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NCERT Solutions Class 12 Maths Chapter 10 Vector Algebra is prepared by the experts in a very concise manner and every concept of our NCERT Solutions is discussed in complete detail. NCERT Solutions Class 12 have complete solutions of intext questions, exercise questions, etc.

  • Vector Algebra – vector algebra is a very important topic of algebra it also helps us understand the algebra of different vector quantities. There are mainly two types of physical quantities namely scalars and vectors. Such quantities which have only magnitude are called the scalar quantities and such quantities which have magnitude, as well as direction, are called the vector quantities. Algebra is one of the important fundamental operations used in the field of mathematics. To express the terms of algebra we use different significant symbols which help us signify different quantities. All these different symbols are also used for expressing equations and formulae to perform the algebraic operation. Vector algebra has many different branches. In vector algebra, we carry out the different algebraic operations based on vector spaces.
  • Some Basic Concepts of Vector – such quantities which have only magnitude are known as scalar quantities. Some examples of scalar quantities are mass, length, speed, area, etc. on the contrary the vector quantities are such quantities that have both magnitude and direction. Example of vector quantities is acceleration, force, weight, momentum, etc. vector algebra have a different application in physics and engineering.
  • Types of Vectors – the vectors are categorized based on their magnitude, direction, and their relationship with different other vectors.  
  •  Zero vectors – such vectors which have 0 magnitudes are called zero vectors. Zero vectors have no magnitude and no direction. Zero vectors are also known as the identity of vectors.

NCERT Solutions Class 12 Maths will help students get a complete understanding of all the concepts.

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NCERT Solutions Class 12 Maths Chapter 10 Vector Algebra

1. Represent graphically a displacement of 40 km, 30° east of north.

Answer:

Here, vector \(\vec {OP}\) represents the displacement of 40 km, 30° East of North.

2. Classify the following measures as scalars and vectors. 

(i) 10 kg 

(ii) 2 metres north-west 

(iii) 40° 

(iv) 40 watt 

(v) 10–19 coulomb 

(vi) 20 m/s2

Answer:

(i) 10 kg is a scalar quantity because it involves only magnitude. 

(ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction. 

(iii) 40° is a scalar quantity as it involves only magnitude. 

(iv) 40 watts is a scalar quantity as it involves only magnitude. 

(v) 10–19 coulomb is a scalar quantity as it involves only magnitude. 

(vi) 20 m/s2 is a vector quantity as it involves magnitude as well as direction.

3. Classify the following as scalar and vector quantities. 

(i) time period 

(ii) distance 

(iii) force 

(iv) velocity 

(v) work done

Answer:

(i) Time period is a scalar quantity as it involves only magnitude. 

(ii) Distance is a scalar quantity as it involves only magnitude. 

(iii) Force is a vector quantity as it involves both magnitude and direction. 

(iv)Velocity is a vector quantity as it involves both magnitude as well as direction. 

(v) Work done is a scalar quantity as it involves only magnitude.

4. In Figure, identify the following vectors.

(i) Coinitial 

(ii) Equal 

(iii) Collinear but not equal

Answer:

(i) Vectors \(\vec a\) and \(\vec d\) are coinitial because they have the same initial point. 

(ii) Vectors \(\vec b\) are \(\vec d\) equal because they have the same magnitude and direction. 

(iii) Vectors \(\vec a\)and \(\vec c\) are collinear but not equal. This is because although they are parallel, their directions are not the same.

5. Answer the following as true or false.

(i) \(\vec a\) and \(-\vec a\) are collinear. 

(ii) Two collinear vectors are always equal in magnitude. 

(iii) Two vectors having same magnitude are collinear. 

(iv) Two collinear vectors having the same magnitude are equal.

Answer:

(i) True. 

Vectors \(\vec a\) and \(-\vec a\) are parallel to the same line. 

(ii) False. 

Collinear vectors are those vectors that are parallel to the same line. 

(iii) False.

Two vectors having the same magnitude need not necessarily be parallel to the same line.

(iv) False.

Only if the magnitude and direction of two vectors are the same, regardless of the positions of their initial points the two vector are said to be equal.

6. Compute the magnitude of the following vectors:

Answer:

The given vectors are:

7. Write two different vectors having same magnitude.

Answer:

Hence\(\vec a\) and \(\vec b\) are two different vectors having the same magnitude. 

The vectors are different because they have different directions.

8. Write two different vectors having same direction.

Answer:

The direction cosines of \(\vec p\) and \(\vec q\)are the same. Hence, the two vectors have the same direction.

9. Find the values of x and y so that the vectors \(2\hat i + 3\hat j \) and \(x\hat i + y\hat j\) are equal

Answer:

The two vectors \(2\hat i + 3\hat j \) and \(x\hat i + y\hat j\) will be equal if their corresponding components are equal. 

Hence, the required values of x and y are 2 and 3 respectively.

10. Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).

Answer:

The vector with the initial point P (2, 1) and terminal point Q (–5, 7) can be given by,

Hence, the required scalar components are –7 and 6 while the vector components are \(-7\hat i \) and \(6\hat j\).

11. Find the sum of the vectors

Answer:

The given vectors are

12. Find the unit vector in the direction of the vector \(\vec a =\hat i + \hat j + 2\hat k\).

Answer:

The unit vector \(\hat a\) in the direction of vector \(\vec a =\hat i + \hat j + 2\hat k\)

13. Find the unit vector in the direction of vector \(\vec{PQ}\), where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.

Answer:

The given points are P (1, 2, 3) and Q (4, 5, 6).

Hence, the unit vector in the direction of \(\vec{PQ}\) is

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14. For given vectors, \(\vec a = 2\hat i - \hat j + 2\hat k\) and \(\vec b= -\hat i + \hat j-\hat k\), find the unit vector in the direction of the vector \(\vec a + \vec b\).

Answer:

The given vectors are \(\vec a = 2\hat i - \hat j + 2\hat k\) and \(\vec b= -\hat i + \hat j-\hat k\) 

Hence, the unit vector in the direction of \((\vec a + \vec b)\) is

15. Find a vector in the direction of vector \(5\hat i - \hat j + 2\hat k\) which has magnitude 8 units.

Answer:

Hence, the vector in the direction of vector \(5\hat i - \hat j + 2\hat k\) which has magnitude 8 units is given by,

16. Show that the vectors \( 2\hat i - 3\hat j + 4\hat k\) and \(-4\hat i + 6\hat j-8\hat k\) are collinear.

Answer:

Hence, the given vectors are collinear.

17. Find the direction cosines of the vector \( \hat i + 2\hat j + 3\hat k\).

Answer:

Hence, the direction cosines of

18. Find the direction cosines of the vector joining the points A (1, 2, –3) and B (–1, –2, 1) directed from A to B.

Answer:

The given points are A (1, 2, –3) and B (–1, –2, 1).

Hence, the direction cosines of \(\vec{AB}\) are

19. Show that the vector \(\hat i + \hat j + \hat k\) is equally inclined to the axes OX, OY, and OZ.

Answer:

Let \(\vec a=\hat i + \hat j + \hat k\) 

Then,

\(|\vec a| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt3\)

Therefore, the direction cosines of

Now, let α, β, and γ be the angles formed by \(\vec a\) with the positive directions of x, y, and z axes. 

Then, we have

Hence, the given vector is equally inclined to axes OX, OY, and OZ.

20. Find the position vector of the mid point of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).

Answer:

The position vector of mid-point R of the vector joining points P (2, 3, 4) and Q (4, 1, – 2) is given by,

21. Show that the points A, B and C with position vectors,

\(\vec a = 3\hat i - 4\hat j - 4\hat k, \vec b= 2\hat i - \hat j + \hat k \;and \; \vec c = \hat i - 3\hat j -5\hat k \)

respectively form the vertices of a right angled triangle.

Answer:

Position vectors of points A, B, and C are respectively given as:

Hence, ABC is a right-angled triangle.

22. In triangle ABC which of the following is not true:

Answer:

On applying the triangle law of addition in the given triangle, we have:

From equations (1) and (3), we have:

Hence, the equation given in alternative C is incorrect. 

The correct answer is C.

23. If \(\vec a\) and \(\vec b\) are two collinear vectors, then which of the following are incorrect: 

A. \(\vec b = \lambda\vec a\), for some scalar λ 

B. \(\vec a = \pm \vec b\)

C. the respective components of \(\vec a\) and \(\vec b\) are proportional

D. both the vectors  \(\vec a\) and \(\vec b\) have same direction, but different magnitudes

Answer:

If  \(\vec a\) and \(\vec b\)are two collinear vectors, then they are parallel. 

Therefore, we have: 

\(\vec b = \lambda\vec a\)(For some scalar λ)

Thus, the respective components of  \(\vec a\) and \(\vec b\) are proportional. 

However, vectors \(\vec a\) and \(\vec b\) can have different directions. 

Hence, the statement given in D is incorrect. 

The correct answer is D.

24. Find the angle between two vectors \(\vec a\) and \(\vec b\) with magnitudes and 2, respectively having \(\vec a.\vec b = \sqrt 6\).

Answer:

It is given that,

Hence, the angle between the given vectors \(\vec a\) and \(\vec b\) is \(\frac\pi4.\)

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25. Find the angle between the vectors \(\hat i - 2\hat j +3 \hat k\) and \(3\hat i - 2\hat j + \hat k\).

Answer:

The given vectors are \(\hat i - 2\hat j +3 \hat k\) and \(3\hat i - 2\hat j + \hat k\).

26. Find the projection of the vector \(\hat i - \hat j\) on the vector \(\hat i + \hat j\).

Answer:

Let 

\(\vec a = \hat i - \hat j\) 

\(\vec b = \hat i + \hat j\) 

Now, projection of vector on is given by,

Hence, the projection of vector \(\vec a\) on \(\vec b\) is 0.

27. Find the projection of the vector \(\hat i + 3\hat j + 7\hat k\) on the vector \(7\hat i - \hat j + 8\hat k\).

Answer:

Let \(\vec a=\hat i + 3\hat j + 7\hat k\) and \(\vec b = 7\hat i - \hat j + 8\hat k\)

Now, projection of vector on is given by,

28. Show that each of the given three vectors is a unit vector:

Also, show that they are mutually perpendicular to each other.

Answer:

Thus, each of the given three vectors is a unit vector.

Hence, the given three vectors are mutually perpendicular to each other.

29. Find 

Answer:

30. Evaluate the product.

\((3\vec a - 5\vec b). (2\vec a + 7 \vec b)\)

Answer:

31. Find the magnitude of two vectors \(\vec a\) and \(\vec b\), having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac12\).

Answer:

Let θ be the angle between the vectors \(\vec a\) and \(\vec b\)

It is given that 

32. Find \(|\vec x|\), if for a unit vector \(\vec a, (\vec x - \vec a). (\vec x + \vec a) = 12\).

Answer:

33. If \(\vec a = 2\hat i + 2\hat j + 3\hat k, \vec b= -\hat i + 2\hat j + \hat k\) and \(\vec c= 3\hat i + \hat j\) are such that \(\vec a + \lambda\vec b\) is perpendicular to \(\vec c\), then find the value of λ.

Answer:

Hence, the required value of λ is 8.

34. Show that:

\(|\vec a|\vec b + |\vec b|\vec a\)  is perpendicular to \(|\vec a|\vec b - |\vec b| \vec a\), for any two nonzero vectors \(\vec a\) and \(\vec b\).

Answer:

Hence, \(|\vec a|\vec b + |\vec b|\vec a\) and \(|\vec a|\vec b - |\vec b| \vec a\) are perpendicular to each other.

35. If \(\vec a.\vec a = 0\) and \(\vec a.\vec b = 0\), then what can be concluded about the vector \(\vec b\)?

Answer:

Hence, vector \(\vec b\) satisfying \(\vec a.\vec b = 0\) can be any vector.

36. If either vector \(\vec a = \vec 0 \;or\;\vec b = \vec 0\), then \(\vec a.\vec b=0\). But the converse need not be true. Justify your answer with an example.

Answer:

We now observe that:

Hence, the converse of the given statement need not be true.

37. If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ ABC. [∠ ABC is the angle between the vectors \(\vec {BA}\) and \(\vec {BC}\)]

Answer:

The vertices of ∆ABC are given as A (1, 2, 3), B (–1, 0, 0), and C (0, 1, 2). 

Also, it is given that ∠ ABC is the angle between the vectors \(\vec {BA}\) and \(\vec {BC}\) .

Now, it is known that:

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38. Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear.

Answer:

The given points are A (1, 2, 7), B (2, 6, 3), and C (3, 10, –1).

Hence, the given points A, B, and C are collinear.

39. Show that the vectors \(2\hat i - \hat j + \hat k, \hat i - 3\hat j - 5\hat k\) and \(3\hat i - 4\hat j - 4\hat k\) form the vertices of a right angled triangle.

Answer:

Let vectors \(2\hat i - \hat j + \hat k, \hat i - 3\hat j - 5\hat k\) and \(3\hat i - 4\hat j - 4\hat k\) be position vectors of points A, B, and C respectively.

Hence, ∆ABC is a right-angled triangle.

40. If \(\vec a\) is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ\(\vec a\) is unit vector if 

(A) λ = 1 

(B) λ = –1 

(C) \(a = |\lambda|\)

(D) \(a = \frac1{|\lambda|}\)

Answer:

Vector λ\(\vec a\) is a unit vector if \(|\lambda \vec a| = 1\).

Hence, vector λ\(\vec a\) is a unit vector if \(a = \frac1{|\lambda|}\)

The correct answer is D.

41. Find

\(|\vec a\times \vec b|,\) if \(\vec a = \hat i - 7\hat j + 7\hat k\) and \(\vec b= 3\hat i - 2\hat j + 2\hat k\).

Answer:

We have,

42. Find a unit vector perpendicular to each of the vector \(\vec a + \vec b \) and \(\vec a- \vec b\), where \(\vec a = 3\hat i + 2\hat j + 2\hat k\) and \(\vec b = \hat i + 2\hat j - 2\hat k\).

Answer:

We have,

Hence, the unit vector perpendicular to each of the vectors \(\vec a + \vec b \) and \(\vec a - \vec b\) is given by the relation,

43. Show that

\((\vec a - \vec b) \times (\vec a + \vec b) = 2(\vec a\times \vec b)\)

Answer:

44. Find λ and µ if \((2\hat i + 6\hat j + 27\hat k)\times (\hat i + \lambda\hat j + \mu\hat k) = \vec 0\).

Answer:

On comparing the corresponding components, we have:

45. Given that 

\(\vec a.\vec b = 0 \) and \(\vec a \times \vec b = \vec 0\) 

What can you conclude about the vectors \(\vec a\) and \(\vec b\)?

Answer:

\(\vec a.\vec b = 0 \)

Then,

(i) Either \(|\vec a| = 0\) or \(|\vec b| = 0\), or \(\vec a \perp \vec b\) (in case \(\vec a\) and \(\vec b\) are non-zero) 

(ii) Either \(|\vec a| = 0\)  or \(|\vec b| = 0\), or \(\vec a \parallel \vec b\)  (in case \(\vec a\) and \(\vec b\) are non-zero) 

But, \(\vec a\) and \(\vec b\) cannot be perpendicular and parallel simultaneously. 

Hence, \(|\vec a| = 0\)  or \(|\vec b| = 0\).

46. If either \(\vec a = \vec 0\) or \(\vec b= \vec 0\), then \(\vec a\times \vec b= \vec0\).

Is the converse true? Justify your answer with an example.

Answer:

Take any parallel non-zero vectors so that \(\vec a\times \vec b= \vec0\).

It can now be observed that:

Hence, the converse of the given statement need not be true.

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47. Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Answer:

The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5), and C (1, 5, 5).

The adjacent sides \(\vec {AB}\) and \(\vec {BC}\) of ∆ABC are given as:

Hence, the area of ∆ABC is \(\frac{\sqrt{61}}{2}\) square units.

48. Find the area of the parallelogram whose adjacent sides are determined by the vector

\(\vec a = \hat i- \hat j + 3\hat k \) and \(\vec b = 2\hat i - 7\hat j + \hat k\).

Answer:

The area of the parallelogram whose adjacent sides are \(\vec a\)and \(\vec b\) is \(|\vec a \times \vec b|\).

Adjacent sides are given as:

Hence, the area of the given parallelogram is 15√2 square units.

49. Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.

Answer:

If \(\vec r\) is a unit vector in the XY-plane, then \(\vec r = cos\theta \hat i + sin\theta\hat j\).

Here, θ is the angle made by the unit vector with the positive direction of the x-axis. 

Therefore, for θ = 30°:

\(\vec r = cos30° \hat i + sin30° \hat j = \frac{\sqrt3}{2}\hat i + \frac12 \hat j \)

Hence, the required unit vector is \(\frac{\sqrt3}{2}\hat i + \frac12 \hat j\).

50. Find the value of x for which \(x (\hat i + \hat j + \hat k)\) is a unit vector.

Answer:

\(x (\hat i + \hat j + \hat k)\) is a unit vector if \(|x (\hat i + \hat j + \hat k)| = 1\) 

Now,

\(|x(\hat i + \hat j + \hat k)| = 1\) 

⇒ \(\sqrt{x^2 + x^2 + x^2} = 1\)

⇒ \(\sqrt{3x^2} = 1\)

⇒ \(\sqrt3 x = 1\)

⇒ \(x = \pm \frac1{\sqrt3}\)

Hence, the required value of x is \(\pm \frac1{\sqrt3}\).

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