Components of a vector

A scalar component, also known as projection, of a vector AB in the positive direction of a straight line C’C is defined as :

AC = | AB | cos θ = AB cos θ AC = | AB | cos θ = AB cos θ

Where θ is the angle that vector AB makes with the specified direction C’C as shown in the figure.

Figure 1

Analytical method

It should be understood that the component so determined is not rooted to the common point “A” or any specific straight line like C’C in a given direction. Matter of fact, component of vector AB drawn on any straight lines like D’D parallel to CC’ are same. Note that the component i.e. projection of the vector is graphically obtained by drawing two perpendicular lines from the ends of the vector on the straight line in the specified direction.

It is clear that scalar component can either be positive or negative depending on the value of angle that the vector makes with the referred direction. The angle lies between the range given by 0 ≤ θ ≤ 180°. This interval means that we should consider the smaller angle between two vectors.

In accordance with the above definition, we resolve a given vector in three components in three mutually perpendicular directions of rectangular coordinate system. Note here that we measure angle with respect to parallel lines to the axes. By convention, we denote components by using the non-bold type face of the vector symbol with a suffix representing direction (x or y or z).

Figure 2

Components of a vector

A B x = AB cos α A B y = AB cos β A B y = AB cos γ A B x = AB cos α A B y = AB cos β A B y = AB cos γ

Where α, β and γ are the angles that vector AB makes with the positive directions of x, y and z directions respectively.

As a vector can be resolved in a set of components, the reverse processing of components in a vector is also expected. A vector is composed from vector components in three directions along the axes of rectangular coordinate system by combining three component vectors. The vector component is the vector counterpart of the scalar component, which is obtained by multiplying the scalar component with the unit vector in axial direction. The vector components of a vector ,a, are a x i a x i , a y j a y j and a z k a z k .

A vector,a, is equal to the vector sum of component vectors in three mutually perpendicular directions of rectangular coordinate system.

where i i , j j and k k are unit vectors in the respective directions.

Magnitude of the sum of two vectors ( a x i + a y j a x i + a y j ) in x and y direction, which are perpendicular to each other, is :

a’ = √ ( a x 2 + a y 2 ) = ( a x 2 + a y 2 ) 1 2 a’ = √ ( a x 2 + a y 2 ) = ( a x 2 + a y 2 ) 1 2

We observe here that resultant vector lies in the plane formed by the two component vectors being added. The resultant vector is, therefore, perpendicular to third component vector. Thus, the magnitude of the sum of vector ( a x i + a y j a x i + a y j ) and third vector , a z k a z k , which are perpendicular to each other, is :

Example 1

Problem : Find the angle that vector 2i + j – k makes with y-axis.

Solution : We can answer this question with the help of expression for the cosine of angle that a vector makes with a given axis. We know that component along y-axis is :

Figure 3: Vector makes an angle with y-axis.

Angle

a y = a cos β ⇒ cos β = a y a a y = a cos β ⇒ cos β = a y a

Here,

a y = 1 a y = 1

and

a = √ ( 2 2 + 1 2 + 1 2 ) a = √ ( 2 2 + 1 2 + 1 2 )

Hence,

⇒ cos β = a y a = 1 √ 6 ⇒ β = cos - 1 1 √ 6 ⇒ cos β = a y a = 1 √ 6 ⇒ β = cos - 1 1 √ 6

Planar components of a vector

The planar components of a vector lies in the plane of vector. Since there are two perpendicular axes involved with a plane, a vector is resolved in two components which lie in the same plane as that of vector. Clearly, a vector is composed of components in only two directions :

a = a cos α i + a cos β j a = a cos α i + a cos β j

From the figure depicting a planar coordinate, it is clear that angle “β” is compliment of angle “α”. If α = θ, then

Figure 4: The direction of a planar vector with respect to rectangular axes can be described by a single angle.

Planar vector

α = θ and β = 90 ° - θ α = θ and β = 90 ° - θ

Putting in the expression for the vector,

From graphical representation, the tangent of the angle that vector makes with x-axis is :

Similarly, the tangent of the angle that vector makes with y-axis is :

Example 2

Problem : Find the unit vector in the direction of a bisector of the angle between a pair of coordinate axes.

Solution : The unit vector along the direction of a bisector lies in the plane formed by two coordinates. The bisector makes an angle of 45° with either of the axes. Hence, required unit vector is :

Figure 5: Unit vector along the bisector.

Unit vector

n = cos θ i + sin θ j = cos 45 ° i + sin 45 ° j ⇒ n = 1 √ 2 x ( i + j ) n = cos θ i + sin θ j = cos 45 ° i + sin 45 ° j ⇒ n = 1 √ 2 x ( i + j )

Note : We may check that the magnitude of the unit vector is indeed 1.

| n | = n = √ { ( 1 √ 2 ) 2 + ( 1 √ 2 ) 2 } = √ 1 = 1 | n | = n = √ { ( 1 √ 2 ) 2 + ( 1 √ 2 ) 2 } = √ 1 = 1

Representation of a vector in component form

The angle involved in determination of components plays important role. There is a bit of ambiguity about angle being used. Actually, there are two ways to write a vector in component form. We shall illustrate these two methods with an illustration. Let us consider an example. Here, we consider a vector OA having magnitude of 10 units. The vector makes 150° and 30° angle with x and y axes as shown in the figure.

Figure 6: Component of a vector

Component of a vector

Using definition of components, the vector OA is represented as :

OA = 10 cos 150 0 i + 10 cos 30 0 j OA = 10 cos 150 0 i + 10 cos 30 0 j ⇒ OA = 10 X - 1 2 i + 10 X 3 2 j ⇒ OA = 10 X - 1 2 i + 10 X 3 2 j ⇒ OA = - 5 i + 5 3 j ⇒ OA = - 5 i + 5 3 j

Note that we use angles that vector makes with the axes to determine scalar components. We obtain corresponding component vectors by multiplying scalar components with respective unit vectors of the axes involved. We can, however, use another method in which we only consider acute angle – irrespective of directions of axes involved. Here, vector OA makes acute angle of 60° with x-axis. While representing vector, we put a negative sign if the component is opposite to the positive directions of axes. We can easily determine this by observing projection of vector on the axes. Following this :

OA = - cos 60 0 i + 10 sin 60 0 j OA = - cos 60 0 i + 10 sin 60 0 j ⇒ OA = - 5 i + 5 3 j ⇒ OA = - 5 i + 5 3 j

Note that we put a negative sign before component along x-direction as projection of vector on x-axis is in opposite direction with respect to positive direction of x-axis. The y - projection, however, is in the positive direction of y-axis. As such, we do not need to put a negative sign before the component. Generally, people prefer second method as trigonometric functions are positive in first quadrant. We are not worried about the sign of trigonometric function at all.

Exercise 1

Write force OA in component form.

Figure 7: Component of a vector.

Component of a vector

Solution

F x = 10 sin 30 0 = 10 X 1 2 = 5 N F x = 10 sin 30 0 = 10 X 1 2 = 5 N F y = 10 cos 30 0 = 10 X 3 2 = 5 3 N F y = 10 cos 30 0 = 10 X 3 2 = 5 3 N F = − F x i − F y j = − 5 i − 5 3 j N F = − F x i − F y j = − 5 i − 5 3 j N

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Exercise 2

Write force OA in component form.

Figure 8: Component of a vector.

Component of a vector

Solution

F x = 10 sin 30 0 = 10 X 1 2 = 5 N F x = 10 sin 30 0 = 10 X 1 2 = 5 N F y = 10 cos 30 0 = 10 X 3 2 = 5 3 N F y = 10 cos 30 0 = 10 X 3 2 = 5 3 N F = F x i − F y j = 5 i − 5 3 j N F = F x i − F y j = 5 i − 5 3 j N

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Vector addition : Algebraic method

Graphical method is meticulous and tedious as it involves drawing of vectors on a scale and measurement of angles. More importantly, it does not allow algebraic operations that otherwise would give a simple solution. We can, however, extend algebraic techniques to vectors, provided vectors are represented on a rectangular coordinate system. The representation of a vector on coordinate system uses the concept of unit vector and component.

Now, the stage is set to design a frame work, which allows vector addition with algebraic methods. The frame work for vector addition draws on two important concepts. The first concept is that a vector can be equivalently expressed in terms of three component vectors :

a = a x i + a y j + a z k b = b x i + b y j + b z k a = a x i + a y j + a z k b = b x i + b y j + b z k

The component vector form has important significance. It ensures that component vectors to be added are restricted to three known directions only. This paradigm eliminates the possibility of unknown direction. The second concept is that vectors along a direction can be treated algebraically. If two vectors are along the same line, then resultant is given as :

When θ = 0°, cosθ = cos 0° = 1 and,

⇒ R = √ ( P 2 + 2 P Q + Q 2 ) = √ { ( P + Q ) 2 } = P + Q ⇒ R = √ ( P 2 + 2 P Q + Q 2 ) = √ { ( P + Q ) 2 } = P + Q

When θ = 180°, cosθ = cos 180° = -1 and,

⇒ R = √ ( P 2 - 2 P Q + Q 2 ) = √ { ( P - Q ) 2 } = P - Q ⇒ R = √ ( P 2 - 2 P Q + Q 2 ) = √ { ( P - Q ) 2 } = P - Q

Thus, we see that the magnitude of the resultant is equal to algebraic sum of the magnitudes of the two vectors.

Using these two concepts, the addition of vectors is affected as outlined here :

1: Represent the vectors (a and b) to be added in terms of components :

a = a x i + a y j + a z k b = b x i + b y j + b z k a = a x i + a y j + a z k b = b x i + b y j + b z k

2: Group components in a given direction :

a + b = a x i + a y j + a z k + b x i + b y j + b z k ⇒ a + b = ( a x + b x ) i + ( a y + b y ) j + ( a z + b z ) k a + b = a x i + a y j + a z k + b x i + b y j + b z k ⇒ a + b = ( a x + b x ) i + ( a y + b y ) j + ( a z + b z ) k

3: Find the magnitude and direction of the sum, using analytical method

Exercises

Exercise 3

Find the angle that vector √3i - j makes with y-axis.

Solution

From graphical representation, the angle that vector makes with y-axis has following trigonometric function :

Figure 9: Vector makes an angle with y-axis.

Angle

tan θ = a x a y tan θ = a x a y

Now, we apply the formulae to find the angle, say θ, with y-axis,

tan θ = - √ 3 1 = - √ 3 = tan 120 ° tan θ = - √ 3 1 = - √ 3 = tan 120 °

⇒ θ = 120 ° ⇒ θ = 120 °

In case, we are only interested to know the magnitude of angle between vector and y-axis, then we can neglect the negative sign,

tan θ’ = √ 3 = tan 60 ° tan θ’ = √ 3 = tan 60 °

⇒ θ’ = 60 ° ⇒ θ’ = 60 °

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Exercise 4

If a vector makes angles α,β and γ with x,y and z axes of a rectangular coordinate system, then prove that :

cos 2 α + cos 2 β + cos 2 γ = 1 cos 2 α + cos 2 β + cos 2 γ = 1

Solution

The vector can be expressed in terms of its component as :

a = a x i + a y j + a z k a = a x i + a y j + a z k

where a x = a cos α ; a y = a cos β ; a z = a cos γ a x = a cos α ; a y = a cos β ; a z = a cos γ .

Figure 10: Vector makes different angles with axes..

Angles

The magnitude of the vector is given by :

a = √ ( a x 2 + a y 2 + a z 2 ) a = √ ( a x 2 + a y 2 + a z 2 )

Putting expressions of components in the equation,

⇒ a = √ ( a 2 cos 2 α + a 2 cos 2 β + a 2 cos 2 γ ) ⇒ a = √ ( a 2 cos 2 α + a 2 cos 2 β + a 2 cos 2 γ )

⇒ √ ( cos 2 α + cos 2 β + cos 2 γ ) = 1 ⇒ √ ( cos 2 α + cos 2 β + cos 2 γ ) = 1

Squaring both sides,

⇒ cos 2 α + cos 2 β + cos 2 γ = 1 ⇒ cos 2 α + cos 2 β + cos 2 γ = 1

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Exercise 5

Find the components of weight of a block along the incline and perpendicular to the incline.

Solution

The component of the weight along the incline is :

Figure 11: Components of weight along the incline and perpendicualt to the incline.

Weight on an incline

W x = mg sin θ = 100 x sin 30 ° = 100 x 1 2 = 50 N W x = mg sin θ = 100 x sin 30 ° = 100 x 1 2 = 50 N

and the component of weight perpendicular to incline is :

W y = mg cos θ = 100 x cos 30 ° = 100 x √ 3 2 = 50 √ 3 N W y = mg cos θ = 100 x cos 30 ° = 100 x √ 3 2 = 50 √ 3 N

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Exercise 6

The sum of magnitudes of two forces acting at a point is 16 N. If the resultant of the two forces is 8 N and it is normal to the smaller of the two forces, then find the forces.

Solution

We depict the situation as shown in the figure. The resultant force R is shown normal to small force F 1 F 1 . In order that the sum of the forces is equal to R, the component of larger force along the x-direction should be equal to smaller force :

Figure 12: The resultant force is perpendicular to smaller of the two forces.

Two forces acting on a point

F 2 sin θ = F 1 F 2 sin θ = F 1

Also, the component of the larger force along y-direction should be equal to the magnitude of resultant,

F 2 cos θ = R = 8 F 2 cos θ = R = 8

Squaring and adding two equations, we have :

F 2 2 = F 1 2 + 64 F 2 2 = F 1 2 + 64

⇒ F 2 2 - F 1 2 = 64 ⇒ F 2 2 - F 1 2 = 64

However, according to the question,

⇒ F 1 + F 2 = 16 ⇒ F 1 + F 2 = 16

Substituting, we have :

⇒ F 2 - F 1 = 64 16 = 4 ⇒ F 2 - F 1 = 64 16 = 4

Solving,

⇒ F 1 = 6 N ⇒ F 2 = 10 N ⇒ F 1 = 6 N ⇒ F 2 = 10 N

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