NCERT Class 12 Mathematics Chapter 10 Vector Algebra

NCERT Class 12 Mathematics Chapter 10 Vector Algebra Solutions, NCERT Solutions For Class 12 Maths, NCERT Class 12 Mathematics Chapter 10 Vector Algebra Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 12 Mathematics Chapter 10 Vector Algebra Question Answer and select needs one.

NCERT Class 12 Mathematics Chapter 10 Vector Algebra

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Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 12 Mathematics Chapter 10 Vector Algebra Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 12 Mathematics Chapter 10 Vector Algebra Solutions for All Subject, You can practice these here.

Vector Algebra

Chapter – 10

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

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2. Classify the following measures as scalars and vectors.

(i) 10 kg.

Ans: 10kg is a scalar.

(ii) 2 meters north-west.

Ans: 2 meters north-west is a vector.

(iii) 40°

Ans: 40° is a scalar.

(iv) 40 watt.

Ans: 40 watts is a scalar.

(v) 10⁻¹⁹ coulomb.

Ans: 10⁻¹⁹  Coulomb is a scalar.

(vi) 20 m/s²

Ans: 20m/s² is a vector

3. Classify the following as scalar and vector quantities.

(i) time period.

Ans: Time period is a scalar.

(ii) distance.

Ans: Distance is a scalar.

(iii) force.

Ans: Force is a vector.

(iv) velocity.

Ans: Velocity is a vector.

(v) work done.

Ans: Work done is a scalar.

4. In Fig 10.6 (a square), identify the following vectors.

(i) Coinitial.

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(ii) Equal.

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(iii) Collinear but not equal.

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5. Answer the following as true or false.

Ans: True.

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

Ans: False.

(iii) Two vectors having same magnitude are collinear.

Ans: False.

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

Ans: False.

1. Compute the magnitude of the following vectors:

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2. Write two different vectors having same magnitude.

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3. Write two different vectors having same direction.

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4. Find the values of x and y so that the vectors 2î + 3ĵ and xî +yĵ are equal.

Ans: It is given that the vectors 2î + 3ĵ and xî + yĵ are equal.

Therefore,

2î + 3ĵ = xî + yĵ

On comparing the components of both sides

⇒ x = 2

⇒ y = 3

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

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

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10. Find a vector in the direction of vector 5î – ĵ + 2k̂ which has magnitude 8 units.

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11. Show that the vectors 2î -3ĵ + 4k̂ and – 4î + 6ĵ – 8k̂ are collinear.

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12. Find the direction cosines of the vector î + 2 ĵ + 3 k̂.

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

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14. Show that the vector î + ĵ + k̂ is equally inclined to the axes OX, OY and OZ.

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15. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are î +2 ĵ – k̂ and – î + ĵ + k̂ respectively, in the ratio 2:1

(i) internally.

(ii) externally.

Ans: The position vector of point R dividing the line segment joining two points P and Q in the ratio m: n is given by:

(i) Internally:

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

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18. In triangle ABC (Fig 10.18), which of the following is not true:

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Hence, the equation given in alternative C is incorrect. 

The correct answer is C.

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2. Find the angle between the vectors hat î -2ĵ +3k̂ and 3î -2ĵ + k̂.

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3. Find the projection of the vector î – ĵ on the vector î + ĵ.

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4. Find the projection of the vector î +3ĵ +7k̂ on the vector 7î – ĵ + 8k̂.

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

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

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So, each of the vector is a unit vector.

Hence,

So, the vectors are mutually perpendicular to each other.

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Ans: Consider the given vectors,

Hence, the value is – 3/2.

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

Ans: we know,

Hence, the angle is cos⁻¹ (10 /√102).

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

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17. Show that the vectors 2î – ĵ + k̂ , î – 3ĵ – 5k̂ and 3î – 4ĵ – 4k̂ form the vertice of a right angled triangle.

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(A) λ = 1

(Β) λ = – 1

(C) a = |λ|

(D) a = 1 / |λ|

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Ans: we have,

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Now,

|a| = 1

4. Show that:

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⇒ î(6μ – 27λ) – ĵ (2 μ – 27) + k̂ (2λ – 6) = 0î + 0ĵ +0k̂

On comparing the corresponding components, we have:

6μ – 27λ = 0

2μ – 27 = 0

2λ – 6 = 0

Now,

2λ – 6 = 0 ⇒ λ = 3

Ans: 

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= î[a₂ (b₃ + c₃) – a₃(b₂ + c₂)] – ĵ [a₁ (b₃ + c₃) – a₃ (b₁ +c₁)] + k̂ [a₁ (b₂ + c₂) – a₂ (b₁ + c₁)] 

= î[a₂ b₃ + a₂ c₃ – a₃ b₂ – a₃ c₂] + ĵ [- a₁ b₃ – a₁ c₃ + a₃ b₁ + a₃ c₁] + k̂ [a₁ b₂ + a₁ c₂ – a₂ b₁ – a₂ c₁] … (1)

On adding (2) and (3), we get:

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

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

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

10. Find the area of the parallelogram whose adjacent sides are determined by the vectors

Ans: The area of the parallelogram whose adjacent sides are:

Adjacent sides are given as:

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

(Α) π / 6

(Β) π / 4

(C) π / 3

(D) π / 2

Ans: 

(A) 1/2

(B) 1

(C) 2

(D) 4

Ans: The position vectors of vertices A, B, C, and D of rectangle ABCD are given as:

Now, it is known that the area of a parallelogram whose adjacent sides are:

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

Ans: 

2. Find the scalar components and magnitude of the vector joining the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂).

Ans: The vector joining the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) can be obtained by,

Hence, the scalar components and the magnitude of the vector joining the given points are respectively {(x₂ – x₁), (y₂ – y₁), (z₂ – z₁)} and √(x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z₁)².

3. A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.

Ans: Let O and B be the initial and final positions of the girl respectively.

Then, the girl’s position can be shown as:

Now, we have:

Hence, the girl’s displacement from her initial point of departure is

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5. Find the value of x for which x ( î + ĵ + k̂ ) is a unit vector.

Ans: x(î + ĵ + k̂) is a unit vector if |x(î + ĵ + k̂)| = 1

Now, 

|x ( î + ĵ + k̂ )| = 1 

⇒ √x² + x² + x² = 1 

⇒ √3x² = 1 

⇒ √3 x = 1 

⇒ x = ± 1/√3

Hence, the required value of x is ± 1/√3

6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors

Ans: We have,

Ans: We have,

8. Show that the points A (1, – 2, – 8), B (5, 0, -2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Ans: The given points are A (1, -2, -8), B (5, 0, -2), and C (11, 3, 7).

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

Now, let point B divide AC in the ratio λ : 1. Then, we have:

⇒ (λ + 1) (5 î – 2 k̂ ) = 11 λ î + 3 λ ĵ +7 λ k̂ + î – 2 ĵ – 8 k̂

⇒ 5 ( λ + 1) î – 2( λ + 1) k̂ = (11 λ + 1) î +(3 λ – 2) ĵ + (7 λ – 8) k̂

On equating the corresponding components, we get:

5(λ + 1) = 11λ + 1

⇒ 5λ + 5 = 11λ + 1

⇒ 6λ = 4

⇒ λ = 4/6 = 2/3

Hence, point B divides AC in the ratio 2:3.

9. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are externally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

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10. The two adjacent sides of a parallelogram are 2 î – 4 ĵ + 5 k̂ and î – 2 ĵ – 3 k̂ . Find the unit vector parallel to its diagonal. Also, find its area.

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Hence, the area of the parallelogram is 11√5 square units.

11. Show that the direction cosines of a vector equally inclined to the axes OΧ, ΟΥ and OZ are

Ans: Let a vector be equally inclined to axes OX, OY, and OZ at angle a. 

Then, the direction cosines of the vector are cos a , cos a , and cos a. 

Now,

cos² α + cos² α + cos² α = 1 

⇒ 3 cos² α = 1 

⇒ cos α = 1/√3

Hence, the direction cosines of the vector which are equally inclined to the axes are 1/√3 ,1/√3 , 1/√3.

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13. The scalar product of the vector î + ĵ + k̂ with a unit vector along the sum of vectors 2î + 4ĵ – 5k̂ and λ î + 2ĵ + 3k̂ is equal to one. Find the value of λ.

Ans: (2 î + 4 ĵ – 5 k̂ ) + ( λ î + 2 ĵ + 3 k̂) 

= (2 + λ) î + 6 ĵ – 2 k̂

Therefore, unit vector along ( 2 î + 4 ĵ – 5 k̂ ) + ( λ î + 2 ĵ + 3 k̂ ) is given as:

⇒ √λ² + 4λ + 44 = λ + 6 

⇒ λ² + 4λ + 44 = (λ + 6)²

⇒ λ² + 4λ + 44 = λ² + 12λ + 36 

⇒ 8λ = 8 

⇒ λ = 1 

Hence, the value of λ is 1.

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Choose the correct answer in Exercises 16 to 19.

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The correct answer is D.

18. The value of î.( ĵ × k̂ ) + ĵ. ( î × k̂ ) + k̂. ( î × ĵ ) is

(A) 0

(B) -1

(C) 1

(D) 3

Ans: î.( ĵ × k̂ ) + ĵ. ( î × k̂ ) + k̂. ( î × ĵ )

= î . î + ĵ . (- ĵ ) + k̂ . k̂

= 1 – ĵ. ĵ + 1

= 1 – 1 + 1

= 1

The correct answer is C.

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