Class 12 - Maths - Vector Algebra

Exercise 10.1

Question 1:

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

From the figure, vector OP represents the displacement of 40 km, 30° east of north.

Question 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

(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.

Question 3:

Classify the following as scalar and vector quantities.

(i) time period              (ii) distance              (iii) force           (iv) velocity           (v) work done

(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.

Question 4:

In Figure, identify the following vectors.

(i) Coinitial                        (ii) Equal                        (iii) Collinear but not equal

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

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

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

directions are not the same.

Question 5:

Answer the following as true or false.

(i) a and –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.

(i) True.

Vectors a and –a are parallel to the same line.

(ii) False.

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

(iii) False.

Exercise 10.2

Question 1:

Compute the magnitude of the following vectors:

a = i + j + k, b = 2i – 7j – 3k, c = i/√3 + j/√3 - k/√3

Given vectors are:

a = i + j + k, b = 2i – 7j – 3k, c = i/√3 + j/√3 - k/√3

Now,

|a| = √{(1)2 + (1)2 + (1)2} = √(1 + 1 + 1) = √3

|b| = √{(2)2 + (-7)2 + (-3)2} = √{4 + 49 + 9} = √62

|c| = √{(1/√3)2 + (1/√3)2 + (1/√3)2} = √{1/3 + 1/3 + 1/3} = 1

Question 2:

Write two different vectors having same magnitude.

Let a = i - 2j + 3k and b = 2i + j – 3k

Now, |a| = √{(1)2 + (-2)2 + (3)2} = √{1 + 4 + 9} = √14

and |b| = √{(2)2 + (1)2 + (-3)2} = √{4 + 1 + 9} = √14

Hence, a and b are two different vectors because they have different directions.

The vectors are different because they have different directions.

Question 3:

Write two different vectors having same direction.

Let a = i + j + k and b = 2i + 2j + 2k

The direction cosine of a are given by

l = 1/√{(1)2 + (1)2 + (1)2} = 1/√3

m = 1/√{(1)2 + (1)2 + (1)2} = 1/√3

n = 1/√{(1)2 + (1)2 + (1)2} = 1/√3

The direction cosine of b are given by

l = 2/√{(2)2 + (2)2 + (2)2} = 2/2√3 = 1/√3

m = 2/√{(2)2 + (2)2 + (2)2} = 2/2√3 = 1/√3

n = 2/√{(2)2 + (2)2 + (2)2} = 2/2√3 = 1/√3

The direction cosines of vectors a and bare the same.

Hence, the two vectors have the same directions.

Question 4:

Find the values of x and y so that the vectors 2i + 3j and xi + yj are equal.

The two vectors will be equal if their corresponding components are equal.

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

Question 5:

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

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

PQ = (-5 - 2)I + (7 - 1)j

=> PQ = -7i + 6j

Hence, the required scalar components are –7 and 6 while the vector components are -7i and

6j.

Question 6:

Find the sum of the vectors a = i – 2j + k, b = -2i + 4j + 5k and c = i – 6j – 7k

Given vectors are: a = i – 2j + k, b = -2i + 4j + 5k and c = i – 6j – 7k

Now, a + b + c = i – 2j + k + -2i + 4j + 5k + i – 6j – 7k

= (1 – 2 + 1)i + (-2 + 4 - 6)j + (1 + 5 - 7)k

= 0.i - 4j + 1.k

= -4j – k

Question 7:

Find the unit vector in the direction of the vector a = i + j + 2k.

The unit vector a in the direction of vector a = i + j + 2k

Now, |a| = √{(1)2 + (1)2 + (2)2} = √{1 + 1 + 4} = √6

So, a^ = a/|a| = (i + j + 2k)/ √6 = i/√6 + j/√6 + 2k/√6

Question 8:

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

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

Now, PQ = (4 - 1)i + (5 - 2)j + (6 - 3)k = 3i + 3j + 3k

|PQ| = √{(3)2 + (3)2 + (3)2} = √{9 + 9 + 9} = 3√3

Hence, the unit vector in the direction of PQ is

PQ/|PQ| = (3i + 3j + 3k)/3√3 = i/√3 + j/√3 + k/√3

Question 9:

For given vectors, a = 2i – j + 2k and b = -i + j - k, find the unit vector in the direction of the vector a + b.

The given vectors are: a = 2i – j + 2k and b = -i + j - k

Now, a + b = (2i – j + 2k) + (-i + j – k) = i + k

|a + b| = √{(1)2 + (1)2} = √2

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

(a + b)/|a + b| = (i + k)/√2 = i/√2 + k/√2

Question 10:

Find a vector in the direction of vector 5i – j + 2k which has magnitude 8 units.

Let a = 5i – j + 2k

|a| = √{(5)2 + (-1)2 + (2)2} = √{25 + 1 + 4} = √30

So, a^ = a/|a| = (5i – j + 2k)/√30

Hence, the vector in the direction of vector 5i – j + 2k which has magnitude 8 units is given by

8a = 8(5i – j + 2k)/√30 = 40i/√30 – 8j/√30 + 16k/√30

Question 11:

Show that the vectors 2i – 3j + 4k and -4i + 6j – 8k are collinear.

Let a = 2i – 3j + 4k and b = -4i + 6j – 8k

Now, b = -4i + 6j – 8k = -2(2i – 3j + 4k) = -2a

=> b = ka, where k = -2

Hence, the given vectors are collinear.

Question 12:

Find the direction cosines of the vector i + 2j + 3k

Let a = i + 2j + 3k

|a| = √{(1)2 + (2)2 + (3)2} = √{1 + 4 + 9} = √14

Hence, the direction cosines of vector a are (1/√14, 2√14, 3/√14)

Question 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.

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

So, AB = (-1 -1)i + (-2 -2)j + (1 + 3)k

=> AB = -2i – 4j + 4k

|AB| = √{(-2)2 + (-4)2 + (4)2} = √{4 + 16 + 16} = √36 = 6

Hence, the direction cosines of AB are (-2/6, -4/6, 4/6) = (-1/3, -2/3, 2/3)

Question 14:

Show that the vector i + j + k is equally inclined to the axes OX, OY, and OZ.

Let a = i + j + k

|a| = √{(1)2 + (1)2 + (1)2} = √3

Therefore, the direction cosines of a are (1/√3, 1/√3, 1/√3)

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

Then, we have

cos α = 1/√3, cos β = 1/√3 and cos γ = 1/√3

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

Question 15:

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are i + 2j – k and –i + j + k respectively, in the ratio 2 : 1                                        (i) internally                                     (ii)externally

The position vector of point R dividing the line segment joining two points P and Q in the ratio

m : n is given by:

1. Internally: (mb + na)/(m + n)
2. Externally: (mb - na)/(m - n)

Position vectors of P and Q are given as:

P = i + 2j – k and Q = –i + j + k

(i) The position vector of point R which divides the line joining two points P and Q internally in

the ratio 2 : 1 is given by,

OR = {2(–i + j + k) + 1(i + 2j – k)}/(2 + 1)

= {(–2i + 2j + 2k) + (i + 2j – k)}/3

= (-i + 4j + k)/3

= -i/3 + 4j/3 + k/3

(ii) The position vector of point R which divides the line joining two points P and Q externally in

the ratio 2 : 1 is given by,

OR = {2(–i + j + k) - 1(i + 2j – k)}/(2 - 1)

= (–2i + 2j + 2k) - (i + 2j – k)

= -3i + 3k

Question 16:

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

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

given by,

OR = {(2i + 3j + 4k) + (4i + j – 2k)}/2

= {(2 + 4)i + (3 + 1)j + (4 - 2)k}/2

= (6i + 4j + 2k)/2

= 3i + 2j + k

Question 17:

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

a = 3i – 4j – 4k, b = 2i – j + k and c = i – 3j – 5k

respectively form the vertices of a right angled triangle.

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

a = 3i – 4j – 4k, b = 2i – j + k and c = i – 3j – 5k

Now,

AB = b – a = (2 - 3)i + (-1 + 4)j + (1 + 4)k = -i + 3j + 5k

BC = c – b = (1 - 2)i + (-3 + 1)j + (-5 - 1)k = -i - 2j - 6k

CA = a – c = (3 - 1)i + (-4 + 3)j + (-4 + 5)k = 2i - j + k

|AB|2 = (-1)2 + (3)2 + (5)2} = 1 + 9 + 25 = 35

|BC|2 = (-1)2 + (-2)2 + (-6)2} = 1 + 4 + 36 = 41

|CA|2 = (2)2 + (-1)2 + (1)2} = 4 + 1 + 1 = 6

Now, |AB|2 + |CA|2 = 35 + 6 = 42 = |BC|2

Hence, ABC is a right-angled triangle.

Question 18:

In triangle ABC which of the following is not true:

1. AB + BC + CA = 0
2. AB + BC - AC = 0
3. AB + BC - CA = 0
4. AB - CB + CA = 0

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

AB + BC = AC  ………1

=> AB + BC = -CA

=> AB + BC + CA = 0  ……2

Hence, the equation given in alternative A is true.

AB + BC = AC

=> AB + BC - AC = 0

Hence, the equation given in alternative B is true.

From equation 2, we get

=> AB - CB + CA = 0

Hence, the equation given in alternative D is true.

Now, consider the equation given in alternative C:

=> AB + BC - CA = 0

=> AB + BC = CA   ………3

From equation 1 and 3, we get

AC = CA

=> AC = -AC

=> AC + AC = 0

=> 2AC = 0

=> AC = 0, which is not true.

Hence, the equation given in alternative C is incorrect.

Therefore, the correct answer is option C.

Question 19:

If a and b are two collinear vectors, then which of the following are incorrect:

1. b = λa, for some scalar λ
2. a = ±b
3. the respective components of a and b are proportional
4. both the vectors a and b have same direction, but different magnitudes

If a and b are two collinear vectors, then they are parallel.

Therefore, we have:

b = λa      (For some scalar λ)

If λ = ±1, the a = ±b

If a = a1i + a2j + a3k and b = b1i + b2j + b3k, then

b = λa

=> b1i + b2j + b3k = λ(a1i + a2j + a3k)

=> b1i + b2j + b3k = (λa1)i + (λa2)j + (λa3)k

=> b1 = λa1, b2 = λa2, b3 = λa3

=> b1/a1 = b2/a2 = b3/a3 = λ

Thus, the respective components of a and b are proportional.

However, vectors a and b can have different directions.

Hence, the statement given in D is incorrect.

So, the correct answer is option D.

Exercise 10.3

Question 1:

Find the angle between two vectors a and b with magnitudes √3 and 2, respectively having    a.b = √6

It is given that,

|a| = √3, |b| = 2 and a.b = √6

Now, we know that a.b = |a||b| cos θ

=> √6 = √3 * 2 * cos θ

=> √3 * √2 = √3 * 2 * cos θ

=> √2/2 = cos θ

=> cos θ = 1/√2

=> cos θ = cos π/4

=> θ = π/4

Hence, the angle between the given vectors a and b is π/4.

Question 2:

Find the angle between the vectors i – 2j +3 k and 3i – 2j + k

The given vectors are a = i – 2j + k and b = 3i – 2j + k.

|a| = √{(1)2 + (-2)2 + (3)2} = √{1 + 4 + 9} = √14

|b| = √{(3)2 + (-2)2 + (1)2} = √{9 + 4 + 1} = √14

Now, a.b = (i – 2j + k).(3i – 2j + k)

= 1 * 3 + (2) * (2) + 3 * 1

= 3 + 4 + 3

= 10

Now, we know that a.b = |a||b| cos θ

=> 10 = √14 * √14 * cos θ

=> 10 = 14 * cos θ

=> 10/14 = cos θ

=> cos θ = 5/7

=> θ = cos-1 (5/7)

Hence, the angle between the given vectors i – 2j +3 k and 3i – 2j + k is cos-1 (5/7).

Question 3:

Find the projection of the vector i - j on the vector i + j.

Let a = i - j and b = i + j

Now, projection of vector a on b is given by,

(a.b)/|b| = {1.1 + (-1).1}/√(1 + 1) = (1 - 1)/ √2 = 0

Hence, the projection of vector a on b is 0.

Question 4:

Find the projection of the vector i + 3j + 7k on the vector 7i - j + 8k.

Let a = i + 3j + 7k and b = 7i - j + 8k

Now, projection of vector a on b is given by,

(a.b)/|b| = {1.1 + 3.(-1) + 7.8}/√{72 + (-1)2 + 82} = (7 – 3 + 56)/ √(49 + 1 + 64) = 60/√114

Hence, the projection of vector i + 3j + 7k on the vector 7i - j + 8k is 60/√114

Question 5:

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

(2i + 3j + 6k)/7, (3i - 6j + 2k)/7, (6i + 2j - 3k)/7

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

Let a = (2i + 3j + 6k)/7 = 2i/7 + 3j/7 + 6k/7

b = (3i - 6j + 2k)/7 = 3i/7 - 6j/7 + 2k/7

c = (6i + 2j - 3k)/7 = 6i/7 + 2j/7 - 3k/7

|a| = √{(2/7)2 + (-6/7)2 + (2/7)2} = √(4/49 + 9/49 + 36/49) = √(49/49) = 1

|b| = √{(3/7)2 + (-6/7)2 + (2/7)2} = √(9/49 + 36/49 + 4/49) = √(49/49) = 1

|c| = √{(6/7)2 + (2/7)2 + (-3/7)2} = √(36/49 + 4/49 + 9/49) = √(49/49) = 1

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

a.b = (2/7) * (3/7) + (3/7) * (-6/7) + (6/7) * (2/7) = 6/49 – 18/49 + 12/49 = 0

b.c = (3/7) * (6/7) + (-6/7) * (2/7) + (2/7) * (-3/7) = 18/49 – 12/49 - 6/49 = 0

c.a = (6/7) * (2/7) + (2/7) * (3/7) + (-3/7) * (6/7) = 12/49 + 6/49 - 18/49 = 0

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

Question 6:

Find |a| and |b| if (a + b).(a - b) = 8 and |a| = 8|b|

Given, (a + b).(a - b) = 8

=> a.a – a.b + b.a – b.b = 8

=> |a|2 - |b|2 = 8

=> (8|b|)2 - |b|2 = 8

=> 64|b|2 - |b|2 = 8

=> 63|b|2 = 8

=> |b|2 = 8/63

=> |b| = √(8/63)

=> |b| = 2√2/3√7           [Magnitude of a vector is non-negative]

Now, |a| = 8|b|

=> |a| = (8 * 2√2)/3√7

=> |a| = 16√2/3√7

Question 7:

Evaluate the product: (3a – 5b).(2a + 7b)

Given, (3a – 5b).(2a + 7b)

= 3a . 2a + 3a . 7b – 5b . 2a – 5b . 7b

= 6a .a + 21a .b – 10a .b – 35b .b

= 6|a|2 + 11a .b – 35|b|2

Question 8:

Find the magnitude of two vectors a and b, having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.

Let θ be the angle between the vectors a and b

Given that |a| = |b|, a.b = 1/2 and θ = 60°   …………1

We know that a.b = |a||b| cos θ

=> 1/2 = |a||a| cos 60°

=> 1/2 = |a||a| * (1/2)

=> |a|2 = 1

=> |a| = 1

Hence, |a| = |b| = 1

Question 9:

Find |x|, if for a unit vector a, (x - a)(x + a) = 12

Given, (x - a)(x + a) = 12

=> x.x + x.a – a.x – a.a = 12

=> |x|2 - |a|2 = 12

=> |x|2 - 1 = 12            [|a| = 1 since a is a unit vetcor]

=> |x|2 = 12 + 1

=> |x|2 = 13

=> |x| = √13

Question 10:

If a = 2i + 2j + 3k, b = -i + 2j + k and c = 3i + j are such that a + λb is perpendicular to c, then find the value of λ.

The given vectors are a = 2i + 2j + 3k, b = -i + 2j + k and c = 3i + j

Now, a + λb = (2i + 2j + 3k) + λ(-I + 2j + k)

= (2 - λ)i + (2 + 2λ)j + (3 + λ)k

If a + λb is perpendicular to c, then

(a + λb).c = 0

=> [(2 - λ)i + (2 + 2λ)j + (3 + λ)k].(3i + j) = 0

=> 3(2 - λ) + 1(2 + 2λ) + 0(3 + λ) = 0

=> 6 - 3λ + 2 + 2λ = 0

=> - λ + 8 = 0

=> λ = 8

Hence, the required value of λ is 8.

Question 11:

Show that: |a|b + |b|a is perpendicular to |a|b - |b|a, for any two non-zero vectors a and b.

= (|a|b + |b|a).( |a|b + |b|a)

= |a|2 b.b - |a||b|b.a + |b||a|a.b - |b|2 a.a

= |a|2 |b|2 - |b|2 |a|2

= 0

Hence, |a|b + |b|a and |a|b - |b|a are perpendicular to each other.

Question 12:

If a.a = 0 and a.b = 0 then what can be concluded about the vector b?

Given that a.a = 0 and a.b = 0

Now, a.a = 0

=> |a|2 = 0

=> |a| = 0

So, a is a zero vector.

Hence, vector b satisfying a.b = 0 can be any vector.

Question 13:

If a, b, c are unit vectors such that a + b + c = 0, find the value of a ⋅ b + b ⋅ c + c ⋅ a

|a + b + c|2 = (a + b + c).(a + b + c)

=> |a + b + c|2 = |a|2 + |b|2 + |c|2 + 2(a.b + b.c + c.a)

=> 0 = 1 + 1 + 1 + 2(a.b + b.c + c.a)

=> 0 = 3 + 2(a.b + b.c + c.a)

=> 2(a.b + b.c + c.a) = -3

=> a.b + b.c + c.a = -3/2

Question 14:

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

Let a = 2i + 4j + 3k and b = 3i + 3j – 6k

Then a.b = (2i + 4j + 3k).(3i + 3j – 6k)

= 2.3 + 4.3 + 3(-6)

= 6 + 12 – 18

= 0

We now observe that:

|a| = √(22 + 42 + 32) = √29

So, |a| ≠ 0

|b| = √{32 + 32 + (-6)2} = √54

So, |b| ≠ 0

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

Question 15:

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 BA and BC]

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 BA and BC.

BA = {1 – (-1)}i + (2 - 0)j + (3 - 0)k = 2i + 2j + 3k

BC = {0 – (-1)}i + (1 - 0)j + (2 - 0)k = i + j + 2k

BA.BC = (2i + 2j + 3k).(i + j + 2k)

= 2 * 1 + 2 * 1 + 3 * 2

= 2 + 2 + 6

= 10

|BA| = √{22 + 22 + 32} = √(4 + 4 + 9) = √17

|BA| = √{12 + 12 + 22} = √(1 + 1 + 2) = √6

Now, BA.BC = |BA||BC| cos (∠ABC)

=> 10 = √17 * √6 * cos (∠ABC)

=> 10 = √102 * cos (∠ABC)

=> cos (∠ABC) = 10/√102

=> ∠ABC = cos-1(10/√102)

Question 16:

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

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

AB = (2 - 1)i + (6 - 2)j + (3 - 7)k = i + 4j - 4k

BC = (3 - 2)i + (10 - 6)j + (-1 - 3)k = i + 4j - 4k

CA = (3 - 1)i + (10 - 2)j + (-1 - 7)k = 2i + 8j - 8k

|AB| = √{12 + 42 + (-4)2} = √(1 + 16 + 16) = √33

|BC| = √{12 + 42 + (-4)2} = √(1 + 16 + 16) = √33

|AC| = √{22 + 82 + 82} = √(4 + 64 + 64) = 2√33

So, |AC| = |AB| + |BC|

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

Question 17:

Show that the vectors 2i – j + k, i – 3j – 5k and 3i – 4j – 4k form the vertices of a right angled triangle.

Let vectors 2i – j + k, i – 3j – 5k and 3i – 4j – 4k be position vectors of points A, B, and C

respectively.

i.e. OA = 2i – j + k, OB = i – 3j – 5k and OC = 3i – 4j – 4k

Now, vectors AB, BC and CA represent the sides of Δ ABC.

AB = (1 - 2)i + (-3 + 1)j + (-5 - 1)k = -i - 2j - 6k

BC = (3 - 1)i + (-4 + 3)j + (-4 + 5)k = 2i - j + k

AC = (2 - 3)i + (-1 + 4)j + (1 + 4)k = -i + 3j + 5k

|AB| = √{(-1)2 + (-2)2 + (-6)2} = √(1 + 4 + 36) = √41

|BC| = √{22 + (-1)2 + 12} = √(4 + 1 + 1) = √6

|AC| = √{(-1)2 + 32 + 52} = √(1 + 9 + 25) = √35

Now, |BC|2 + |AC|2 = 6 + 36 = 41 = |AB|2

Hence, ∆ABC is a right-angled triangle.

Question 18:

If is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λa is unit vector if

(A) λ = 1                   (B) λ = –1                    (C) a = |λ|              (D) a = 1/|λ|

Vector λa is a unit vector if

|λa| = 1

Now, |λa| = 1

=> |λ||a| = 1

=> |a| = 1/|λ|

=> a = 1/|λ|            [since |a| = a]

Hence, vector λa is a unit vector if a = 1/|λ|.

The correct answer is option D.

Exercise 10.4

Question 1:

Find: |a * b| if a = i – 7j + 7k and b = 3i – 2j + 2k

Given, a = i – 7j + 7k and b = 3i – 2j + 2k

Now, a * b =   i       j       k

1     -7     7

3      -2   -2

= i(-14 + 14) + j(2 - 21) + k(-2 + 21)

= 19j + 19k

So, |a * b|= √{192 + 192} = √(2 * 192 ) = 19√2

Question 2:

Find a unit vector perpendicular to each of the vector a + b and a - b, where a = 3i + 2j + 2k and  b = i + 2k – 2k.

Given, a = 3i + 2j + 2k and b = i + 2k – 2k

So, a + b = 4i + 4j and a – b = 2i + 4k

Now, (a + b) * (a - b) =   i        j        k

4       4       0

2       0       4

= i(16) – j(16) + k(-8)

= 16i – 16j – 8k

So, |(a + b)*(a - b)| = √{162 + (-16)2 + (-8)2

= √(256 + 256 + 64)

= √576

= 24

Hence, the unit vector perpendicular to each of the vectors a + b and a - b is given by the

relation,

= ± {(a + b) * (a - b)}/|(a + b) * (a - b)|

= ±(16i – 16j – 8k)/24

= ±(2i – 2j –k)/3

= ±2i/3 ± 2i/3 ± k/3

Question 3:

If a unit vector makes an angle π/3 with i, π/4 with j and an acute angle θ with k, then find θ and hence, the compounds of a.

Let the unit vector a have (a1, a2, a3) components i.e.

a = a1i + a2j + a3k

Since a is a unit vector, so |a| = 1.

Also, it is given that a makes angles π/3 with i, π/4 with j and an acute angle θ with k

Then we have

cos π/3 = a1/|a|

=> 1/2 = a1               [Since |a| = 1]

cos π/4 = a2/|a|

=> 1/√2 = a2               [Since |a| = 1]

Also, cos θ = a3/|a|

=> a3 = cos θ              [Since |a| = 1]

Now, |a| = 1

=> √(a12 + a22 + a32) = 1

=> a12 + a22 + a32 = 1

=> (1/2)2 + (1/√2)2 + (cos θ)2 = 1

=> 1/4 + 1/2 + cos2 θ = 1

=> 3/4 + cos2 θ = 1

=> cos2 θ = 1 – 3/4

=> cos2 θ = 1/4

=> cos θ = 1/2

=> θ = π/3

Hence, a3 = π/3

Therefore, θ = π/3 and the component of a are (1/2, 1/√2, 1/2).

Question 4:

Show that (a - b) * (a + b) = 2(a * b)

Given, (a - b) * (a + b) = (a - b) * a + (a - b) * b

= a * a – b * a + a * b – b * b

= 0 + a * b + a * b – 0

= 2(a * b)

Question 5:

Find λ and μ if (2i + 6j + 27k) * (i + λj + μk) = 0

(2i + 6j + 27k) * (i + λj + μk) = 0

=>    i         j            k

2        6           27   = 0i + 0j + 0k

1        λ            μ

=> i(6μ - 27 λ) – j(2μ - 27) + k(2λ - 6) = 0i + 0j + 0k

On comparing the corresponding components, we get:

6μ - 27 λ = 0

2μ – 27 = 0

2λ – 6 = 0

Now, 2μ – 27 = 0

=> μ = 27/2

2λ – 6 = 0

=> λ = 3

Hence, λ = 3 and μ = 27/2

Question 6:

Given that a.b = 0 and a * b = 0. What can you conclude about the vectors a and b?

Given, a.b = 0

Then,

(i) Either |a| = 0 or |b| = 0 or a Ʇ b (in case a and b are non-zero)

Again a * b = 0

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

But, a and b cannot be perpendicular and parallel simultaneously.

Hence,|a| = 0 or |b| = 0.

Question 7:

Let the vectors a, b, c given as a1i + a2j + a3k, b1i + b2j + b3k, c1i + c2j + c3k, then show that:           a * (b + c) = a * b + a * c

Given, a = a1i + a2j + a3k, b = b1i + b2j + b3k, c = c1i + c2j + c3k

b + c = (b1 + c1)i + (b2 + c2)j + (b3 + c3)k

Now, a * (b + c) =        i                  j                 k

a1              a2                a3

b1 + c1     b2 + c2        b3 + c3

= {a2(b3 + c3) – a3(b2 + c2)}i - {a1(b3 + c3) – a3(b1 + c1)}j + {a1(b2 + c2) - a2(b1 + c1)}k

=> a * (b + c) = {a2b3 + a2c3) – a3b2 - a3c2}i + {-a1b3 - a1c3 + a3b1 + a3c1}j

+ {a1b2 + a1c2 - a2b1 – a2c1)}k   ……………………1

Now, a * b =                i                  j                 k

a1              a2                a3

b1               b2               b3

= {a2b3 – a3b2}i + {a3b1 – a1b3}j + {a1b2 - a2b1}k   …………..2

Now, a * c =                i                  j                 k

a1              a2                a3

c1               c2               c3

= {a2c3 – a3c2}i + {a3c1 – a1c3}j + {a1c2 - a2c1}k   …………..3

On adding equation 2 and 3, we get

(a * b) + (a * c) = {a2b3 – a3b2 + a2c3 – a3c2}i + { a3b1 – a1b3 + a3c1 – a1c3}j

+ {a1b2 - a2b1 + a1c2 - a2c1}k        …………………..4

From equation 1 and 4, we get

a * (b + c) = a * b + a * c

Hence, the given result is proved.

Question 8:

If either a = 0 or b = 0, then a * b = 0. Is the converse true?    Justify your answer with an example.

Take any parallel non-zero vectors so that a * b = 0

Let a = 2i + 3j + 4k, b = 4i + 6j + 8k

Now, a * b =                i                j                k

2              3               4

4               6               8

= {24 - 24}i - {16 - 16}j + {12 - 12}k

= 0

It can now be observed that:

|a| = √{22 + 32 + 42} = √(4 + 9 + 16) = √29

So, a ≠ 0

|b| = √{42 + 62 + 82} = √(16 + 36 + 64) = √116

So, b ≠ 0

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

Question 9:

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

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

The adjacent sides AB and BC of ∆ABC are given as:

AB = (2 - 1)i + (3 - 1)j + (5 - 2)k = i + 2j + 3k

BC = (1 - 2)i + (5 - 3)j + (5 - 5)k = -i + 2j

Area of ∆ABC = (1/2) * |AB * BC|

Now, AB * BC=           i                j                k

1              2                3

-1             2                0

= {-6}i - {3}j + {2 + 2}k

= -6i – 3j + 4k

|AB * BC| = √{(-6)2 + (-3)2 + 42} = √(36 + 9 + 16) = √61

Hence, the area of ∆ABC is √61/2 units.

Question 10:

Find the area of the parallelogram whose adjacent sides are determined by the vector    a = i – j + 3k and b = 2i – 7j + k

The area of the parallelogram whose adjacent sides are a and b is |a * b|.

Adjacent sides are given as: a = i – j + 3k and b = 2i – 7j + k

Now, a * b =                i                j                k

1             -1               3

2               -7                1

= {-1 + 21}i - {-1 + 21}j + {-7 + 2}k

= 20i + 5j – 5k

|a * b| = √{(20)2 + (5)2 + (-5)2} = √(400 + 25 + 25) = √450 = 15√2

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

Question 11:

Let the vectors and be such that |a| = 3 and |b| = √2/3, the a * b is a unit vector, if the angle between a and b is                                                                                   (A) π/6                         (B) π/4                       (C) π/3                          (D) π/2

It is given that |a| = 3 and |b| = √2/3

We know that a * b = |a||b| sin θ n^, where is a unit vector perpendicular to both a and b and

θ is the angle between a and b.

Now, a * b is a unit vector if |a * b| = 1

=> |a||b| |sin θ n^| = 1

=> |a||b| |sin θ| = 1

=> 3 * (√2/3) * sin θ = 1

=> sin θ = 1/√2

=> sin θ = sin π/4

=> θ = π/4

Hence, a * b is a unit vector if the angle between a and b is π/4.

So, the correct answer is option B.

Question 12:

Area of a rectangle having vertices A, B, C, and D with position vectors –i + j/2 + 4k, i + j/2 + 4k, i – j/2 + 4k and –i – j/2 + 4k respectively is                                              (A) 1/2                                   (B) 1                                (C) 2                                     (D) 4

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

OA = –i + j/2 + 4k, OB = i + j/2 + 4k, OC = i – j/2 + 4k and OD = –i – j/2 + 4k

The adjacent sides AB and BC of the given rectangle are given as:

AB = (1 + 1)i + (1/2 – 1/2)j + (4 - 4)k = 2i

BC = (1 - 1)i + (-1/2 – 1/2)j + (4 - 54)k = -j

Now, AB * BC =          i                j                k

2              0                0

0             -1                0

= {-2}k

= -2k

|AB * BC| = √{(-2)2} = √4 = 2

Now, it is known that the area of a parallelogram whose adjacent sides are a and b is |a * b|

Hence, the area of the given rectangle is |AB * BC| = 2 square units.

So, the correct answer is option C.

Miscellaneous Exercise 10

Question 1:

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

If r is a unit vector in the XY-plane, then r = cos θ i + sin θ j

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

Therefore, for θ = 30°

So, r = cos 30° i + sin 30° j = (√3/2)i + (1/2)j

Hence, the required unit vector is (√3/2)i + (1/2)j

Question 2:

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

The vector joining the points P(x1, y1, z1) and Q(x2, y2, z2) is

PQ = Position vector of Q - Position vector of P

= (x2 – x1)i + (y2 – y1)j + (z2 – z1)k

|PQ| = √{(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2}

Hence, the scalar components and the magnitude of the vector joining the given points

are {(x2 – x1), (y2 – y1), (z2 – z1)} and √{(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2} respectively .

Question 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.

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:

OA = -4i

AB = i|AB| cos 60° + j|AB| sin 60°

= i(3 * 1/2) + j(3 * √3/2)

= 3i/2 + 3√3j/2

By the triangle law of vector addition, we have:

OB = OA + AB

= (-4i) + (3i/2 + 3√3j/2)

= (-4i + 3i/2) + 3√3j/2

= (-4 + 3/2)i + 3√3j/2

= -5i/2 + 3√3j/2

Hence, the girl’s displacement from her initial point of departure is -5i/2 + 3√3j/2

Question 4:

If a = b + c, then is it true that |a| = |b| + |c|? Justify your answer.

In ΔABC, let CB = a, CA = b and AB = c (as shown in the figure)

Now, by the triangle law of vector addition, we have

a = b + c

It is clearly known that |a|, |b| and |c|represent the sides of ∆ABC.

Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the

third side.

So, |a| < |b| + |c|

Hence, it is not true that |a| = |b| + |c|

Question 5:

Find the value of x for which x(I + j + k) is a unit vector.

x(I + j + k) is a unit vector if |x(I + j + k)| = 1

=> √(x2 + x2 + x2) = 1

=> √(3x2) = 1

=> x√3 = 1

=> x = 1/√3

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

Question 6:

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors a = 2i + 3j – k and b = i – 2j + k.

Given, a = 2i + 3j – k and b = i – 2j + k

Let c be the resultant of a and b.

Then, c = a + b = (2 + 1)i + (3 – 2)j + (-1 + 1)k = 3i + j

Now, |c| = √{(3)2 + (1)2} = √(9 + 1) = √10

So, c^ = c/|c| = (3i + j)/√10

Hence, the vector of magnitude 5 units and parallel to the resultant of vectors a and b is

±5 . c^ = ±5 . (3i + j)/√10 = ±3√10i/2 ±√10j/2

Question 7:

If a = i + j + k, b =2 i - j + 3k and c = i - 2j + k find a unit vector parallel to the vector 2a – b + 3c.

Given, a = i + j + k, b =2 i - j + 3k and c = i - 2j + k

Now, 2a – b + 3c = 2(i + j + k) – (2i - j + 3k) + 3(i - 2j + k)

= 2i + 2j + 2k – 2i + j - 3k + 3i - 6j + 3k

= 3i - 3j + 2k

|2a – b + 3c| = √{(3)2 + (-3)2 + (2)2} = √(9 + 9 + 4) = √22

Hence, the unit vector along 2a – b + 3c is

(2a – b + 3c)/|2a – b + 3c| = (3i - 3j + 2k)/√22 = 3i/√22 – 3j/√22 + 2k/√22

Question 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.

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

AB = (5 - 1)i + (0 + 2)j + (-2 + 8)k = 4i + 2j + 6k

BC = (11 - 5)i + (3 - 0)j + (7 + 2)k = 6i + 3j + 9k

AC = (11 - 1)i + (3 + 2)j + (7 + 8)k = 10i + 5j + 15k

|AB| = √{(4)2 + (2)2 + (6)2} = √(16 + 4 + 36) = √56 = 2√14

|BC| = √{62 + 32 + 92} = √(36 + 9 + 81) = √126 = 3√14

|AC| = √{(10)2 + 52 + 152} = √(100 + 25 + 225) = √350 = 5√14

Now, |AC| = |AB| + |BC|

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

Now, let the point B divide AC in the ratio k : 1

Then, we have OB = (k* OC + OA)/(k + 1)

=> 5i – 2k = {k* (11i + 3j + 7k) + (i – 2j – 8k)}/(k + 1)

=> (5i – 2k)(k + 1) = k* (11i + 3j + 7k) + (i – 2j – 8k)

=> 5(k + 1)i – 2(k + 1)k = (11k + 1)i + (3k - 2)j + (7k - 8)k

On equating the corresponding components, we get

5(k + 1) = 11k + 1

=> 5k + 5 = 11k + 1

=> 6k = 4

=> k = 4/6 = 2/3

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

Question 9:

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2a + b) and (a – 3b) externally in the ratio 1: 2. Also, show that P is the mid-point of the line segment RQ.

It is given that OP = 2a + b, OQ = a – 3b

It is given that point R divides a line segment joining two points P and Q externally in the ratio

1 : 2. Then, on using the section formula, we get:

OR = {2(2a + b) – (a – 3b)}/(2 - 1) = 4a + 2b – a + 3b = 3a + 5b

Therefore, the position vector of point R is 3a + 5b

Position vector of the mid-point of RQ = (OQ + OR)/2

= {(a – 3b) + (3a + 5b)}/2

= 3a + b = OP

Hence, P is the mid-point of the line segment RQ.

Question 10:

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

Adjacent sides of a parallelogram are given as: a = 2i – 4j + 5k and b = i – 2j – 3k

Thus, the diagonal of a parallelogram is given by a + b

Now, a + b = (2i – 4j + 5k) + (i – 2j – 3k) = 3i – 6j + 2k

Thus, the unit vector parallel to the diagonal is

(a + b)/|a + b| = (3i – 6j + 2k)/√{32 + (-6)2 + 22}

= (3i – 6j + 2k)/√{9 + 36 + 4}

= (3i – 6j + 2k)/√49

= (3i – 6j + 2k)/7

= 3i/7 – 6j/7 + 2k/7

So, area of parallelogram ABCD = |a * b|

Now, a * b =                i                j                k

2             -4               5

1               -2             -3

= {12 + 10}i - {-6 - 5}j + {-4 + 4}k

= 22i + 11j

|a * b| = √{(22)2 + (11)2 } = 11√(4 + 1) = 11√5

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

Question 11:

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are 1/√3, 1/√3, 1/√3

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

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

Now, cos2 α + cos2 α + cos2 α = 1

=> 3 cos2 α = 1

=> cos2 α = 1/3

=> cos α = 1/√3

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

and 1/√3.

Question 12:

Let a = i + 4j + 2k, b = 3i – 2j + 7k and c = 2i – j + 4k. Find a vector d which is perpendicular to both a and b, and c.d = 15.

Let d = d1i + d2j + d3k

Since d is perpendicular to both a and b

So, d . a = 0

=> (d1i + d2j + d3k).( i + 4j + 2k) = 0

=> d1 + 4d2 + 2d3 = 0   ………….1

and d . b = 0

=> (d1i + d2j + d3k).( 3i – 2j + 7k) = 0

=> 3d1 - 2d2 + 7d3 = 0   ………..2

Also, it is given that:

c . d = 15

=> 2d1 - d2 + 4d3 = 15   ………..3

On solving equation 1, 2, and 3, we get:

d1 = 160/3, d2 = -5/3, d3 = -70/3

So, d = 160i/3 - 5j/3 - 70k/3 = (160i - 5j - 70k)/3

Hence, the required vector is (160i - 5j - 70k)/3.

Question 13:

The scalar product of the vector i + j + k with a unit vector along the sum of vectors 2i + 4j – 5k and λi + 2j + 3k is equal to one. Find the value of λ.

(2i + 4j – 5k) + (λi + 2j + 3k) = (2 + λ)i + 6j - 2k

Therefore, the unit vector along (2 + λ)i + 6j - 2k is given as:

{(2 + λ)i + 6j - 2k}/√{(2 + λ)2 + 62 + (-2)2}

= {(2 + λ)i + 6j - 2k}/√(λ2 + 4λ + 4 + 36 + 4)

= {(2 + λ)i + 6j - 2k}/√(λ2 + 4λ + 44)

Scalar product of (i + j + k) with this unit vector is 1.

=> (i + j + k) . {(2 + λ)i + 6j - 2k}/√(λ2 + 4λ + 44) = 1

=> {(2 + λ) + 6 - 2}/√(λ2 + 4λ + 44) = 1

=> λ + 6 = √(λ2 + 4λ + 44)

=> (λ + 6)2 = λ2 + 4λ + 44

=> λ2 + 12λ + 36 = λ2 + 4λ + 44

=> 12λ - 4λ = 44 – 36

=> 8λ = 8

=> λ = 1

Hence, the value of λ is 1.

Question 14:

If a, b and c are mutually perpendicular vectors of equal magnitudes, show that the vector        a + b + c is equally inclined to a, b and c.

Since a, b and c are mutually perpendicular vectors, we have

a.b = b.c = c.a = 0

It is given that: |a| = |b| = |c|

Let vector a + b + c be inclined to a, b and c at angles θ1, θ2 and θ3 respectively.

Then, we have:

cos θ1 = {(a + b + c).a}/{|a + b + c||a|}

= (a.a + b.a + c.a)/{|a + b + c||a|}

= |a|2/{|a + b + c||a|}                              [Since b.a = c.a = 0]

= |a|/|a + b + c|

cos θ2 = {(a + b + c).b}/{|a + b + c||b|}

= (a.b + b.b + c.b)/{|a + b + c||b|}

= |b|2/{|a + b + c||b|}                              [Since a.b = c.b = 0]

= |b|/|a + b + c|

cos θ3 = {(a + b + c).c}/{|a + b + c||c|}

= (a.c + b.c + c.c)/{|a + b + c||c|}

= |c|2/{|a + b + c||c|}                              [Since a.c = b.c = 0]

= |c|/|a + b + c|

Since, |a| = |b| = |c|,

So, cos θ1= cos θ2 = cos θ3

=> θ1= θ2 = θ3

Hence, the vector a +b + c is equally inclined to a, b and c.

Question 15:

Prove that (a + b).(a + b) = |a|2 + |b|2 if and only if a, b are perpendicular, given a ≠ 0, b ≠ 0.

Given, (a + b).(a + b) = |a|2 + |b|2

=> a.a + a.b + b.a + b.b = |a|2 + |b|2

=> |a|2 + 2a.b + |b|2 = |a|2 + |b|2

=> 2a.b = 0

=> a.b = 0

So, a and b are perpendicular.

Question 16:

If θ is the angle between two vectors a and b, then a.b ≥ 0 only when

(A) 0 < θ < π/2               (B) 0 ≤ θ ≤ π/2               (C) 0 < θ < π              (D) 0 ≤ θ ≤ π

Let θ be the angle between two vectors a and b.

Then, without loss of generality, a and b are non-zero vectors so that |a| and |b| are positive.

We know that a.b = |a||b| cos θ

Given, a.b ≥ 0

=> |a||b| cos θ ≥ 0

=> cos θ ≥ 0                      [Since |a| and |b| are positive]

=> 0 ≤ θ ≤ π/2

Hence, a.b ≥ 0 when 0 ≤ θ ≤ π/2

So, the correct answer is option B.

Question 17:

Let a and b be two unit vectors and θ is the angle between them. Then a + b is a unit vector if

(A) θ = π/4                  (B) θ = π/3                  (C) θ = π/2                       (D) θ = 2π/3

Let a and b be two unit vectors and θ be the angle between them.

Then, |a| = |b| = 1

Now, a + b is a unit vector if |a + b| = 1

=> (a + b)2 = 1

=> (a + b).(a + b) = 1

=> a.a + a.b + b.a + b.b = 1

=> |a|2 + 2a.b + |b|2 = 1

=> 1 + 2|a||b| cos θ + 1 = 1

=> 1 + 2.1.1 cos θ + 1 = 1

=> 1 + 2 cos θ + 1 = 1

=> 2 cos θ = -1

=> cos θ = -1/2

=> cos θ = cos 2π/3

=> θ = 2π/3

Hence, a + b is a unit vector if θ = 2π/3

The correct answer is option D.

Question 18:

The value of is i.(j * k) + j.(i * k) + k.(i * j) is

(A) 0                                  (B) –1                               (C) 1                                 (D) 3

Given, i.(j * k) + j.(i * k) + k.(i * j)

= i.i + j.(-j) + k.k

= 1 - j.j + 1

= 1 – 1 + 1

= 1

Hence, the correct answer is option C.

Question 19:

If θ be the angle between two vectors a and b, then |a.b| = |a * b|, when θ is equal to

(A) 0                               (B) π/4                    (C) π/2                      (D) π

Let θ be the angle between two vectors a and b.

Then, without loss of generality, and are non-zero vectors, so that |a| and |b| are positive.

Now, |a.b| = |a * b|

=> |a||b| cos θ = |a||b| sin θ

=> cos θ = sin θ

=> tan θ = 1

=> tan θ = tan π/4

=> θ = π/4

Hence, |a.b| = |a * b| when θ is equal to π/4

The correct answer is option B.