Vector Algebra
Introduction
In this chapter we will come to know about some interesting facts of vectors and how they are used in our day-to-day life.
Following are some examples where vectors are used.
- Motion of aeroplane is based on the concept of vectors. When the motion of the aeroplane is considered both speed and in which direction it flies in the sky is also taken into account.
- The landing instructions which a pilot receives from air traffic controller.
- Suppose when a ball is thrown let it be thrown with some speed (for e.g.:-5m/s) and the direction in which it is thrown is let north-east .This shows we are considering a vector as the ball has both speed and direction.
- Vectors are to derive results in engineering and science. Fluid mechanics, static, Electrical Engineering etc.
- In game of cricket the bowler bowls the ball with some speed and in some direction and also the batsman hits the ball with some speed and in particular direction. So the concept of vectors is involved
- A motor can be either used for pumping water or in a car. Motors work on the principle of torque which is a vector quantity.
- In the game of carom board the striker is hit with some speed and in some direction.
Vectors
Vectors are the physical quantities which have both direction as well as magnitude. Vectors are represented by an arrow. The point A (as shown in the figure) from where the vector \(|\overrightarrow {AB} |\) starts is called its initial point. The point B where it ends is called its terminal point.
The distance between initial and terminal points of a vector is called the magnitude (or length) of the vector, denoted by \(|\overrightarrow {AB} |\) , or \(|\overrightarrow {a} |\) or |a|.
The arrow indicates the direction of the vector.
Scalar & Vector
Consider the following measures and to classify them as scalar and vectors.
Right handed rectangular coordinate system
Right hand thumb rule is applied to understand the orientation conventions for vectors in three dimensions. Suppose if we fold our 4 fingers as shown in the diagram and move from X to Y axis then the thumb gives the direction of Z-axis.
Position Vector
Consider a point P in space, having coordinates (x, y, z) with respect to the origin O (0, 0, 0). Then, the vector \(|\overrightarrow {AB} |\) having O and P as its initial and terminal points, respectively, is called position vector of the point P with respect to O.
The value of \(|\overrightarrow {OP} | = \sqrt {{x^2} + {y^2} + {z^2}} \).
Direction Angle & Direction Cosines
Consider a position vector \(\overrightarrow {OP} \) (or \(\overrightarrow r \)) of a point P(x, y, z) as shown in the figure. The angles α, β and γ made by the vector \(\overrightarrow r \) with the (+)ve directions of x,y and z-axes. The cosine values of these angles, i.e., cos α, cos β and cos γ are known as direction cosines of the vector \(\overrightarrow r \). They are denoted by (l, m and n). Therefore, cos α = l, cosβ = m and cos γ = n.
Consider the right angled triangle OAP,
\( \Rightarrow \cos \alpha = \left( {\dfrac{x}{r}} \right)\) where (\(r = \,|\overrightarrow r |\)). Similarly from right angled triangles(OBP) and (OCP) we get \(\cos \beta = \left( {\dfrac{y}{r}} \right)\) and \(\cos \gamma = \left( {\dfrac{z}{r}} \right)\) . The coordinates of the point P can also be written as (lr, mr, nr). The lr, mr, nr are called as direction ratios of vector \(\overrightarrow r \). The direction ratios are denoted as a ,b and c respectively.
In general l2 + m2 + n2 = 1.
Problem: -
Write two different vectors having same direction.
Answer: -
Consider \(\overrightarrow a = \left( {\widehat i + \widehat j + \widehat k} \right)\) and \(\overrightarrow b = \left( {2\widehat i + 2\widehat j + 2\widehat k} \right)\). The direction cosines of \(\overrightarrow a \) are given by,
\(l = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }}\)
\(m = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }}\)
\(n = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }}\)
The direction cosines of \(\overrightarrow b \) are given by,
\(l = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{1}{{\sqrt 3 }}\)
\(m = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{1}{{\sqrt 3 }}\)
\(n = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{1}{{\sqrt 3 }}\)
The direction cosines of \(\overrightarrow a \) and \(\overrightarrow b \) are the same. Hence, the two vectors have the same direction.
Problem:-
Find the direction cosines of the vector î +2ĵ + 3k̂
Answer:-
Let \(\overrightarrow a \) = î +2ĵ + 3k̂.
Therefore \(|\overrightarrow a |\,\, = \dfrac{1}{{\sqrt {{1^2} + {2^2} + {3^2}} }} = \dfrac{1}{{\sqrt {1 + 4 + 9} }} = \dfrac{1}{{\sqrt {14} }}.\)
Hence, the direction cosines of \(\overrightarrow a \) are \(\dfrac{1}{{\sqrt {14} }},\,\,\dfrac{2}{{\sqrt {14} }},\,\,\dfrac{3}{{\sqrt {14} }}.\)
Types of Vectors
- Zero Vector
- Unit Vector
- Coinitial Vector
- Collinear Vector
- Equal Vectors
- Negative Vectors
- Free Vectors
Zero Vector:-
A vector, whose initial and terminal points coincide, is called a zero vector or (null vector). It is denoted by \(\overrightarrow 0 \). Zero vector cannot be assigned a definite direction as it has zero magnitude.
Unit Vector:-
A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The unit vector in the direction of a given vector \(\overrightarrow a \) is denoted by \({\widehat a}\).
Coinitial Vectors:-
Two or more vectors having same initial point are called Coinitial vectors.
Collinear Vectors:-
Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions.
For example: - Consider 3 vectors as shown in the figure, they all are parallel to each other but their magnitudes are different as well as the directions. But they are said to be collinear vectors because they are parallel to each other.
Equal Vectors:-
Two vectors are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points.
For example: - Consider 2 vectors whose magnitudes and their directions are same irrespective of origin, then they are known as equal vectors.
Negative of a Vector:-
A vector whose magnitude is same as that of a given vector but direction is opposite to that of it is called negative of the given vector.
Free Vectors:-
Vectors that don’t change even if it is displaced in parallel direction without changing its magnitude and direction are called free vectors.
Problem:- Represent graphically a displacement of 40 km, 30° west of south.
Answer: -
The vector \(\overrightarrow {OP} \) represents the required displacement.
Addition of Vectors
Consider a scenario when a man moves from A to B and then from B to C (as shown in the fig).
Then the net displacement made by a man from point A to point C, is given by the vector \(\overrightarrow {AC} \) and expressed as:-
\(\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} \)
This is known as the triangle law of addition.
Head to tail vector addition
To add two vectors, position them so that the initial point of one coincides with the terminal of the other.
Consider two vectors \(\overrightarrow a \) and \(\overrightarrow b \) and move them parallel to each other such that tail of \(\overrightarrow b \) touches the head of \(\overrightarrow a \). Then the resultant vector will be given as vector (\(\overrightarrow a + \overrightarrow b \)). Also \(\overrightarrow {AC} = - \,\overrightarrow {CA} \) from equation (\(\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} \)).
\(\overrightarrow {AA} = \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = 0\). This implies when we consider the sides of the triangle in order, it leads to zero resultant as both initial and terminal points coincides with each other. This is shown in the figure (ii).
Constructing a vector as shown in the figure (iii) \(\overrightarrow {BC'} \) and the magnitude of \(\overrightarrow {BC'} \) is same as \(\overrightarrow {BC} \) but the direction is opposite to the vector \(\overrightarrow {BC} \), i.e.
\(\overrightarrow {BC'} = - \,\overrightarrow {BC} \)
By applying triangle law of addition we have, \(\overrightarrow {AC'} = \overrightarrow {AB} + \overrightarrow {BC} \).
\[ = \overrightarrow {AB} + ( - \,\overrightarrow {BC} )\]
\[\overrightarrow {AC'} = \overrightarrow a - \,\overrightarrow b \]
Therefore vector \(\overrightarrow {AC'} \) is said to represent the difference of \(\overrightarrow a \) and \(\overrightarrow b \).
Parallelogram law of vector addition:-
If we have two vectors (\(\overrightarrow a \) and \(\overrightarrow a \)) represented by the two adjacent sides of a parallelogram in magnitude and direction, then their sum (\(\overrightarrow a + \overrightarrow b \)) is represented in magnitude and direction by the diagonal of the parallelogram through common point.
By using parallelogram law of vector, it will be given as: \(\overrightarrow {OC} = \overrightarrow {OA} + \overrightarrow {AC} \) (Fig)
In triangle OAC applying triangle law of addition,
\(\overrightarrow {OC} = \overrightarrow {OA} + \overrightarrow {AC} \)
Therefore, \(\overrightarrow {OC} = \overrightarrow {a} + \overrightarrow {b} \)
Again applying triangle law of addition in triangle OCB,
\(\overrightarrow {OC} = \overrightarrow {OB} + \overrightarrow {CB} \)
\(\overrightarrow {OC} = \overrightarrow {a} + \overrightarrow {b} \)
Hence proved.
Problem:- Find the sum of the vectors \(\overrightarrow a = \left( {\widehat i - 2\widehat j + \widehat k} \right)\) and \(\overrightarrow b = \left( { - 2\widehat i - 4\widehat j + 5\widehat k} \right)\) and \(\overrightarrow c = \left( {\widehat i - 6\widehat j - 7\widehat k} \right)\).
Answer:-
The given vectors are \(\overrightarrow a = \left( {\widehat i - 2\widehat j + \widehat k} \right)\) and \(\overrightarrow b = \left( { - 2\widehat i - 4\widehat j + 5\widehat k} \right)\) and \(\overrightarrow c = \left( {\widehat i - 6\widehat j - 7\widehat k} \right)\).
Therefore \(\overrightarrow a + \overrightarrow b + \overrightarrow c = (1 - 2 + 1)\widehat i + ( - 2 + 4 - 6)\widehat j + (1 + 5 - 7)\widehat k\)
\( = \left( {0.\widehat i - 4\widehat j - 1\widehat k} \right)\)
\({ = - 4\widehat j - \widehat k}\) .
Properties of vector addition
Commutative property: - For any two vectors \(\overrightarrow a \) and \(\overrightarrow b \) if we add (\(\overrightarrow a + \overrightarrow b \)) or (\(\overrightarrow b + \overrightarrow b \) ) ; the result will be same i.e. (\(\overrightarrow a + \overrightarrow b \)) = (\(\overrightarrow a + \overrightarrow b \)).
Associative property:-
For any three vectors \(\overrightarrow a \), \(\overrightarrow b \) and \(\overrightarrow c \) then:
\((\overrightarrow a + \overrightarrow b ) + \overrightarrow c = \overrightarrow a + (\overrightarrow b + \overrightarrow c )\)
If we add first \(\overrightarrow a \) and \(\overrightarrow b \) and then add the result to \(\overrightarrow c \) it will be same if we first add (\(\overrightarrow b \) and \(\overrightarrow c \)) and then adding the result the vector \(\overrightarrow a \).
Additive Identity:- Suppose if add \(\overrightarrow 0 \) to any vector \(\overrightarrow a \) then we will get the vector \(\overrightarrow a \).
(\(\overrightarrow a \) + \(\overrightarrow 0 \)) = (\(\overrightarrow 0 \) + \(\overrightarrow a \)) = \(\overrightarrow a \).
Therefore \(\overrightarrow 0 \) is the additive identity for the vector addition.
Multiplication of a Vector by a Scalar:-
Let \(\overrightarrow a \) be any vector and λ be a scalar. Multiplication of vector \(\overrightarrow a \) by a scalar λ is denoted as : \(\lambda \overrightarrow a \). \(\lambda \overrightarrow a \) is also a vector which has the direction same or opposite to that of vector \(\overrightarrow a \).
The magnitude of vector (\(\lambda \overrightarrow a \) is given as : \(|\lambda \overrightarrow a |\, = |\lambda |\,\,|\overrightarrow a |\)
Geometrically can be shown as:-
Additive Inverse: -
If we add a vector \(\overrightarrow a \) with itself but in opposite direction then the resultant is \(\overrightarrow a \). If λ (scalar) = – 1, then \(\lambda \overrightarrow a = - \overrightarrow a \), the magnitude of negative vector is same as \(\overrightarrow a \) but direction is opposite to \(\overrightarrow a \). The vector \(\left( { - \overrightarrow a } \right)\) is called the negative (or additive inverse) of vector \(\overrightarrow a \). It is given as \(\overrightarrow a \) + (\( - \overrightarrow a \)) = (\( - \overrightarrow a \)) + \(\overrightarrow a \) = \(\overrightarrow 0 \).
Unit Vector in a direction
Unit vector is a vector of magnitude 1. It is denoted by \({\widehat a}\).
Let λ(scalar) \( = \left( {\dfrac{1}{{|\overrightarrow a |}}} \right)\), then \(|\lambda \overrightarrow a |\, = |\lambda |\,\,|\overrightarrow a |\). \( \Rightarrow \left( {\dfrac{1}{{|\overrightarrow a |}}} \right)|\overrightarrow a |\, = 1\). Therefore \(\lambda \overrightarrow a \) represents the unit vector in the direction of \(\overrightarrow a \).
Unit vector can be written as: \(\widehat a = \left( {\dfrac{1}{{|\overrightarrow a |}}} \right) \times \overrightarrow a \).
Components of a vector
Unit Vector
Consider points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on x-axis y-axis and z-axis, respectively. \(|\overrightarrow {OA} |\, = 1,\,\,|\overrightarrow {OB} |\, = 1\) and \(|\overrightarrow {OC} |\, = 1\) , this shows all the 3 vectors having the same magnitude 1.
The vectors \(\overrightarrow {OA} ,\,\,\,\overrightarrow {OB} \,\,\,and\,\,\overrightarrow {OC} \) are known as unit vectors along the axes OX, OY and OZ respectively and they are denoted by î , ĵ and k̂ respectively.
Problem:-
Find the unit vector in the direction of the vector \({\overrightarrow a }\) = (î + ĵ +2 k̂).
Answer:
The unit vector n̂ in the direction of vector \({\overrightarrow a }\) = (î + ĵ +2 k̂).
\(|\overrightarrow a | = \sqrt {{1^2} + {1^2} + {2^2}} = \sqrt 6 \).
Therefore \(\widehat a = \left( {\dfrac{{\overrightarrow a }}{{|\overrightarrow a |}}} \right)\)
\( \Rightarrow \widehat a = \dfrac{{\widehat i + \widehat j + 2\widehat k}}{{\sqrt 6 }} = \dfrac{1}{{\sqrt 6 }}\widehat i + \dfrac{1}{{\sqrt 6 }}\widehat j + \dfrac{2}{{\sqrt 6 }}\widehat k\)
Problem:-
Find the unit vector in the direction of vector a + b, \(\overrightarrow a = (2\widehat i - \widehat j + 2\widehat k)\) and \(\overrightarrow b = (- \widehat i + \widehat j - \widehat k)\).
Answer:-
The given vectors are \(\overrightarrow a = (2\widehat i - \widehat j + 2\widehat k)\) and \(\overrightarrow b = (- \widehat i + \widehat j - \widehat k)\).
Therefore \(\overrightarrow a + \overrightarrow b = (2 - 1)\widehat i + ( - 1 + 1)\widehat j + (2 - 1)\widehat k\)\( = (1)\widehat i + (0)\widehat j + (1)\widehat k\)
= î + k̂
\(|\overrightarrow a + \overrightarrow b | = \sqrt {{1^2} + {1^2}} = \sqrt 2 \)\(\dfrac{{\overrightarrow a + \overrightarrow b }}{{|\overrightarrow a + \overrightarrow b |}} = \dfrac{{\widehat i + \widehat k}}{{\sqrt 2 }} = \dfrac{{\widehat i}}{{\sqrt 2 }} + \dfrac{{\widehat k}}{{\sqrt 2 }}\)
Position Vector:-
Consider a position vector \(\overrightarrow {OP} \) of a point P(x, y, z) as shown in the figure. Let P1 = perpendicular from P onto the plane XOY .This shows P1P is parallel to the z-axis. Also from the figure we can see that the î , ĵ and k̂ are the unit vectors along x, y and z-axes respectively.
By the definition of the coordinates of P, \(\overrightarrow {{P_1}P} = \overrightarrow {OR} = z\widehat k\) . Similarly, \(\overrightarrow {Q{P_1}} = \overrightarrow {OS} = y\widehat j\) and \(\overrightarrow {OQ} = x\widehat i\) .
Therefore \(\overrightarrow {O{P_1}} = \overrightarrow {OQ} + \overrightarrow {Q{P_1}} = x\widehat i + y\widehat j\) and \(\overrightarrow {OP} = \overrightarrow {O{P_1}} + \overrightarrow {{P_1}P} = x\widehat i + y\widehat j + z\widehat k\).
Hence, the position vector of P with reference to O is given as \(\overrightarrow {OP} \) or \(\overrightarrow r = x\widehat i + y\widehat j + z\widehat k\). Where r = position vector OP. This form of any vector is known as component form. x, y and z are called scalar components of vector \(\overrightarrow r \). x î + yĵ + z k̂ are called vector components of vector \(\overrightarrow r \) along the 3 respective axes.
The length of any vector r⃗ = x î + yĵ + z k̂ is given as | \(\overrightarrow r \)| = \(\sqrt {{x^2} + {y^2} + {z^2}} \).
Problem: - Compute the magnitude of the following vectors: \(\overrightarrow a = \widehat i + \widehat j + \widehat k\) and \(\overrightarrow b = 2 \widehat i - 7 \widehat j - 3 \widehat k\).
Answer:
The given vectors are: \(\overrightarrow a = \widehat i + \widehat j + \widehat k\) and \(\overrightarrow b = 2 \widehat i - 7 \widehat j - 3 \widehat k\).
\(|\overrightarrow a | = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3 \)
\(|\overrightarrow b | = \sqrt {{{(2)}^2} + {{( - 7)}^2} + {{( - 3)}^2}} = \sqrt {62} \)
Write two different vectors having same magnitude.
Answer: -
Consider \(\overrightarrow a = \widehat i - 2\widehat j + 3\widehat k\) and \(\overrightarrow b = 2\widehat i + \widehat j - 3\widehat k\).
It can be observed that:
\(|\overrightarrow a | = \sqrt {{{(1)}^2} + {{( - 2)}^2} + {{(3)}^2}} = \sqrt {1 + 4 + 9} = \sqrt {14} \) and
\(|\overrightarrow b | = \sqrt {{{(2)}^2} + {{(1)}^2} + {{( - 3)}^2}} = \sqrt {4 + 1 + 9} = \sqrt {14} \).
Hence, \(\overrightarrow a \) and \(\overrightarrow b \) are two different vectors having the same magnitude. The vectors are different because they have different directions.
Components of a vector: Vectors Operations
Let there be two vectors given by \(\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\) and \(\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\)
- Sum of vectors a and b is given by: \(\overrightarrow a + \overrightarrow b = ({a_1} + {a_2})\widehat i + ({b_1} + {b_2})\widehat j + ({a_3} + {b_3})\widehat k\)
- Difference of vectors a and b is given by: \(\overrightarrow a - \overrightarrow b = ({a_1} - {a_2})\widehat i + ({b_1} - {b_2})\widehat j + ({a_3} - {b_3})\widehat k\)
- Vectors a & b are equal if and only if a1 = b1, a2 = b2 and a3 = b3.
- Multiplication of vector by Scalar \(\lambda \overrightarrow a = (\lambda {a_1})\widehat i + (\lambda {a_2})\widehat j + (\lambda {a_3})\widehat k\)
Distributive law
Let \(\overrightarrow a \) and \(\overrightarrow b \) be any two vectors, and k and m be any scalars. Then
\[k\overrightarrow a + m\overrightarrow a = (k + m)\overrightarrow a \]
\[km\overrightarrow a = (km)\overrightarrow a \]
\[k(\overrightarrow a + \overrightarrow b ) = k\overrightarrow a + k\overrightarrow b \]
Collinear Vectors
For any value of λ, the vector (λa) is always collinear to the vector \(\overrightarrow a \).
Two vectors \(\overrightarrow a \) and \(\overrightarrow a \) are collinear if and only if there exists a nonzero scalar λ such that \(\overrightarrow b = \lambda \overrightarrow a \).
\( \Rightarrow {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k = \lambda ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k)\) where \(\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\) and \(\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\)
\( \Rightarrow {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k = (\lambda {a_1})\widehat i + (\lambda {a_2})\widehat j + (\lambda {a_3})\widehat k\)
\( \Rightarrow {b_1} = \lambda {a_1},{b_2} = \lambda {a_2},{b_3} = \lambda {a_3}\)
Therefore, \(\left( {\dfrac{{{b_1}}}{{{a_1}}}} \right) = \left( {\dfrac{{{b_2}}}{{{a_2}}}} \right) = \left( {\dfrac{{{b_3}}}{{{a_3}}}} \right) = \lambda \)
Note:-
- If \(\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\), then a1, a2 and a3 are also called as direction ratios of \(\overrightarrow a \).
- If l, m, n are direction cosines of a vector, then \(l\widehat i + m\widehat j + n\widehat k = (\cos \alpha )\widehat i + (\cos \beta )\widehat j + (\cos \gamma )\widehat k\) is the unit vector in the direction of that vector, where α, β and γ are the angles which the vector makes with x, y and z axes respectively.
Problem:-
Show that the vectors \(2\widehat i - 3\widehat j + 4\widehat k\) and \( - 4\widehat i + 6\widehat j - 8\widehat k\) are collinear.
Answer:-
Let \(\overrightarrow a = 2\widehat i - 3\widehat j + 4\widehat k\) and \(\overrightarrow b = - 4\widehat i + 6\widehat j - 8\widehat k\).
It is observed that \(\overrightarrow b = - 4\widehat i + 6\widehat j - 8\widehat k = - 2(2\widehat i - 3\widehat j + 4\widehat k) = - 2\overrightarrow a \)
Therefore \(\overrightarrow b = \lambda \overrightarrow a \) where λ = - 2
Hence, the given vectors are collinear.
Problem:-
Find a vector in the direction of vector \(5\widehat i - \widehat j + 2\widehat k\) which has magnitude 8 units.
Answer:-
Let \(\overrightarrow a = 5\widehat i - \widehat j + 2\widehat k\), So, \(|\overrightarrow a |\,\, = \sqrt {{5^2}{{( - 1)}^2} + {2^2}} = \sqrt {25 + 1 + 4} = \sqrt {30} \)
Therefore \(\widehat a = \left( {\dfrac{{\overrightarrow a }}{{|\overrightarrow a |}}} \right) = \dfrac{{5\widehat i - \widehat j + 2\widehat k}}{{\sqrt {30} }}\)
Hence, the vector in the direction of vector \({5\widehat i - \widehat j + 2\widehat k}\) which has magnitude 8 units is given by,
Problem:-
Show that the vector \(\widehat i + \widehat j + \widehat k\) is equally inclined to the axes OX, OY, and OZ.
Answer:-
Let \(\overrightarrow a = \widehat i + \widehat j + \widehat k\). Then, \(|\overrightarrow a |\,\, = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3 \).
Therefore, the direction cosines of \(\overrightarrow a \) are \(\dfrac{1}{{\sqrt 3 }}\), \(\dfrac{1}{{\sqrt 3 }}\), \(\dfrac{1}{{\sqrt 3 }}\).
Now, let α, β and γ be the angles formed by with the positive directions of x, y, and z axes.
\(\cos \alpha = \dfrac{1}{{\sqrt 3 }},\,\,\cos \beta = \dfrac{1}{{\sqrt 3 }},\,\,\cos \gamma = \dfrac{1}{{\sqrt 3 }}\)
Hence, the given vector is equally inclined to axes OX, OY and OZ.
Vectors joining two points
Let P1(x1, y1, z1) and P2(x2, y2, z2) are initial point and terminal point respectively, then the vector joining P1 and P2 is given by vector \(\overrightarrow {{P_1}{P_2}} \).
Joining the points P1 and P2 with the origin O, and applying triangle law of addition. Then from triangle OP1P2(figure) : \(\overrightarrow {O{P_1}} + \overrightarrow {{P_1}{P_2}} = \overrightarrow {O{P_2}} \).
The vector will be always specified as: (Terminal point – Initial point).
This implies \(\overrightarrow {{P_1}{P_2}} = \overrightarrow {O{P_2}} - \overrightarrow {O{P_1}} \).
Therefore \(\overrightarrow {{P_1}{P_2}} = \left( {{x_2}\widehat i + {y_2}\widehat j + {z_2}\widehat k} \right) - \left( {{x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k} \right) = ({x_2} - {x_1})\widehat i + ({y_2} - {y_1})\widehat j + ({z_2} - {z_1})\)
The magnitude of \(\overrightarrow {{P_1}{P_2}} \) is given as: \(|\overrightarrow {{P_1}{P_2}} | = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \)
Problem:-
Find the vector with initial point (2, 1) and terminal point (-5, 7).
Answer:-
\(\overrightarrow {AB} = ( - 5 - 2)\widehat i + (7 - 1)\widehat j = - 7\widehat i + 6\widehat j\)
Problem:-
Find the direction cosines of the vector joining the points A (1, 2, – 3) and B (– 1, – 2, 1).
Answer:-
The given points are A (1, 2, – 3) and B (– 1, – 2, 1). 
Therefore \(\overrightarrow {AB} = ( - 1 - 1)\widehat i + ( - 2 - 2)\widehat j + (1 + 3)\widehat k = - 2\widehat i - 4\widehat j + 4\widehat k\)
Therefore \(|\overrightarrow {AB} | = \sqrt {{{( - 2)}^2} + ( - 4) + {4^2}} = \sqrt {4 + 16 + 16} = \sqrt {36} = 6\)Hence, the direction cosines of \(\overrightarrow {AB} \) are \(\dfrac{{ - 2}}{6}\), \(\dfrac{{ - 4}}{6}\), \(\dfrac{{4}}{6}\) = \(\dfrac{{ - 1}}{3}\), \(\dfrac{{ - 2}}{3}\), \(\dfrac{{2}}{3}\).
Problem:-
Find the unit vector in the direction of vector \(\overrightarrow {PQ} \), where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
Answer:-
The given points are P (1, 2, 3 ) and Q (4, 5, 6 ). 
Therefore \(\overrightarrow {PQ} = (4 - 1)\widehat i + (5 - 2)\widehat j + (6 - 3)\widehat k = 3\widehat i + 3\widehat j + 3\widehat k\)
\(|\overrightarrow {PQ} | = \sqrt {{3^2} + {3^2} + {3^2}} = \sqrt {27} = 3\sqrt 3 \)
Hence, the unit vector in the direction of \(\overrightarrow {PQ} \) is
\(\dfrac{{\overrightarrow {PQ} }}{{|\overrightarrow {PQ} |}} = \dfrac{{3\widehat i + 3\widehat j + 3\widehat k}}{{3\sqrt 3 }} = \dfrac{3}{{3\sqrt 3 }}\widehat i + \dfrac{3}{{3\sqrt 3 }}\widehat j + \dfrac{3}{{3\sqrt 3 }}\widehat k\)
\( = \dfrac{1}{{\sqrt 3 }}\widehat i + \dfrac{1}{{\sqrt 3 }}\widehat j + \dfrac{1}{{\sqrt 3 }}\widehat k\)
Problem:-
Show that the points A, B and C with position vectors, \(\overrightarrow a = 3\widehat i - 4\widehat j - 4\widehat k,\,\,\,\overrightarrow b = 2\widehat i - \widehat j + \widehat k\) and \(\overrightarrow c = \widehat i - 3\widehat j - 5\widehat k\) respectively form the vertices of a right angled triangle.
Answer:-
Position vectors of points A, B, and C are respectively given as:
\(\overrightarrow a = 3\widehat i - 4\widehat j - 4\widehat k,\,\,\,\overrightarrow b = 2\widehat i - \widehat j + \widehat k,\,\,\,\overrightarrow c = \widehat i - 3\widehat j - 5\widehat k\)
Therefore \(\overrightarrow {AB} = (\overrightarrow b - \overrightarrow a ) = (2 - 3)\widehat i + ( - 1 + 4)\widehat j + (1 + 4)\widehat k = - \widehat i + 3\widehat j + 5\widehat k\)
\(\overrightarrow {BC} = (\overrightarrow c - \overrightarrow b ) = (1 - 2)\widehat i + ( - 3 + 1)\widehat j + ( - 5 - 1)\widehat k = - \widehat i - 2\widehat j - 6\widehat k\)
\(\overrightarrow {CA} = (\overrightarrow a - \overrightarrow c ) = (3 - 1)\widehat i + ( - 4 + 3)\widehat j + ( - 4 + 5)\widehat k = 2\widehat i - \widehat j + \widehat k\)
Therefore
\(|\overrightarrow {AB} {|^2}\) = (-1)2+ (3)2+ (5)2 = 35
\(|\overrightarrow {BC} {|^2}\) = (-1)2+ (-2)2+ (-6)2 = 41
\(|\overrightarrow {CA} {|^2}\) = (2)2+ (-1)2+ (1)2 = 6
Therefore \(|\overrightarrow {AB} {|^2}\) + \(|\overrightarrow {CA} {|^2}\) = 36 + 6 = \(|\overrightarrow {BC} {|^2}\)
Hence, ABC is a right-angled triangle.
Section Formula
Let P and Q be two points represented by the position vectors \(\overrightarrow {OP} \) and \[\overrightarrow {OQ} \) respectively w.r.t. to origin.
Then the line segment which joins the points P and Q can be divided by the third point R in two ways :- Internally and Externally.
Case 1:- When R divides PQ internally ,
If R divides \(\overrightarrow {PQ} \) such that m (\(\overrightarrow {QR} \)) = n (\[\overrightarrow {PR} \)),
Where m and n are positive scalar quantities.
Suppose R divides \(\overrightarrow {PQ} \) internally in the ratio m : n.
Consider triangles ORQ and OPR (from the figure):
\(\overrightarrow {RQ} = \overrightarrow {OQ} - \overrightarrow {OR} = \overrightarrow b - \overrightarrow r \) (Equation(1))
\(\overrightarrow {PR} = \overrightarrow {OR} - \overrightarrow {OP} = \overrightarrow r - \overrightarrow a \) (Equation (2))
Therefore from (1) and (2) after simplification
\(m(\overrightarrow b - \overrightarrow r ) = n(\overrightarrow r - \overrightarrow a ) \Rightarrow \overrightarrow r = \dfrac{{m\overrightarrow b + n\overrightarrow a }}{{m + n}}\)
Hence the position vector of the point R which divides P and Q internally in the ratio (m:n) is given by: \(\overrightarrow {OR} = \dfrac{{m\overrightarrow b + n\overrightarrow a }}{{m + n}}\)Case 2:-When R divides PQ externally,
Let we have 2 vectors P and R and the point R is present outside the \(\overrightarrow {PQ} \).
The \(\overrightarrow {PQ} \) is divided externally in the ratio (m : n).
Therefore \(\dfrac{{\overrightarrow {PQ} }}{{\overrightarrow {QR} }} = \dfrac{m}{n}\) (equation (1))
From the figure we have : \(\overrightarrow a + \overrightarrow {PR} = \overrightarrow r \)
\( \Rightarrow \overrightarrow {PR} = \overrightarrow r - \overrightarrow a \).
Also \(\overrightarrow b + \overrightarrow {QR} = \overrightarrow r \)
\( \Rightarrow \overrightarrow {QR} = \overrightarrow r - \overrightarrow b \)
From equation(1) : \(\dfrac{{\overrightarrow r - \overrightarrow a }}{{\overrightarrow r - \overrightarrow b }} = \dfrac{m}{n}\)
On simplification we will get \(\overrightarrow r = \dfrac{{m\overrightarrow b - n\overrightarrow a }}{{m - n}}\)
Hence the position vector of the point R which divides P and Q externally in the ratio (m:n) is given by : \(\overrightarrow OR = \dfrac{{m\overrightarrow b - n\overrightarrow a }}{{m - n}}\)
Problem:-
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (î + 2ĵ - k̂) and (-î + ĵ + k̂) respectively, in the ration 2 : 1.
(i) Internally (ii) Externally 
Answer:
The position vector of point R dividing the line segment joining two points, P and Q in the ratio m: n is given by:
Internally: \(\dfrac{{m\overrightarrow b + n\overrightarrow a }}{{m + n}}\)
Externally: \(\dfrac{{m\overrightarrow b - n\overrightarrow a }}{{m - n}}\)
Position vectors of P and Q are given as:
\(\overrightarrow {OP} = \widehat i + 2\widehat j - \widehat k\) and \[\overrightarrow {OQ} = - \widehat i + \widehat j + \widehat 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,
\(\overrightarrow {OR} = \dfrac{{2( - \widehat i + \widehat j + \widehat k) + 1(\widehat i + 2\widehat j - \widehat k)}}{{2 + 1}} = \dfrac{{ - 2\widehat i + 2\widehat j + 2\widehat k + \widehat i + 2\widehat j - \widehat k}}{3}\)(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,
\(\overrightarrow {OR} = - 3\widehat i + 3\widehat k\)
Problem:-
Find the position vector of the midpoint of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).
Answer:-
The position vector of mid-point R of the vector joining points
P (2, 3, 4) and Q (4, 1, –2) is given by,
\(\overrightarrow {OR} = \frac{{(2\widehat i + 3\widehat j + 4\widehat k) + (4\widehat i + \widehat j - 2\widehat k)}}{2} = \frac{{6\widehat i + 4\widehat j + 2\widehat k}}{2}\)
\(\overrightarrow {OR} = 3\widehat i + 2\widehat j + \widehat k\)
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