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NCERT exemplar Class 12 Maths solutions chapter 10 Vector Algebra helps in answering questions like What is your weight? How should a footballer hit the ball to score a goal? The answers to both these questions would be very different. The answer to the former could be 50 kg, a quantity which has the presence of only one value. The second question's answer is quantities that consist of muscle strength, the direction of the foot and the ball, the air current, and the other players' distance. Such quantities are called Vectors and are discussed in detail in NCERT exemplar Class 12 Maths solutions chapter 10.
Question:1
Find the unit vector in the direction of sum of vectors
Answer:
We have,
Since, unit vector is needed to be found in the direction of the sum of vectors and .
So, add vectors and .
Let,
Substituting the values of vectors and .
We know that a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:
Thus, unit vector in the direction of sum of vectors is .
Question:2(i)
If find the unit vector in the direction of
Answer:
We have,
(i). We need to find the unit vector in the direction of .
First, let us calculate .
As we have,
We can easily multiply vector by a scalar by multiplying similar components, that is, vector’s magnitude by the scalar’s magnitude.
We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:
Now we know the value of , so just substitute the value in the above equation.
Thus, unit vector in the direction of is
Question:2(ii)
If find the unit vector in the direction of
Answer:
We need to find the unit vector in the direction of
First, let us calculate .
As we have,
Then multiply equation (a) by 2 on both sides,
We can easily multiply vector by a scalar by multiplying similar components, that is, vector’s magnitude by the scalar’s magnitude.
Subtract (b) from (c). We get
We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:
Question:3
Answer:
We have,
Coordinates of P is (5, 0, 8).
Coordinates of Q is (3, 3, 2).
So,
Position vector of P is given by,
To find unit vector in the direction of PQ, we need to find position vector of PQ.
Position vector of PQ is given by,
We know that a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:
Thus, unit vector in the direction of PQ is
Question:4
Answer:
We have been given that,
Position vector of A
Thus, position vector of point C is
Question:5
Answer:
Let the points be A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3).
Let us find the position vectors of these points.
Assume that O is the origin.
Position vector of A is given by,
Know that, two vectors are said to be collinear, if they lie on the same line or parallel lines.
Since, A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3) are collinear, we can say that:
Sum of modulus of any two vectors will be equal to the modulus of third vector.
This means, we need to find
To find :
Position vector of B-Position vector of A
= Position vector of C-Position vector of A
Substitute values of from (i), (ii) and (iii) respectively. We get,
Thus, value of k is -2.
Question:6
A vector is inclined at equal angles to the three axes. If the magnitude of is units, find .
Answer:
Given that,
Magnitude of =
Also, given that
Vector is equally inclined to the three axes.
This means, direction cosines of the unit vector will be same. The direction cosines are (l, m, n).
The direction cosines of a vector are simply the cosines of the angles between the vector and the three coordinate axes.
We know the relationship between direction cosines is,
Also, we know that is represented in terms of direction cosines as,
Question:7
Answer:
Given that,
Magnitude of vector = 14
Also, direction ratios = 2 : 3 : -6
Also can be defined as,
Know that, the direction cosines of a vector are the cosines of the angles between the vector and the three coordinate axes.
∴, the direction cosines l, m and n are
Since, makes an acute angle with x-axis, then k will be positive.
Thus, the direction cosines (l, m, n) are ; and the components of are (4,6,-12)
Question:8
Find a vector of magnitude 6, which is perpendicular to both the vectors and
Answer:
Let the vectors be and , such that
We need to find a vector perpendicular to both the vectors and
Any vector perpendicular to both and can be given as,
Let
As we know, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
So,
Question:10
If . Interpret the result geometrically?
Answer:
Given that,
Now, let us interpret the result graphically.
Let there be a parallelogram ABCD.
Here, and .
And AB and AD sides are making angle θ between them.
Area of parallelogram is given by,
Area of parallelogram = Base × Height
So from the diagram, area of parallelogram ABCD can be written as,
Since, parallelogram on the same base and between the same parallels are equal in area, so we have
Question:11
Find the sine of the angle between the vectors
Answer:
We have
Let these vectors be represented as,
Comparing the vectors, we get
To find sine of the angle between the vectors, we can first find out cosine of the angle between them.
Cosine of the angle between is given by,
By algebraic identity, we have
Thus, sine of the angle between the vectors is .
Question:12
If A, B, C, D are the points with position vectors
respectively, find the projection of along .
Answer:
Given are points, A, B, C and D.
Let O be the origin.
We have,
Position vector of A
Multiply numerator and denominator by √21.
Question:13
Answer:
We have,
The coordinates of points A, B and C are (1, 2, 3), (2, -1, 4) and (4, 5, -1) respectively.
We need to find the area of this triangle ABC.
We have the formula given as,
Thus, area of yriangle ABC is .
Question:14
Answer:
We have,
Given:
There are more than 1 parallelogram, and their bases can be taken as common and they are between same parallels.
To Prove:
These parallelograms whose bases are same and are between the same parallel sides have equal area.
Proof:
Let ABCD and ABFE be two parallelograms on the same base AB and between same parallel lines AB and DF.
Here,
AB || DC and AE || BF
We can represent area of parallelogram ABCD as,
Now, area of parallelogram ABFE can be represented as,
Area of parallelogram ABFE
⇒Area of parallelogram ABFE …(ii)
From equation (i) and (ii), we can conclude that
Area of parallelogram ABCD = Area of parallelogram ABFE
Thus, parallelogram on same base and between same parallels are equal in area.
Hence, proved.
Question:15
Answer:
Given:
a, b, c are magnitudes of the sides opposite to the vertices A, B, C respectively.
⇒ AB = c, BC = a and CA = b
To Prove:
In triangle ABC,
Construction: We have constructed a triangle ABC and named the vertices according to the question.
Note the height of the triangle, BD.
If ∠BAD = A
Then, BD = c sin A
Proof:
Here, components of c which are:
c sin A
c cos A
are drawn on the diagram.
Using Pythagoras theorem which says that,
(hypotenuse)^{2} =(perpendicular)^{2} + (base)^{2}
Take triangle BDC, which is a right-angled triangle.
Here,
Hypotenuse = BC
Base = CD
Perpendicular = BD
We get,.
Hence proved
Question:16
Answer:
Let are vertices of a triangle ABC.
Also, we get
Position vector of A
Position vector of B
Position vector of C
We need to show that,
gives the vector are of the triangle.
We know that,
Vector area of triangle ABC is given as,
Thus, shown.
We know that, two vectors are collinear if they lie on the same line or parallel lines.
For to be collinear, area of the ?ABC should be equal to 0.
⇒ Area of ?ABC = 0
Thus, this is the required condition for to be collinear.
Now, we need to find the unit vector normal to the plane of the triangle.
Let be the unit vector normal to the plane of the triangle.
Thus, unit vector normal to the plane of the triangle is
Question:17
Answer:
We have,
Let ABCD be a parallelogram.
In ABCD,
And since, AD || BCWe need to show that,
We know that,Vector area of parallelogram ABCD is given by,
Area of parallelogram ABCD
Hence, shown.
Now, we need to find the area of parallelogram whose diagonals are
We have already derived the relationship between area of parallelogram and diagonals of parallelogram, which is
Question:18
If find a vector .
Answer:
Given that,
Comparing Left Hand Side and Right Hand Side, we get
From coefficient of ⇒ z-y = 0 …(i)
From coefficient of ⇒ -(z-x) = 1
⇒ x-z = 1 …(ii)
From coefficient of ⇒ y-x = -1
⇒ x-y = 1 …(iii)
Now, add equations (ii) and (iii), we getAdd equations (iv) and (v), we get
Question:19
The vector in the direction of the vector that has magnitude 9 is
Answer:
C)
Given is the vector
Let this vector be , such that
Let us first find the unit vector in the direction of this vector .
We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
Unit vector in the direction of the vector is given as,
Therefore,
We have found unit vector in the direction of the vector , but we need to find the unit vector in the direction of but also with the magnitude 9.
We have the formula:
Vector in the direction of with a magnitude of 9
Question:20
The position vector of the point which divides the join of points in the ratio 3 : 1 is
Answer:
D)
We are given points
Let these points be
.
Also, given in the question that,
A point divides AB in the ratio of 3 : 1.
Let this point be C.
⇒ C divides AB in the ratio = 3 : 1
We need to find the position vector of C.
We know the position vector of a point C dividing the line segment joining the points P and Q, whose position vectors are p and q in the ratio m : n internally is given by,
According to the question, here
m : n = 3 : 1
⇒ m = 3 and n = 1
.
Question:21
The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively is
Answer:
C)
Let initial point be A(2,5,0) and terminal point be B(-3,7,4).So, the required vector joining A and B is the vector .
Question:22
The angle between two vectors and with magnitudes and 4, respectively, and is
Answer:
Answer :(B)
Given that,
Let θ be the angle between vector a and b.
Question:23
Find the value of λ such that the vectors are orthogonal
A. 0
B. 1
C.
D.
Answer:
D)
Given that, are orthogonal.
Question:24
The value of λ for which the vectors are parallel is
A.
B.
C.
D.
Answer:
Answer:(A)
Given that, are parallel
Question:25
The vectors from origin to the points A and B are respectively, then the area of triangle OAB is
A. 340
B.
C.
D.
Answer:
Answer :(D)
Given that, vector from origin to the point A, and vector from origin to the point B,
Question:29
If are unit vectors such that , then the value of is
A. 1
B. 3
C.
D. None of these
Answer:
Answer :(C)
Question:30
Answer:
Answer :(A)
Let θ be the angle between
From figure we can see that, is the projection of
Question:31
If are three vectors such that and then value of is
A. 0
B. 1
C. –19
D. 38
Answer:
Answer :(C)
Given that, and
Question:32
If and , then the range of is
A. [0, 8]
B. [–12, 8]
C. [0, 12]
D. [8, 12]
Answer:
Answer :(C)
Given that, and ,
Question:33
The number of vectors of unit length perpendicular to the vectors is
A. one
B. two
C. three
D. infinite
Answer:
Answer :(B)
Given that ,
Now, a vector which is perpendicular to both is given by
Hence, there are two unit length perpendicular to the .
Question:34
Fill in the blanks
The vector bisects the angle between the non-collinear vectors and if ________
Answer:
Let and are two non-collinear vectors.
Let bisects the angle between and .
Thus, the vector bisects the angle between the non-collinear vectors and if they are equal.
Question:35
Fill in the blanks
If and for some non-zero vector , then the value of is _________
Answer:
Given that, and for some non-zero vector
Question:37
Fill in the blanks
The values of k for which is parallel to holds true are _______.
Answer:
Given that,
⇒ k cannot be equal to , otherwise it will become null vector and then it will not be parallel to .
Since, k is along the direction of and not in its opposite direction.
Question:42
True and False
Position vector of a point P is a vector whose initial point is origin.
Answer:
True
Explanation:
Consider a point P in space, having coordinates (x, y, z) with respect to the origin O(0, 0, 0). Then, the vector having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O.
Question:43
True and False
If , then the vectors and are orthogonal.
Answer:
True
Explanation:
Given that,
On squaring both the sides, we get
Hence, and are orthogonal.
Question:45
True and False
If and are adjacent sides of a rhombus, then .
Answer:
False
Explanation:
Given that, are perpendicular to each other.
But, adjacent sides of rhombus are not perpendicular.
NCERT exemplar Class 12 Maths chapter 10 solutions would help achieve academic excellence, be it in the 12 board exams or additional exams, get rid of doubts in fundamental concepts, and promote intellectual curiosity. The students can also use NCERT exemplar Class 12 Maths solutions chapter 10 PDF download, to access the solutions offline.
Sub-topics Covered in NCERT exemplar solutions for Class 12 Maths chapter 10 Vector Algebra
The sub-topics that are covered under the Class 12 Maths NCERT exemplar solutions chapter 10 Vector Algebra are:
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Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | Vector Algebra |
Chapter 11 | |
Chapter 12 | |
Chapter 13 |
This basic maths chapter covers the topics like vectors, properties, vector by scalar, types of vector etc.
Students appearing for the 12 board exams and those who are preparing from entrance exams for engineering can make use of NCERT exemplar Class 12 Maths solutions chapter 10.
One can use NCERT exemplar Class 12 Maths solutions chapter 10 pdf download by an online webpage to pdf converter
Our experts have covered all the questions and have given NCERT exemplar solutions for Class 12 Maths chapter 10 for main exercise of the chapter.
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hello,
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I hope this was helpful!
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