of A
$=\left (\hat{i}+3\hat{j} \right )-\left (4\hat{i}-\hat{j} \right )\\ =\hat{i}+3\hat{j} -4\hat{i}+\hat{j} \\ AB=-3\hat{i}+4j$
So,
$\left |\vec{AB} \right |=(-3)^{2}+(4)^{2}\\ =\sqrt{9+16}=\sqrt{25}\\\left |\vec{AB} \right |=5$

Algebra of vector exercise 22.5 question 4 (ii)

Answer:

$\vec{AB}=4\hat{i}-8\hat{j}\\ \left |\vec{AB} \right |=4\sqrt{5}$
Hint: $\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$
Position vector $\vec{AB}$ is given x, y
$\vec{AB}$ = position vector of B - position vector of A
Given:
A=(-6,3) and B(-2,-5)
Solution:
We know position vector of a point (x, y) is given by $x\hat{i} +y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are unit vector in
x and y direction
$\vec{A}=-6\hat{i}+3\hat{j}\\ \vec{B}=-2i-5\hat{j}$
Then the position vector $\vec{AB}$ is given by
$\vec{AB}$ = position vector of B - position vector of A
$=\left (2\hat{i}-5\hat{j} \right )-\left (-6\hat{i}+3\hat{j} \right )\\ =-2\hat{i}-5\hat{j} +6\hat{i}-3\hat{j} \\ =4\hat{i}-8\hat{j}\\ \left |\vec{AB} \right |=\sqrt{4^{2}+(-8)^{2}}=\sqrt{16+64}\\ =\sqrt{80}\\ \left |\vec{AB} \right |=4\sqrt{5}$ -

Algebra of Vectors exercise 22.5 question 5
Answer:

(-1,-2)
Hint: Position vector $\vec{AB}$ is given $\vec{AB}$ = position vector B- position vector of A
Given: Coordinate of A (-1, 3)
Coordinates of B (-2,1)
Solution:
$\vec{A}= -\hat{i}+3\hat{j}$ & $\vec{B}= -2\hat{i}+\hat{j}$
We know position vector of point (x, y) is given by $(x\hat{i}+y\hat{j})$
Let 0 be the origin
Let P(x,y) be the second point
Then $\vec{P}$ is the tip of the position vector $\vec{OP}$ of the point p
We have: $\vec{OP}=x\hat{i}+y\hat{j}$
position vector of B- position vector of A
$\vec{AB}$ = position vector of B= position vector of A
$=(-2\hat{i}+\hat{j})-(-\hat{i}+3\hat{j})\\ =-2\hat{i}+\hat{j}+\hat{i}-3\hat{j}\\ =-\hat{i}-2\hat{j}$
$\vec{OP}=\vec{AB}$ given
$x\hat{i}+y\hat{j}=-\hat{i}-2\hat{j}$
then x=-1 and y= -2
Hence coordinate of the required point is (-1,-2)

Algebra of Vectors exercise 22.5 question 6
Answer:

the coordinates of D is (-4,-3)
Hint: Position vectors $\vec{AB}$ is given by = position vector of B -position vector of A
Given;
Coordinates of A=(-2,-1)
Coordinates of B=(3,0)
Coordinates of C=(1,-2)
Let the coordinates of D is (x , y)
Since ABCD is a parallelogram,
AB=DC
We have
$\vec{AB}=\vec{DC}$
As we know position vector of point (x, y) is given by $x\hat{i}=y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are position vector of x, and y direction.
$\vec{A}=-2\hat{i}-\hat{j}\\ \vec{B}=3\hat{i}+0\hat{j}\\ \vec{C}=\hat{i}-2\hat{j}\\ \vec{D}=x\hat{i}+y\hat{j}$
Then the position vector $\vec{AB}$ is given by
$\vec{AB}$ = position vector of B - position vector of A
$\left (3\hat{i}+0\hat{j} \right )-\left (-2\hat{i}-\hat{j} \right )\\ 3\hat{i}+0\hat{j} +2\hat{i}+\hat{j} \\ \vec{AB}=5\hat{i}+\hat{j}$
Then we find $\vec{DC}$
$\vec{DC}$ = position vector C- position vector of D
$=\left (\hat{i}-2\hat{j} \right )-\left (x\hat{i}+y\hat{j} \right )\\ =\hat{i}-x\hat{i}-2\hat{j}+y\hat{j} \\ =\hat{i}\left ( 1-x \right )+\hat{j}\left ( -2-y \right )$
We have $\vec{AB}=\vec{DC}$
Then:
$5\hat{i}+\hat{j}=\left ( 1-x \right )\hat{i}+\left ( -2-y \right )\hat{j}$
Then comparing coefficient of $\hat{i}$ and $\hat{j}$
$\begin{matrix} 5=1-x & 1=-2-y\\ x=1-5 &y=-2-1 \\ x=-4 & y=-3 \end{matrix}$
Hence the coordinates of D is (-4, -3)

Algebra of Vectors exercise 22.5 question 7
Answer:

$\hat{i}-5\hat{j}$
Hint: position vector of point (x , y) is given by where $\hat{i}$ and $\hat{j}$ are the position in x an y direction.
Given:
Point $A(3,4),B(5,-6),C(4,-1)$
Find $\vec{a}+2\vec{b}-3\vec{c}$
Solution:
As we know position vector of point (x, y) is given by $\left (x\hat{i}+y\hat{j} \right )$
Where $\hat{i}$ and $\hat{j}$ are the position vector in x and y direction
Then,
$\vec{A}=\left (3\hat{i}+4\hat{j} \right )\\ \vec{B}=\left (5\hat{i}+(-6)\hat{j} \right )=5\hat{i}-6\hat{j}\\ \vec{C}=\left (4\hat{i}+(-1)\hat{j} \right )=4\hat{i}-\hat{j}\\ 2\vec{B}=2\left (5\hat{i}-6\hat{j} \right )=10\hat{i}-12\hat{j}\\ 3\vec{C}=3\left (4\hat{i}-\hat{j} \right )=12\hat{i}-3\hat{j}$
$\vec{a}+2\vec{b}-3\vec{c}$ Substitute value of $\vec{a},2\vec{b},3\vec{c}$ in equation $\vec{a}+2\vec{b}-3\vec{c}$
$=3\hat{i}+4\hat{j}+10\hat{i}-12\hat{j}-\left ( 12\hat{i}-3\hat{j} \right )\\ =3\hat{i}+4\hat{j}+10\hat{i}-12\hat{j}- 12\hat{i}+3\hat{j}\\ =\hat{i}\left ( 3+10-12 \right )+\hat{j}\left ( 4-12+3 \right )\\ =\hat{i}\left ( 1 \right )+\hat{j}\left ( -5 \right )\\ =\hat{i}-5\hat{j}$

Algebra of Vectors exercise 22.5 question 8
Answer:

(9, -4)
Hint: position vector $\vec{AB}$ is given by $\vec{AB}$ =position vector of B- position vector of A
Given: coordinates of A(4, -1) , a $\vec{a}$ is the position vector whose tip i is (5, -3)
Solution:
Let O be the origin of point (0, 0)
Position P(5, -3) be the tip of the position vector $\vec{a}$
Then the $\vec{a}$ , $\vec{OP}$ =Position vector of p- position vector of o.
As we know position vector of point (x, y) is given by $\left ( x\hat{i} + y\hat{j}\right )$ where $\hat{i}$ and $\hat{j}$ are the unit vectors.
Then
$\vec{P}=5\hat{i}+-3\hat{j}=5\hat{i}-3\hat{j}\\ \vec{O}=0\hat{i}+0\hat{j} \\ \vec{OP}=5\hat{i}-3\hat{j}-0\hat{i}+0\hat{j}\\ \vec{OP}=5\hat{i}-3\hat{j}-0\hat{i}-0\hat{j}\\ =5-0\hat{i}+-3-0\hat{j}\\ \vec{OP}=5\hat{i}-3\hat{j}$
Let the coordinate of B be (x,y) and A has coordinate (4,-1)
Therefore
$\vec{A}=4\hat{i}+-1\hat{j}=4\hat{i}-\hat{j}\\ \vec{B}=x\hat{i}+y\hat{j}$
AB= $\vec{AB}$ Position vectors of B – position vector of A
$\vec{AB}=x\hat{i}+y\hat{j}-\left (4\hat{i}-\hat{j} \right )\\ =x\hat{i}+y\hat{j}-4\hat{i}+\hat{j}\\ =\left (x-4 \right )\hat{i}+\hat{j}\left (y+1 \right )$
Now,
$\vec{AB}=\vec{a}\\ \left ( x-4 \right )\hat{i}+\left ( y+1 \right )\hat{j}=5\hat{i}-3\hat{j}$
Comparing coefficient of $\hat{i}$ & $\hat{j}$
$x-4=5\\ x=5+4\\ x=9\\ x+1=-3\\ y=-3-1\\ y=-4$
Hence, the coordinate of B are (9,-4)

Algebra of Vector exercise 22.5 question 9
Answer:

the magnitude of AB and AC is equal
Hence the points $2\hat{i},-\hat{i}-4\hat{j}$ and $-\hat{i}+4\hat{j}$ form an isosceles triangle
Hint: the magnitude of $\left |\vec{a} \right |=\sqrt{x^{2}+y^{2}}$
Position vector $\vec{AB}$ is given by the $\vec{AB}$ = position vector of B - position vector of A
Given: the points A,B,C with position vectors $\vec{a},\vec{b},\vec{c}$ respectively
Also,
$\vec{a}=2\hat{i}\\ \vec{b}=-\hat{i}-4\hat{j}\\ \vec{c}=-\hat{i}+4\hat{j}$
Solution:
Now we find the position vector $\vec{AB}$
Then,
$\vec{AB}=\vec{b}-\vec{a}\\ =\left (-\hat{i}-4\hat{j} \right )-2\hat{i}\\ =-\hat{i}-4\hat{j}-2\hat{i}\\ \vec{AB}=-3\hat{3}-4\hat{j}$
Now
$\left |\vec{AB} \right |=\sqrt{x^{2}+y^{2}}$
Here x=-3 and y= -4
$=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5$
$\vec{BC}$ =Position vector of C – Position vector of B
$=\left ( -\hat{i}+4\hat{j} \right )-\left ( -\hat{i}-4\hat{j} \right )\\ = -\hat{i}+4\hat{j}+\hat{i}+4\hat{j} \\ =0\hat{i}+8\hat{j}\\ =8\hat{j}$
$\vec{AC}$ =Position vector of C – position vector of A
$=\left ( -\hat{i}+4\hat{j} \right )-\left ( 2\hat{i} \right )\\ = -\hat{i}+4\hat{j}-\hat{j} \\ =-3\hat{i}+4\hat{j}\\$
Now
$\left |\vec{AC} \right |=\sqrt{x^{2}+y^{2}}$
Here x=-3 and y=4
$=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5$
Since, the magnitude of AB and AC are equal
Hence, the points $2\hat{i},-\hat{i}-4\hat{j}$ and $-\hat{i}+4\hat{j}$ form an isosceles triangle

Algebra of Vector exercise 22.5 question 10

Answer: $\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$
Hint: a unit vector parallel to
$\vec{a}=\hat{a}=\frac{\vec{a}}{\left | \vec{a} \right |}\\ \left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}$
Given: $\vec{a}=\hat{i}+\sqrt{3}\hat{j}$
Solution:
We can find the magnitude of $\vec{a}$
$\left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}$
Here x = 1 and $y=\sqrt{3}$
$\left |\vec{a} \right |=1^{2}+(\sqrt{3})^{2}\\ =\sqrt{1+3}=4\\ \left |\vec{a} \right |=2$
Unit vector parallel to $\vec{a}=\hat{a}=\frac{\vec{a}}{\left | \vec{a} \right |}\\$
$=\frac{1}{2}\left ( \hat{i}+\sqrt{3}\hat{j} \right )\\ =\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$

Algebra of Vector exercise 22.5 question 11
Answer:

$\lambda=8$ and $\mu =-5$
Hint: here we use sector formula
$\frac{\left ( mx_{2}+nx_{1} \right )}{m+n},\frac{\left ( my_{2}+ny_{1} \right )}{m+n}$
Where m,n are the given ratios
Given: the position vectors of points A, B and C are
$\vec{A}=\lambda \hat{i}+3\hat{j}\\\vec{B}=12\hat{i}+\mu \hat{j}\\\vec{C}=11\hat{i}-3\hat{j}$
Solution:
It is given that C, divides the line segment joining A and B in the ratio 3:1
$\begin{aligned} &11 \hat{\imath}-3 \hat{\jmath}=3 \times(12 \hat{\imath}+\mu \hat{\jmath})+1(\lambda \hat{\imath}+3 \hat{\jmath}) \\ &11 \hat{\imath}-3 \hat{\jmath}=\frac{36 \hat{\imath}+3 u \hat{\jmath}+\lambda \hat{\imath}+3 \hat{\jmath}}{4} \\ \end{aligned}$
$\begin{aligned}&11 \hat{\imath}-3 \hat{\jmath}=\frac{(36+\lambda) \hat{\imath}+(3 \mu+3) \hat{\jmath}}{4} \\ &4(11 \hat{\imath}-3 \hat{\jmath})=(36+\lambda) \hat{\imath}+(3 u+3) \hat{\jmath} \\ &44 \hat{\imath}-12 \hat{\jmath}=(36+\lambda) \hat{\imath}+(3 u+3) \hat{\jmath} \end{aligned}$
Equating the corresponding components, we get;
$36+\lambda=44\\ \lambda=44-36\\ \lambda=8$
and
$3\mu+3=-12\\ 3\mu=-12-3=-15\\ \mu=\frac{-15}{3}=-5\\ \mu=-5$
Thus, the values of $\lambda$ and $\mu$ are 8 and -5 respectively

Algebra of vectors exercise 22.5, question 12 (i)
Answer:

: The components of $\vec{P}$ along x axis is a vector of magnitude 3 along the positive direction of x axis
The component of $\vec{P}$ along y axis is a vector of magnitude 2 along the positive direction of Y axis
$3\hat{i},2\hat{j}$
Hint: position vector of point (x, y) is given by $x\hat{i}+y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are the unit vector in x and y direction
Given: P(3,2)
Solution:
As we know, position vector of a point (x, y) is given by $x\hat{i}+y\hat{j}$
Then,
Position vector of a point (3, 2) is $\vec{p}=3\hat{i}+2\hat{j}$
The components of $\vec{P}$ along x axis is a vector of magnitude 3 along the positive direction of x axis
The component of $\vec{P}$ along y axis is a vector of magnitude 2 along the positive direction of Y axis

Algebra of vectors exercise 22.5, question 12 (ii)
Answer:

The components of $\vec{Q}$ along x axis is a vector of magnitude 5 along the negative direction of x axis
The component of $\vec{Q}$ along y axis is a vector of magnitude 1 along the positive direction of Y axis
$-5\hat{i},\hat{j}$
Hint: position vector of point (x, y) is given by $x\hat{i}+y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are the unit vector in x and y direction
Given:Q=(-5,1)
Solution:
As we know, position vector of a point (x, y) is given by $x\hat{i}+y\hat{j}$
So
$\vec{Q}=-5\hat{i}+1\hat{j}\\ \vec{Q}=-5\hat{i}+\hat{j}\\$
The components of $\vec{Q}$ along x axis is a vector of magnitude 5 along the negative direction of x axis
The component of $\vec{Q}$ along y axis is a vector of magnitude 1 along the positive direction of Y axis

Algebra of vectors exercise 22.5, question 12 (iii)
Answer:

The components of $\vec{R}$ along x axis is a vector of magnitude 11 along the negative direction of x axis
The component of $\vec{R}$ along y axis is a vector of magnitude 9 along the negative direction of Y axis
$-11\hat{i},-9\hat{j}$
Hint: position vector of point (x, y) is given by $x\hat{i}+y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are the unit vector in x and y direction
Given:R=(-11,-9)
Solution:
As we know, position vector of a point (x, y) is given by $x\hat{i}+y\hat{j}$
So
$\vec{R}=-11\hat{i}+(-9)\hat{j}\\ \vec{R}=-11\hat{i}-9\hat{j}\\$
The components of $\vec{R}$ along x-axis is a vector of magnitude 11 along the negative direction of x axis
The component of $\vec{R}$ along y-axis is a vector of magnitude 9 along the negative direction of Y axis

Algebra of vectors exercise 22.5, question 12 (iv)
Answer:

The components of $\vec{S}$ along x axis is a vector of magnitude 4 along the positive direction of x axis
The component of $\vec{S}$ along y axis is a vector of magnitude 3 along the negative direction of Y axis
$4\hat{i},-3\hat{j}$
Hint: position vector of point (x, y) is given by $x\hat{i}+y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are the unit vector in x and y direction
Given:S=(-4,-3)
Solution:
As we know, position vector of a point (x, y) is given by $x\hat{i}+y\hat{j}$
So
$\vec{S}=4\hat{i}+(-3)\hat{j}\\ \vec{S}=4\hat{i}-3\hat{j}\\$
The components of $\vec{S}$ along x-axis is a vector of magnitude 4 along the positive direction of x-axis
The component of $\vec{S}$ along y-axis is a vector of magnitude 3 along the negative direction of Y-axis

Class 12th RD Sharma Chapter 22 Exercise 22.5 Solutions are crafted by a team of experts who always thrive to make perfect solutions for the students. The benefits of grabbing these solutions are:

Crafted by experts:

RD Sharma Class 12th Exercise 22.5 solutions are created by a team of experts who take the utmost care to format perfect answers for students according to the latest marking pattern and CBSE guidelines. Every student must practice RD Sharma Class 12th Exercise 22.5 to score well without difficulty.

2. The best source of preparation:

Because of their high quality, RD Sharma Class 12th Exercise 22.5 solutions are the best to use. The most recent NCERT and CBSE updates make it an excellent choice. Furthermore, our experts are familiar with the marking scheme and can provide solutions accordingly, making it easier to get good grades.

3. Questions from the NCERT:

The NCERT is the teacher's favorite book from where they ask the questions, so it is necessary to learn NCERT concepts and solve questions. Rd Sharma Class 12th Chapter 22 Exercise 22.5 solutions help students understand and learn concepts, as well as answer questions about RD Sharma Class 12 Solutions Chapter 22 Exercise 22.5.

4. Different approaches to answering questions

As we know that a team of subject experts is behind these solutions, a student will find alternatives to solve a single question. As a result, the experts have devised novel approaches to answering a single question, making us unique.

5. Performance as a benchmark

The solutions help students perform better on exams. RD Sharma Class 12 Solutions Ex. 22.5 covers all of the essential questions found in exams, allowing students to establish a benchmark score.

6. Free of cost:

Students will find all of these solutions for free at Career360, the best website to visit for all of the benefits associated with board exam questions and competitive exams. With Career360, a student will not find exams scary at all. RD Sharma solutions These solutions have benefited thousands of students, making Career360 the ideal choice for all students.

RD Sharma Chapter wise Solutions

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

Frequently Asked Question (FAQs)

1. How many questions are there in each exercise of RD Sharma Solutions for Class 12 Maths Chapter 22?

The number of questions in each exercise of RD Sharma Solutions for Class 12 Maths Chapter 22 ais

Exercise 22.1 - 24 questions

Exercise 22.2 - 13 questions

Exercise 22.3 - 7 questions

Exercise 22.4 - 6 questions

Exercise 22.5 - 16 questions

Exercise 22.6 - 20 questions

Exercise 22.7 - 14 questions

Exercise 22.8 - 18 questions

Exercise 22.9 - 16 questions

MCQ - 27 questions

VSA - 52 questions

FBQ - 23 questions

2. What is a cross product in the Algebra of vectors?

The multiplication sign(x) between two vectors denotes a cross product. In a three-dimensional system, it is a binary vector operation. If P and Q are two vectors independent of each other, the result of their cross product (P x Q) is perpendicular to both vectors and normal to the plane containing both vectors. It is symbolized by;

P x Q = |P| |Q| sin θ

3. What do you mean by scalar and vector quantities?

Scalar quantities are those quantities that have only magnitude with no direction whereas vector quantities have both magnitude and direction.

4. Where can I find RD Sharma solutions for Class 12 Mathematics online?

At Career360, all the solutions for RD Sharma are present and you can visit the website for downloading the PDF free of cost. The material is a reliable source for preparation.

5. What are the basic concepts of vectors?

Position vectors and Direction Cosines are the basic concepts of vectors. You can visit Career36the 0 for the latest updated solutions for RD Sharma.