RD Sharma Class 12 Exercise 22.5 Algebra of vectors Solutions Maths - Download PDF Free Online
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.
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
5 Hints: Given: Position vector of a point (-4,-3) be Find: We know position vector of a point(x,y) is given by where and are units vectors in x and y directions. Now we need to find magnitude of i.e Here, x=-4 and y=-3 thus
Hints: Given: Position vector of point (12,n) be Find: We know position vector of a point(x,y) is given by where and are units vectors in x and y directions. Then: Here, x=12 and y=n [given] Squaring both sides
Hint: Position vector is given x, y = 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 where and are unit vector in x and y direction Then the position vector is given by = position vector of B - position vector of A So,
Hint: Position vector is given x, y = 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 where and are unit vector in x and y direction Then the position vector is given by = position vector of B - position vector of A -
(-1,-2) Hint: Position vector is given = position vector B- position vector of A Given: Coordinate of A (-1, 3) Coordinates of B (-2,1) Solution: & We know position vector of point (x, y) is given by Let 0 be the origin Let P(x,y) be the second point Then is the tip of the position vector of the point p We have: position vector of B- position vector of A = position vector of B= position vector of A given then x=-1 and y= -2 Hence coordinate of the required point is (-1,-2)
the coordinates of D is (-4,-3) Hint: Position vectors 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 As we know position vector of point (x, y) is given by where and are position vector of x, and y direction. Then the position vector is given by = position vector of B - position vector of A Then we find = position vector C- position vector of D We have Then: Then comparing coefficient of and Hence the coordinates of D is (-4, -3)
Hint: position vector of point (x , y) is given by where and are the position in x an y direction. Given: Point Find Solution: As we know position vector of point (x, y) is given by Where and are the position vector in x and y direction Then, Substitute value of in equation
(9, -4) Hint: position vector is given by =position vector of B- position vector of A Given: coordinates of A(4, -1) , 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 Then the , =Position vector of p- position vector of o. As we know position vector of point (x, y) is given by where and are the unit vectors. Then Let the coordinate of B be (x,y) and A has coordinate (4,-1) Therefore AB= Position vectors of B – position vector of A Now, Comparing coefficient of & Hence, the coordinate of B are (9,-4)
the magnitude of AB and AC is equal Hence the points and form an isosceles triangle Hint: the magnitude of Position vector is given by the = position vector of B - position vector of A Given: the points A,B,C with position vectors respectively Also, Solution: Now we find the position vector Then, Now Here x=-3 and y= -4 =Position vector of C – Position vector of B =Position vector of C – position vector of A Now Here x=-3 and y=4 Since, the magnitude of AB and AC are equal Hence, the points and form an isosceles triangle
and Hint: here we use sector formula Where m,n are the given ratios Given: the position vectors of points A, B and C are Solution: It is given that C, divides the line segment joining A and B in the ratio 3:1 Equating the corresponding components, we get; and Thus, the values of and are 8 and -5 respectively
: The components of along x axis is a vector of magnitude 3 along the positive direction of x axis The component of along y axis is a vector of magnitude 2 along the positive direction of Y axis Hint: position vector of point (x, y) is given by where and 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 Then, Position vector of a point (3, 2) is The components of along x axis is a vector of magnitude 3 along the positive direction of x axis The component of along y axis is a vector of magnitude 2 along the positive direction of Y axis
The components of along x axis is a vector of magnitude 5 along the negative direction of x axis The component of along y axis is a vector of magnitude 1 along the positive direction of Y axis Hint: position vector of point (x, y) is given by where and 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 So The components of along x axis is a vector of magnitude 5 along the negative direction of x axis The component of along y axis is a vector of magnitude 1 along the positive direction of Y axis
The components of along x axis is a vector of magnitude 11 along the negative direction of x axis The component of along y axis is a vector of magnitude 9 along the negative direction of Y axis Hint: position vector of point (x, y) is given by where and 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 So The components of along x-axis is a vector of magnitude 11 along the negative direction of x axis The component of along y-axis is a vector of magnitude 9 along the negative direction of Y axis
The components of along x axis is a vector of magnitude 4 along the positive direction of x axis The component of along y axis is a vector of magnitude 3 along the negative direction of Y axis Hint: position vector of point (x, y) is given by where and 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 So The components of along x-axis is a vector of magnitude 4 along the positive direction of x-axis The component of 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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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.
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