NCERT Class 12 Mathematics Chapter 10 Vector Algebra

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NCERT Class 12 Mathematics Chapter 10 Vector Algebra

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Vector Algebra

Chapter – 10

Exercise 10.1

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.

EXERCISE 10.2

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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EXERCISE 10.3

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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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EXERCISE 10.4

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

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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

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(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:

Miscellaneous Exercise on Chapter 10

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

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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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