Example 1

Problem : Three radial vectors OA, OB and OC act at the center of a circle of radius “r” as shown in the figure. Find the magnitude of resultant vector.

Figure 1: Three radial vectors OA, OB and OC act at the center of a circle of radius “r”.

Sum of three vectors

Solution : It is evident that vectors are equal in magnitude and is equal to the radius of the circle. The magnitude of the resultant of horizontal and vertical vectors is :

R’ = √ ( r 2 + r 2 ) = √ 2 r R’ = √ ( r 2 + r 2 ) = √ 2 r

The resultant of horizontal and vertical vectors is along the bisector of angle i.e. along the remaining third vector OB. Hence, magnitude of resultant of all three vectors is :

R’ = OB + R’ = r + √ 2 r = ( 1 + √ 2 ) r R’ = OB + R’ = r + √ 2 r = ( 1 + √ 2 ) r

Example 2

Problem : Under what condition does the magnitude of the resultant of two vectors of equal magnitude, is equal in magnitude to either of two equal vectors?

Solution : We know that resultant of two vectors is represented by the closing side of a triangle. If the triangle is equilateral then all three sides are equal. As such magnitude of the resultant of two vectors is equal to the magnitude of either of the two vectors.

Figure 2: Resultant of two vectors

Two vectors

Under this condition, vectors of equal magnitude make an angle of 120° between them.

Example 3

Problem : If three vectors are represented by three sides of a triangle, then find the resultant vector.

Solution : Let us consider any two vectors in sequence like AB and BC. According to triangle law of vector addition, the resultant vector is represented by the third closing side in the opposite direction. It means that :

Figure 3: Three vectors are represented by three sides.

Three vectors

⇒ AB + BC = AC ⇒ AB + BC = AC

Adding vector CA on either sides of the equation,

⇒ AB + BC + CA = AC + CA ⇒ AB + BC + CA = AC + CA

The right hand side of the equation is vector sum of two equal and opposite vectors, which evaluates to zero. Hence,

Figure 4: The resultant of three vectors represented by three sides is zero.

Three vectors

⇒ AB + BC + CA = 0 ⇒ AB + BC + CA = 0

Note : If the vectors represented by the sides of a triangle are force vectors, then resultant vector is zero. It means that three such forces represented by the sides of a triangle is a balanced force system.

Example 4

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

Example 5

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 6: 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

Example 6

Problem : 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 7: 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 °

Example 7

Problem : At what angle does two vectors a+b and a-b act so that the resultant is √ ( 3 a 2 + b 2 ) √ ( 3 a 2 + b 2 ) .

Solution : The magnitude of resultant of two vectors is given by :

Figure 8: The angle between the sum and difference of vectors.

Angle

R = √ { ( a + b ) 2 + ( a - b ) 2 + 2 ( a + b ) ( a - b ) cos θ } R = √ { ( a + b ) 2 + ( a - b ) 2 + 2 ( a + b ) ( a - b ) cos θ }

Substituting the expression for magnitude of resultant as given,

⇒ √ ( 3 a 2 + b 2 ) = √ { ( a + b ) 2 + ( a - b ) 2 + 2 ( a + b ) ( a - b ) cos θ } ⇒ √ ( 3 a 2 + b 2 ) = √ { ( a + b ) 2 + ( a - b ) 2 + 2 ( a + b ) ( a - b ) cos θ }

Squaring on both sides, we have :

⇒ ( 3 a 2 + b 2 ) = { ( a + b ) 2 + ( a - b ) 2 + 2 ( a + b ) ( a - b ) cos θ } ⇒ ( 3 a 2 + b 2 ) = { ( a + b ) 2 + ( a - b ) 2 + 2 ( a + b ) ( a - b ) cos θ }

⇒ cos θ = ( a 2 - b 2 ) 2 ( a 2 - b 2 ) = 1 2 = cos 60 ° ⇒ cos θ = ( a 2 - b 2 ) 2 ( a 2 - b 2 ) = 1 2 = cos 60 °

⇒ θ = 60 ° ⇒ θ = 60 °

Example 8

Problem : 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 9: 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

Example 9

Problem : If a and b be two vectors, then prove

| a + b | ≤ | a | + | b | | a + b | ≤ | a | + | b |

Solution : We know that vectors A, B and their sum A+B is represented by three side of a triangle OAC. Further we know that a side of triangle is always less than the sum of remaining two sides. It means that :

Figure 10: Sum of two vectors

Two vectors

OC < OA + OC ⇒ | a + b | < | a | + | b | OC < OA + OC ⇒ | a + b | < | a | + | b |

There is one possibility, however, that two vectors A and B are collinear and act in the same direction. In that case, magnitude of their resultant will be equal to the sum of the magnitudes of individual vector.

OC = OA + OB ⇒ | a + b | = | a | + | b | OC = OA + OB ⇒ | a + b | = | a | + | b |

Combining two results, we have :

| a + b | ≤ | a | + | b | | a + b | ≤ | a | + | b |

Example 10

Problem : If a and b be two vectors, then prove

| a - b | ≥ | a | - | b | | a - b | ≥ | a | - | b |

Solution : We know that vectors A, B and their difference A-B are represented by three side of a triangle OAE. Further we know that a side of triangle is always less than the sum of remaining two sides. It means that sum of two sides is greater than the third side :

Figure 11: Difference of two vectors

Two vectors

OE + AE > OA ⇒ OE > OA - AE ⇒ | a - b | > | a | - | b | OE + AE > OA ⇒ OE > OA - AE ⇒ | a - b | > | a | - | b |

There is one possibility, however, that two vectors A and B are collinear and act in the same direction. In that case, magnitude of their difference will be equal to the difference of the magnitudes of individual vector.

⇒ OE = OA - AE ⇒ | a - b | = | a | - | b | ⇒ OE = OA - AE ⇒ | a - b | = | a | - | b |

Combining two results, we have :

| a - b | ≥ | a | - | b | | a - b | ≥ | a | - | b |

Example 11

Problem : If resultant of three vectors a, b and c is zero (null vector), then prove that

a sin α = b sin β = c sin γ a sin α = b sin β = c sin γ

Figure 12: Three non-collinear vectors.

Three vectors

where α, β and γ be the angle between the remaining pairs of vectors.

Solution : If the resultant of three vectors is zero, then they are represented by three sides of a triangle in magnitude and direction.

Figure 13: Three vectors are represented by three sides of a triangle.

Three vectors

Considering the magnitude of vectors and applying sine law of triangle, we have :

AB sin BCA = BC sin CAB = CA sin ABC AB sin BCA = BC sin CAB = CA sin ABC

⇒ AB sin ( π - α ) = BC sin ( π - β ) = CA sin ( π - γ ) ⇒ AB sin ( π - α ) = BC sin ( π - β ) = CA sin ( π - γ )

⇒ AB sin α = BC sin β = CA sin γ ⇒ AB sin α = BC sin β = CA sin γ

This relationship is also known as Lami’s theorem. It is important to note that the ratio involves exterior (outside) angles – not the interior angles of the triangle. Also, the angle associated with the magnitude of a vector in the individual ratio is the included angle between the remaining vectors (refer the figure in the question).

Example 12

Problem : Find the components of weight along the incline and perpendicular to the incline.

Solution : The component of the weight along the incline is :

Figure 14: 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

Example 13

Problem : 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 15: 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