Product Of Three Vectors

i.

Let, be unit vectors in the direction of positive X-axis, Y-axis, Z-axis respectively.

Hence,

To Prove :

Formulae :

a) Dot Products :

i)

ii)

b) Cross Products :

i)

ii)

iii)

c) Scalar Triple Product :

Now,

(i)

…………

= 1 …………

………… eq(1)

(ii)

…………

= 1 …………

………… eq(2)

(iii)

…………

= 1 …………

………… eq(3)

From eq(1), eq(2) and eq(3),

Hence Proved.

Notes :

1. A cyclic change of vectors in a scalar triple product does not change its value i.e.

2. Scalar triple product of unit vectors taken in a clockwise direction is 1, and that of unit vectors taken in anticlockwise direction is -1

ii.

Let, be unit vectors in the direction of positive X-axis, Y-axis, Z-axis respectively.

Hence,

To Prove :

Formulae :

a) Dot Products :

i)

ii)

b) Cross Products :

i)

ii)

iii)

c) Scalar Triple Product :

Answer :

(i)

…………

= -1 …………

………… eq(1)

(ii)

…………

= -1 …………

………… eq(2)

(iii)

…………

= -1 …………

………… eq(3)

From eq(1), eq(2) and eq(3),

Hence Proved.

Notes :

1. A cyclic change of vectors in a scalar triple product does not change its value i.e.

2. Scalar triple product of unit vectors taken in a clockwise direction is 1, and that of unit vectors taken in anticlockwise direction is -1

i. and

Given Vectors :

1)

2)

3)

To Find :

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

For given vectors,

= 6 + 5 – 21

= - 10

ii. and

Given Vectors :

1)

2)

3)

To Find :

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

For given vectors,

= 6 + 15 – 28

= - 7

iii. and

Given Vectors :

1)

2)

3)

To Find :

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

For given vectors,

= - 2 + 6

= 4

i.

Given :

Coterminous edges of parallelopiped are where,

To Find : Volume of parallelepiped

Formulae :

1) Volume of parallelepiped :

If are coterminous edges of parallelepiped,

Where,

Then, volume of parallelepiped V is given by,

2) Determinant :

Answer :

Volume of parallelopiped with coterminous edges

= 1(-1) -1(-2) + 1(3)

= -1+2+3

= 4

Therefore,

ii.

Given :

Coterminous edges of parallelopiped are where,

To Find : Volume of parallelepiped

Formulae :

1) Volume of parallelepiped :

If are coterminous edges of parallelepiped,

Where,

Then, volume of parallelepiped V is given by,

2) Determinant :

Answer :

Volume of parallelopiped with coterminous edges

= -3(-36) -7(36) + 5(-24)

= 108 – 252 – 120

= -264

As volume is never negative

Therefore,

iii.

Given :

Coterminous edges of parallelopiped are where,

To Find : Volume of parallelepiped

Formulae :

1) Volume of parallelepiped :

If are coterminous edges of parallelepiped,

Where,

Then, volume of parallelepiped V is given by,

2) Determinant :

Answer :

Volume of parallelopiped with coterminous edges

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

= 2 + 4 + 6

= 12

Therefore,

iv.

Given :

Coterminous edges of parallelopiped are where,

To Find : Volume of parallelepiped

Formulae :

1) Volume of parallelepiped :

If are coterminous edges of parallelepiped,

Where,

Then, volume of parallelepiped V is given by,

2) Determinant :

Answer :

Volume of parallelopiped with coterminous edges

= 6(10) + 0 + 0

= 60

Therefore,

i. and

Given Vectors :

To Prove : Vectors are coplanar.

i.e.

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

For given vectors,

= 1(3) + 2(-6) + 3(3)

= 3 – 12 +9

= 0

Hence, the vectors are coplanar.

Note : For coplanar vectors ,

ii. and

Given Vectors :

To Prove : Vectors are coplanar.

i.e.

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

For given vectors,

= 1(4) – 3(6) + 1(14)

= 4 – 18 + 14

= 0

Hence, the vectors are coplanar.

Note : For coplanar vectors ,

iii. and

Given Vectors :

To Prove : Vectors are coplanar.

i.e.

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

For given vectors,

= 2(2) + 1(16) + 2(-10)

= 4 + 16 -20

= 0

Hence, the vectors are coplanar.

Note : For coplanar vectors ,

i. .. and

Given : Vectors are coplanar.

Where,

To Find : value of

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

As vectors are coplanar

…………eq(1)

For given vectors,

…………eq(2)

From eq(1) and eq(2),

ii. and

Given : Vectors are coplanar.

Where,

To Find : value of

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

As vectors are coplanar

…………eq(1)

For given vectors,

…………eq(2)

From eq(1) and eq(2),

iii. .. and

Given : Vectors are coplanar.

Where,

To Find : value of

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

As vectors are coplanar

…………eq(1)

For given vectors,

…………eq(2)

From eq(1) and eq(2),

Given Vectors :

To Find :

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

For given vectors,

= - 34 + 14 + 5

= - 15

Given :

1) Coterminous edges of parallelepiped are

2) Volume of parallelepiped,

V = 546 cubic unit

To Find : value of

1) Volume of parallelepiped :

If are coterminous edges of parallelepiped,

Where,

Then, volume of parallelepiped V is given by,

2) Determinant :

Answer :

Given volume of parallelepiped,

V = 546 cubic unit ………eq(1)

Volume of parallelopiped with coterminous edges

cubic unit ………eq(2)

From eq(1) and eq(2)

Given Vectors :

To Prove : Vectors are parallel to same plane.

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

Vectors will be parallel to the same plane if they are coplanar.

For vectors to be coplanar,

Now, for given vectors,

= 1(4) – 3(6) + 1(14)

= 4 – 18 + 14

= 0

Hence, given vectors are parallel to the same plane.

Given : vectors are coplanar. Where,

To Prove : c² = ab

Formulae :

1) Scalar Triple Product:

If

Then,

2) Determinant :

Answer :

As vectors are coplanar

………eq(1)

For given vectors,

= a.(- c) – a.(b - c) + c(c)

= – ac – ab + ac + c²

= - ab + c²

………eq(2)

From eq(1) and eq(2),

- ab + c² = 0

Therefore,

Hence proved.

Note : Three vectors are coplanar if and only if

Given :

Let A, B, C & D be four points with position vectors .

Therefore,

To Prove : Points A, B, C & D are coplanar.

Formulae :

1) Vectors :

If A & B are two points with position vectors ,

Where,

then vector is given by,

2) Scalar Triple Product:

If

Then,

3) Determinant :

Answer :

For given position vectors,

Vectors are given by,

………eq(1)

………eq(2)

………eq(3)

Now, for vectors

= 2(21) – 4(15) + 6(3)

= 42 – 60 + 18

= 0

Hence, vectors are coplanar.

Therefore, points A, B, C & D are coplanar.

Note : Four points A, B, C & D are coplanar if and only if

Given :

Let A, B, C & D be four points with position vectors .

Therefore,

To Prove : Points A, B, C & D are coplanar.

Formulae :

1) Vectors :

If A & B are two points with position vectors ,

Where,

then vector is given by,

2) Scalar Triple Product:

If

Then,

3) Determinant :

Answer :

For given position vectors,

Vectors are given by,

………eq(1)

………eq(2)

………eq(3)

Now, for vectors

= -10(112) – 12(-84) + 4(28)

= -1120 + 1008 + 112

= 0

Hence, vectors are coplanar.

Therefore, points A, B, C & D are coplanar.

Note : Four points A, B, C & D are coplanar if and only if

Given :

Let, A, B, C & D be four points with given position vectors

To Find : value of λ

Formulae :

1) Vectors :

If A & B are two points with position vectors ,

Where,

then vector is given by,

2) Scalar Triple Product:

If

Then,

3) Determinant :

Answer :

For given position vectors,

Vectors are given by,

………eq(1)

………eq(2)

………eq(3)

Now, for vectors

= -2(- λ – 10) – 3(13) + 1(28 – 5λ)

= 2λ + 20 – 39 + 28 – 5λ

= 9 – 3λ

………eq(4)

Four points A, B, C & D are coplanar if and only if

………eq(5)

From eq(4) and eq(5)

9 – 3λ = 0

3λ = 9

Given :

Let, A, B, C & D be four points with given position vectors

To Find : value of λ

Formulae :

1) Vectors :

If A & B are two points with position vectors ,

Where,

then vector is given by,

2) Scalar Triple Product:

If

Then,

3) Determinant :

Answer :

For given position vectors,

Vectors are given by,

………eq(1)

………eq(2)

………eq(3)

Now, for vectors

= -2(2 + 2λ) – 0 + 2(4)

= - 4 - 4λ + 8

= 4 – 4λ

………eq(4)

Four points A, B, C & D are coplanar if and only if

………eq(5)

From eq(4) and eq(5)

4 – 4λ = 0

4λ = 4

Given Points :

A ≡ (4, 5, 1)

B ≡ (0, -1, -1)

C ≡ (3, 9, 4)

D ≡ (-4, 4, 4)

To Prove : Points A, B, C & D are coplanar.

Formulae :

4) Position Vectors :

If A is a point with co-ordinates (a₁, a₂, a₃)

then its position vector is given by,

5) Vectors :

If A & B are two points with position vectors ,

Where,

then vector is given by,

6) Scalar Triple Product:

If

Then,

7) Determinant :

Answer :

For given points,

A ≡ (4, 5, 1)

B ≡ (0, -1, -1)

C ≡ (3, 9, 4)

D ≡ (-4, 4, 4)

Position vectors of above points are,

Vectors are given by,

………eq(1)

………eq(2)

………eq(3)

Now, for vectors

= 4(15) – 6(21) + 2(33)

= 60 – 126 + 66

= 0

Hence, vectors are coplanar.

Therefore, points A, B, C & D are coplanar.

Note : Four points A, B, C & D are coplanar if and only if

Given :

Points A, B, C & D are coplanar where,

A ≡ (3, 2, 1)

B ≡ (4, λ, 5)

C ≡ (4, 2, -2)

D ≡ (6, 5, -1)

To Find : value of λ

Formulae :

1) Position Vectors :

If A is a point with co-ordinates (a₁, a₂, a₃)

then its position vector is given by,

2) Vectors :

If A & B are two points with position vectors ,

Where,

then vector is given by,

3) Scalar Triple Product:

If

Then,

4) Determinant :

Answer :

For given points,

A ≡ (3, 2, 1)

B ≡ (4, λ, 5)

C ≡ (4, 2, -2)

D ≡ (6, 5, -1)

Position vectors of above points are,

Vectors are given by,

………eq(1)

………eq(2)

………eq(3)

Now, for vectors

= - 1(9) – (2 - λ).(7) – 4(3)

= - 9 – 14 + 7λ – 12

= 7λ – 35

………… eq(4)

But points A, B, C & D are coplanar if and only if

………… eq(5)

From eq(4) and eq(5)

7λ – 35 = 0

Two vectors are equal if and only if their corresponding components are equal.

Thus, the given vectors and are equal if and only if

x = 3, - y = 2, - z = 1

x = 3, y = - 2, z = - 1

x + y + z = 3 + (- 2) + (- 1) = 3 - 3 = 0

Since, these two vectors are equal, therefore comparing these two vectors we get,

x = 3 , - y = 2 , - z = 1

⇒x = 3,y = - 2,z = - 1

∴x + y + z = 3 + ( - 2) + ( - 1) = 3 - 2 - 1 = 0

Ans:x + y + z = 0

The sum of vectors is

Let the unit vector in the direction of be , then

Let be the sum of the vectors

⇒

⇒

⇒

|| = (4² + 3² + ( - 12)²)^1/2

⇒|| = (16 + 9 + 144)^1/2 = (169)^1/2 = 13

a unit vector in the direction of the sum of the vectors is given by:

Ans:

If the scalar product (dot product) is zero, two non - zero vectors are perpendicular.

2 - 2λ + 3 = 0

- 2λ = 5

Since these two vectors are perpendicular the dot product of these two vectors is zero.

i.e.: = 0

⇒

⇒2 + λ×( - 2) + 3 = 0

⇒5 = 2 λ

⇒ λ = 5/2

Ans: λ = 5/2

Since these two vectors are parallel

∴

⇒

⇒

Ans:

Two nonzero vectors are parallel if their vector product (cross product) is zero,

On comparing with right hand side,

6 + 18p = 0

(You can solve using - 6p - 2)

Projection of vector on vector

So we first calculate the magnitude of vector b and the scalar product of a vector and .

Projection of vector on vector = 4 (1)

Putting the values from above in equation (1),

2λ = 10

λ = 5

projection of a on b is given by:

|| = (2² + 6² + 3²)^1/2

⇒|| = (4 + 36 + 9)^1/2 = (49)^1/2 = 7

a unit vector in the direction of the sum of the vectors is given by:

Now it is given that:

⇒

⇒2 λ + 6 + (3×4) = 28

⇒ λ = (28 - 12 - 6)/2

⇒ λ = 10/2 = 5

Ans: λ = 5

As vector and is perpendicular, . So, using

(Negative value not considered as magnitude is positive).

Since a and b vectors are perpendicular .

⇒

Now,

||² = ||² + ||² + 2||||cos

⇒13² = 5² + ||² + 0 …( cos)

⇒||² = 169 - 25 = 144

⇒|| = 12

Ans:|| = 12

⇒||² - ||² = 15

⇒||² = ||² + 15

Now , a is a unit vector,

⇒|| = 1

⇒||² = 1² + 15

⇒||² = 16

⇒|| = 4

Ans: || = 4

(

(Using , commutative law)

(As magnitude of unit vector is 1)

Now ,

⇒

Ans:

Now ,

⇒

Ans:

projection of a on b is given by:

∴ the projection of the vector along the vector _{is :}

Ans: the projection of the vector along the vector _is:1

Projection of vector on vector .

= 1

projection of a on b is given by:

|| = (2² + 6² + 3²)^1/2

⇒|| = (4 + 36 + 9)^1/2 = (49)^1/2 = 7

a unit vector in the direction of the sum of the vectors is given by:

Ans: the projection of the vector on the vector

Projection of vector on vector

We will first find vector product of and then scalar product of that with .

= 2.3 + 1.(5) + 3.(- 7)

= 6 + 5 - 21

= - 10

∴

∴ = (2×3) + (1×5) + (3× - 7)

= 6 + 5 - 21 = - 10

Ans: - 10

|| = (2² + ( - 3)² + 6²)^1/2

⇒|| = (4 + 9 + 36)^1/2 = (49)^1/2 = 7

a unit vector in the direction of the sum of the vectors is given by:

a vector in the direction of which has magnitude 21 units.

= 21

Ans:

First, we find the unit vector in the direction of a given vector,

Now vector in the direction of the given vector and with magnitude 21 is

For perpendicular vectors scalar product is zero.

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

6 - 3λ + 2 + 2λ = 0

8 - λ = 0

λ = 8

⇒

Since

⇒ = 0

⇒

⇒(2 - λ)×3 + (2 + 2 λ)×1 = 0

⇒6 + 2 - 3 λ + 2 λ = 0

⇒ λ = 8

Ans: λ = 8

|| = (1² + ( - 2)² + 2²)^1/2

⇒|| = (1 + 4 + 4)^1/2 = (9)^1/2 = 3

a unit vector in the direction of the sum of the vectors is given by:

a vector in the direction of which has magnitude 15 units.

= 15

Ans:

First, we find the unit vector in the direction of a given vector,

Now vector in the direction of the given vector and with magnitude 15 is

First, we will find vector, then we will find a unit vector in the given direction,

Vector with magnitude 6 in the direction of the vector is

∴ (

⇒(

LET, (

|| = (1² + ( - 2)² + 2²)^1/2

⇒|| = (1 + 4 + 4)^1/2 = (9)^1/2 = 3

a unit vector in the direction of the sum of the vectors is given by:

a vector of magnitude 6 units which is parallel to the vector is:

Ans:

Projection of vector on vector

)

= 0

So, projection of vector on other is 0.

projection of a on b is given by:

|| = (1² + 1² + 0²)^1/2

⇒|| = (1 + 1)^1/2 = (2)^1/2

a unit vector in the direction of the sum of the vectors is given by:

Ans: the projection of the vector on the vector

|| =

|| = ₂

Since,

Substituting the given values we get:

⇒

⇒

⇒

⇒ θ = 45^° =

Ans: θ = 45^° =

Using scalar product, we can find the angle between two vectors.

∴

∴| = (0² + 19² + 19²)^1/2 = (2×19²)^1/2 = 19√2

Ans: ∴| = 19√2

|| = ₁

|| = ₂

Since,

Substituting the given values we get:

⇒

⇒

⇒

⇒ θ = 60^° =

Ans: θ = 60^° =

Using vector product, we can calculate the angle between vectors.

It is given that:

⇒

Since sinθ and cosθ cannot be 0 simultaneously ∴| = 0

Conclusion: when

Then | = 0

As , using scalar product and vector product.

Now

As cosθ and sinθ cannot be 0 simultaneously So, then either vector a is o or b is 0.

If the vector product is zero, two vectors are parallel.

On comparing with the right hand side, we have

9λ - 6 = 0

It is given that

⇒

⇒

⇒

Ans: λ = 2/3

According to the right hand coordinate system,

Then putting values in the equation

= 1 - 1 + 1 = 1

We know that:

= 1

∴

Ans: = 1

Scalar triple product geometrically represents the volume of the parallelepiped whose three coterminous edges are represented by .i.e. V =

∴ V = = 2(4 - 2) - ( - 3)(2 - ( - 3)) + 4( - 2 - 6) = 4 + 15 - 32 = | - 13| =

13 cubic units.

Ans:13 cubic units.

The volume of parallelepiped

= (- 3).(- 1) - 2.4).3 - (2.(- 1) - 1.4).(- 2) + (2.2 - 1.(- 3)).

= (3 - 8).3 + (- 2 - 4).2 + (4 - (- 3)).2

= - 15 - 12 + 14

= - 27 + 14

= - 13

Volume of parallelepiped

If three planes lie in a single plane, then the volume of parallelepiped will be zero. So, planes are coplanar if

The volume of parallelepiped

= (- 2. - 2 - 4.4)4 - (- 2. - 2 - 4. - 2) - 2 + (4. - 2 - (- 2). - 2) - 2

= (4 - 16)4 + (4 + 8)2 - (- 8 - 4)2

= - 48 + 24 - (- 24)

= - 48 + 48 = 0

So, planes are coplanar.

If are coplanar then = 0

L.H.S = - 2( - 8 - 4) + 2(4 + 8) + 4(4 - 16) = 24 + 24 - 48 = 0 = R.H.S

∴L.H.S = R.H.S

Hence proved that the vectors

Are coplanar.

Given that the vector product is zero.

On comparing with the right hand side, we have

6μ - 27λ = 0

2μ - 27 = 0

2λ - 6 = 0

It is given that

⇒

⇒

⇒2 λ - 6 = 0

⇒ λ = 6/2 = 3

⇒2 μ - 27 = 0

⇒ μ = 27/2

Ans: λ = 3 , μ = 27/2

It is given that:

⇒

⇒sinθ = cosθ

⇒tanθ = 1

⇒

Ans:

We have

Equating both

When the two vectors are parallel or collinear, they can be added in a scalar way because the angle between them is zero degrees,they are I the same or opposite direction.

Therefore when two vectors are either parallel or collinear then

As, magnitude of a vector cannot be zero (leaving zero vector)

1 - cosθ = 0

Cosθ = 1

θ = 0˚

So, vectors are either parallel or collinear.

Direction cosines of a vector l, m, n are related to each other as

Now given that equally inclined to three axes with an angle of θ. Then direction cosines l, m, n are

l = m = n = cosθ

Putting values of direction cosines in equation,

Cos²θ + Cos²θ + Cos²θ = 1

3Cos²θ = 1

Let the inclination with:

x - axis =

y - axis =

z - axis =

The vector is equally inclined to the three axes.

⇒

Direction cosines

We know that:cos² α + cos² β + cos² γ = 1

⇒ cos² α + cos² α + cos² α = 1 …(

⇒3 cos² α = 1

⇒

Ans:

So direction ratios are 3, - 4, - 6.

Let P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) be the two points then Direction ratios of line joining P and Q i.e. PQ are x₂ - x₁,y₂ - y₁,z₂ - z

Here, P is(1, 5, 4) and Q is (4, 1, - 2)

Direction ratios of PQ are:(4 - 1),(1 - 5),( - 2 - 4) = 3, - 4, - 6

Ans: the direction ratios of _are: 3, - 4, - 6

Let the inclination with:

x - axis =

y - axis =

z - axis =

Direction cosines

For a vector

Ans:

The direction cosines and direction ratios are related as

, where a, b, c are direction ratios and r is magnitude.

Now direction ratios are 1, 2, 3 respectively and magnitude of vector is

Putting the values

It is given that are unit vectors ,as well as is also a unit vector

⇒|| = || = || = 1

Since the modulus of a unit vector is unity.

Now,

||² = ||² + ||² + 2||||cosθ

⇒1² = 1² + 1² + 2×1×1× cosθ

⇒ cosθ = (1 - 1 - 1)/2

⇒

⇒

Ans:

1 - 1 - 1 = 2cosθ

- 1 = 2cosθ

Tip – A vector in the direction of another vector is given by and the unit vector is given by

So, a vector parallel to is given by where is an arbitrary constant.

Now,

Hence, the required unit vector

Given vector

Property : The unit vector corresponding to the vector

Therefore the unit vector corresponding to the vector

is

Given - Two adjacent sides of a triangle are represented by the vectors and

To find – Area of the triangle

Formula to be used - where and

Tip – Area of triangle and magnitude of a vector is given by

Hence,

i.e. the area of the parallelogram = sq. units

Tip – A vector in the direction of another vector is given by and the unit vector is given by

So, a vector parallel to is given by where is an arbitrary constant.

Now,

Hence, the required unit vector

Given - Two adjacent sides of a triangle are represented by the vectors and

To find – Area of the triangle

Formula to be used - where and

Tip – Area of triangle and magnitude of a vector is given by

Hence,

i.e. the area of the parallelogram = sq. units

Formula to be used – The direction cosines of a vector is given by .

Hence, the direction cosines of the vector is given by

Formula to be used – The direction cosines of a vector is given by .

Hence, the direction cosines of the vector is given by

Given vector

Property: for the vector ,

1) Direction ratios dr’s are a,b,c

2) Direction cosines dc’s are

Therefore the dc’s of the vector ,,

=,,

= ,,

Given A(1,2,-3) and B(-1,-2,1)

Property: The position vector of the of the vector joining two points (x₁,y₁,z₁)and (x₂,y₂,z₂) is

So, the position vector of the line joining A and B is

Property: for the vector , Direction cosines dc’s are

Therefore the Dc’s of the vector ,,

= ,,

=,,

=

=

Given - A(1, 2, -3) and B(-1, -2, 1) are the end points of a vector

Tip – If P(a₁,b₁,c₁) and Q(a₂,b₂,c₂) be two points then the vector is represented by

Hence,

Formula to be used – The direction cosines of a vector is given by .

Hence, the direction cosines of the vector is given by

Given - A(1, 2, -3) and B(-1, -2, 1) are the end points of a vector

Tip – If P(a₁,b₁,c₁) and Q(a₂,b₂,c₂) be two points then the vector is represented by

Hence,

Formula to be used – The direction cosines of a vector is given by .

Hence, the direction cosines of the vector is given by

Given - A vector makes angle α, β and γ with the x-axis, y-axis and z-axis respectively.

To Find - (sin²α + sin²β + sin²γ)

Formula to be used - cos²α + cos²β + cos²γ=1

Hence,

sin²α + sin²β + sin²γ

=(1-cos²α) +(1-cos²β) +(1-cos²γ)

= 3–(cos²α + cos²β + cos²γ)

=3-1

=2

Given - A vector makes angle α, β and γ with the x-axis, y-axis and z-axis respectively.

To Find - (sin²α + sin²β + sin²γ)

Formula to be used - cos²α + cos²β + cos²γ=1

Hence,

sin²α + sin²β + sin²γ

=(1-cos²α) +(1-cos²β) +(1-cos²γ)

= 3–(cos²α + cos²β + cos²γ)

=3-1

=2

Given α , β and γ are the angles made by the vector with X,Y and z axes respectively

are the direction cosines .

As we know that if are the direction cosines , then the relation between them is

We also know that

So we can write ((

⇒

⇒

Tip – Magnitude of a vector is given by

A unit vector is a vector whose magnitude = 1.

Formula to be used -

Hence, magnitude of will be given by

= 1 i.e a unit vector

Given vector

UNIT VECTOR: the vector with magnitude as 1.

Property: The magnitude of the vector

The magnitude of the given vector is

=

=

=1

As the magnitude of the given vector is 1, it is a UNIT VECTOR.

Tip – Magnitude of a vector is given by

A unit vector is a vector whose magnitude = 1.

Formula to be used -

Hence, magnitude of will be given by

= 1 i.e a unit vector

Given vector is

Property: for the vector , Direction cosines dc’s are

Therefore the dc’s of the given vector is

=

=

=

Let the angle made by the vector with the Z axis be γ.

we got that the cosine of the angle γ is

⇒

⇒

⇒ γ=

Formula to be used – The direction cosines of a vector is given by .

Hence, the direction cosines of the vector is given by

The direction cosine of z-axis = i.e. where is the angle the vector makes with the z-axis.

Formula to be used – The direction cosines of a vector is given by .

Hence, the direction cosines of the vector is given by

The direction cosine of z-axis = i.e. where is the angle the vector makes with the z-axis.

Given

And

Let angle between the vectors and be θ

Using the dot product property of the vectors,

Substituting the given values in the equation,

⇒

⇒

⇒θ=

Given - and are vectors such that and and

To find – Angle between and .

Formula to be used -

Hence, i.e.

Given - and are vectors such that and and

To find – Angle between and .

Formula to be used -

Hence, i.e.

Given

Given

And

Let angle between the vectors and be θ

Using the dot product property of the vectors,

Substituting the given values in the equation,

⇒

⇒

⇒

⇒

⇒θ=

Given - and are vectors such that and

To find – Angle between and .

Formula to be used -

Hence, i.e.

Given - and are vectors such that and

To find – Angle between and .

Formula to be used -

Hence, i.e.

Given - and

To find – Angle between and .

Formula to be used -

Tip – Magnitude of a vector is given by

Here, )=3+4+3=10

Hence, i.e.

Given - and

To find – Angle between and .

Formula to be used -

Tip – Magnitude of a vector is given by

Here, )=3+4+3=10

Hence, i.e.

Given vectors and

Magnitude |

Magnitude of |

Property:

(

Then

= (1 x 3)+(-2 x -2)+(3 x 1)

=3+4+3

= 10

Let angle between the vectors and be θ

Using the dot product property of the vectors,

Substituting the given values in the equation,

⇒

⇒

⇒θ=

Given vectors and

(

= -8+3+5

=0

As (, then the cosine of angle between the vectors and is 0.

⇒

⇒ .

Given - and

To find – Angle between and .

Formula to be used - where and are two vectors

Tip – Magnitude of a vector is given by

Here,

and

Hence, i.e.

Given - and

To find – Angle between and .

Formula to be used - where and are two vectors

Tip – Magnitude of a vector is given by

Here,

and

Hence, i.e.

Given - and

To find – Angle between and .

Formula to be used - where and are two vectors

Tip – Magnitude of a vector is given by

Here,

and

Hence, i.e.

Given vectors and

Let the vector be

|

Let the vector be

|

= (5 7) + 0 -(4 1)

=35-4

=31

Let angle between the vectors and be θ

Using the dot product property of the vectors,

Substituting the given values in the equation,

⇒=

⇒θ=

Given - and

To find – Angle between and .

Formula to be used - where and are two vectors

Tip – Magnitude of a vector is given by

Here,

and

Hence, i.e.

Given - and

To find – Value of

Formula to be used - where and are two vectors

Tip – For perpendicular vectors, i.e. i.e. the dot product=0

Hence,

i.e.

Given - and

To find – Value of

Formula to be used - where and are two vectors

Tip – For perpendicular vectors, i.e. i.e. the dot product=0

Hence,

i.e.

Given vectors and

Also given that

Let the angle between the vectors and be θ.

⇒ θ=

=

⇒

So, (

⇒ (2 3) + (4 -2) + (-1=0

⇒ 6-8-=0

⇒

Given vectors and

Property:

Projection of the vector on is =

Therefore the projection of on is

=

=

=

Given -

To find – Projection of on i.e.

Formula to be used - where and are two vectors

Tip – If and are two vectors, then the projection of on is defined as

Magnitude of a vector is given by

So,

Given -

To find – Projection of on i.e.

Formula to be used - where and are two vectors

Tip – If and are two vectors, then the projection of on is defined as

Magnitude of a vector is given by

So,

Given |

Squaring on both the sides,

⇒

⇒ 4.=0

⇒=0

⇒

Given -

Tip – If and are two vectors then

Hence,

i.e.

So,

Given -

Tip – If and are two vectors then

Hence,

i.e.

So,

Given - and are two mutually perpendicular unit vectors i.e.

To Find –

Formula to be used - where and are two vectors

Tip -

Hence,

Given - and are two mutually perpendicular unit vectors i.e.

To Find –

Formula to be used - where and are two vectors

Tip -

Hence,

Given are mutually perpendicular unit vectors

⇒|

And angle between the vectors is and =0

Asking to find (3

Multiplying ,

=(35)- (36) ( + (25)(- (26)

= 15( [reason: dot product is commutative i.e,

=15-8(-12

=15-12 [reason:=0]

= 3

Given vectors and

Also given

As they are perpendicular,=0

⇒ ().(

⇒ (3 1) + (1) + (-2 -3) =0

⇒ 3++6=0

⇒

Given - and

To find – Value of