RD Sharma Class 12 Exercise 22.5 Algebra of vectors Solutions Maths - Download PDF Free Online
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.
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:
5Hints: $\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$
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:
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:
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:
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:
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:
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:
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 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 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 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 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:
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RD Sharma Chapter wise Solutions
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
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