Mathematics Addition of Vectors, Multiplication of a Vector by a Scalar And Section formula For CBSE-NCERT

### Topic covered

♦ Multiplication of a Vector by a Scalar
♦ Components of a vector
♦ Vector joining two points
♦ Section formula

=> A vector vec(AB) simply means the displacement from a point A to the point B. Now consider a situation that a girl moves from A to B and then from B to C (Fig).
=> The net displacement made by the girl from point A to the point C, is given by the vector AC and expressed as

vec(AC) = vec(AB) + vec (BC)

This is known as the triangle law of vector addition.

In general, if we have two vectors veca and vecb (Fig(i)), then to add them, they are positioned so that the initial point of one coincides with the terminal point of the other (Fig (ii)).

=> For example, in Fig (ii), we have shifted vector vecb without changing its magnitude and direction, so that it’s initial point coincides with the terminal point of vec a.
=> Then, the vector veca + vecb, represented by the third side AC of the triangle ABC, gives us the sum (or resultant) of the vectors vec a and vec b i.e., in triangle ABC (Fig (ii)), we have

vec(AB) + vec (BC) = vec(AC)

Now again, since vec(AC) = - vec (CA) , from the above equation, we have

vec(AB) + vec (BC) + vec (CA) = vec ("AA") = 0

=> This means that when the sides of a triangle are taken in order, it leads to zero resultant as the initial and terminal points get coincided (Fig(iii)).

color{blue}{"Difference Of Vector"}Now, construct a vector vec (BC) so that its magnitude is same as the vector vec (BC) , but the direction opposite to that of it (Fig(iii)), i.e.,

vec(BC') = - vec (BC)

=> Then, on applying triangle law from the Fig 10.8 (iii), we have

color{red}{vec(AC') = vec (AB) + vec (BC') = vec (AB) + ( - vec(BC) ) = vec a - vec b}

The vector vec (AC) is said to represent the difference of vec a  and vec b

=> If we have two vectors vec a and vec b represented by the two adjacent sides of a parallelogram in magnitude and direction (Fig), then their sum vec a + vec b is represented in magnitude and direction by the diagonal of the parallelogram through their common point. This is known as the parallelogram law of vector addition.

color{red} "Note" From Fig , using the triangle law, one may note that

vec(OA) + vec (AC) = vec (OC)

or  vec(OA) + vec (OB) = vec (OC) since vec(AC) = vec (OB)

which is parallelogram law.

color{blue} "Property 1" For any two vectors vec a  and vecb

vec a + vec b = vec b + vec a ....... (Commutative property)

color {red} " Proof" Consider the parallelogram ABCD (Fig). Let vec (AB) = veca and vec (BC) = vec b , then using the triangle law, from triangle ABC, we have

vec(AC) = vec a + vec b

=> Now, since the opposite sides of a parallelogram are equal and parallel, from Fig 10.10, we have, vec(AD) = vec(BC) = vec b and vec(DC) = vec(AB) = vec a .

Again using triangle law, from triangle ADC, we have

vec(AC) = vec(AD) + vec (DC) = vec b + vec a

Hence  vec a + vec b = vec b + vec a

color{blue} "Property 2" For any three vectors veca, vecb and vecc

(vec a + vec b ) + vec c = vec a + ( vec b + vec c)

color{red} "Proof " Let the vectors veca, vecb and vecc be represented by vec(PQ), vec(QR) and vec(RS), respectively, as shown in Fig 10.11(i) and (ii).

Then veca + vecb = vec(PQ) + vec(QR) = vec(PR)

and vecb + vecc = vec(QR) + vec(RS) = vec(QS)

So (veca +vec b) + vecc = vec(PR) + vec(RS) = vec(PS)

and veca + (vecb + vecc) = vec(PQ) + vec(QS) = vec(PS)

Hence (veca + vecb) + vecc= veca + (vecb + vecc)

color {blue} "Remark" The associative property of vector addition enables us to write the sum of three vectors veca ,vecb , vecc as veca + vecb + vecc  without using brackets.

Note that for any vector veca, we have

veca + vec0 = vec0 + veca = a

Here, the zero vector vec0 is called the additive identity for the vector addition.

### Multiplication of a Vector by a Scalar

=> Let veca be a given vector and λ a scalar.
=> Then the product of the vector veca by the scalar λ, denoted as λveca , is called the multiplication of vector veca by the scalar λ.
=>Note that, λveca is also a vector, collinear to the vector veca . The vector λveca has the direction same (or opposite) to that of vector veca according as the value of λ is positive (or negative). Also, the magnitude of vector λ veca is |λ| times the magnitude of the vector veca , i.e.,

| lamda veca | = | lamda | | vec a |

A geometric visualisation of multiplication of a vector by a scalar is given in Fig

=> When λ = –1, then λveca = −veca, which is a vector having magnitude equal to the magnitude of veca and direction opposite to that of the direction of veca . The vector – veca is called the negative (or additive inverse) of vector veca and we always have

veca + (-veca ) = ( - vec a ) + vec a = vec 0

=> Also, if lamda = 1/(|vec a |), provided veca ≠ 0, i.e. veca is not a null vector, then

| lamda vec a | = | lamda | | vec a | = 1/ (| vec a | ) ( | vec a | ) = 1

=> So, λveca represents the unit vector in the direction of veca . We write it as

"unit vector a" = 1/( | vec a| ) vec a

color{red} "Note" For any scalar k , kvec0 = vec 0

### Components of a vector

=> Let us take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and z-axis, respectively. Then, clearly

|vec(OA)| =1 , |vec (OB)| = 1 and |vec(OC)| = 1

=> The vectors vec(OA), vec(OB) and vec(OC) , each having magnitude 1, are called unit vectors along the axes OX, OY and OZ, respectively, and denoted by hati, hat j  and hat k , respectively (Fig 10.13).

=> Now, consider the position vector vec(OP) of a point P(x, y, z) as in Fig.

=> Let P_1 be the foot of the perpendicular from P on the plane XOY. We, thus, see that P_1 P is parallel to z-axis. As hat i , hat j  and hat k are the unit vectors along the x, y and z-axes, respectively, and by the definition of the coordinates of P, we have vec(P_1P) = vec(OR) = zhatk.

Similarly, (QP_1) = vec(OS) = y hatj

and vec(OQ )= x hati
.
=> Therefore, it follows that vec(OP_1) = vec(OQ) + vec(QP_1) = x hat i + y hatj

and vec(OP) = vec(OP_1) + vec(P_1P) = xhati + yhatj + zhatk

=> Hence, the position vector of P with reference to O is given by

vec(OP) (or vecr ) = xhati + yhatj + zhatk

=> This form of any vector is called its component form. Here, x, y and z are called as the scalar components of vecr , and xhati, yhatj and zhatk are called the vector components of vecr along the respective axes.

● The length of any vector vecr = xhat i + y hat j + z hatk , is readily determined by applying the Pythagoras theorem twice.
We note that in the right angle triangle OQP1 (Fig)

|vec(OP_1) | = sqrt ( |vec(OQ)|^2 + | vec (QP_1)|^2 ) = sqrt (x^2 +y^2)

and in the right angle triangle OP_1P, we have

|vec(OP_1) | = sqrt(|vec(OP_1)|^2 + | vec(P_1P)|^2 ) = sqrt((x^2 + y^2 ) + z^2)

=> Hence, the length of any vector vec r = x hat i + y hat j + z hat k is given by

color{orange}{| vec r | = | x hati + y hatj + z hat k | = sqrt ( x^2 + y^2 + z^2)}

\color{green} ✍️ If vec a and vec b are any two vectors given in the component form a_1 hat i + a_2 hat j + a_3 hat k and  b_1 hat i + b_2 hat j + b_3 hat k  respectively, then

(i) the sum (or resultant) of the vectors veca and vecb is given by

veca + vecb = (a_1 + b_1 )hat i + (a_2 + b_2 )hat j + (a_3 + b_3 ) hat k

(ii) the difference of the vector veca and vecb is given by

vec a − vec b = (a_1 − b_1 )hat i + (a_2 − b_2 ) hatj + (a_3 − b_3 ) hat k

(iii) the vectors veca and vecb are equal if and only if

a_1 = b_1, a_2 = b_2 and a_3 = b_3

(iv) the multiplication of vector veca by any scalar λ is given by

λveca = (λa_1 ) hat i + (λa_2 ) hatj + (λa_3 )hat k

The addition of vectors and the multiplication of a vector by a scalar together give the following distributive laws:

\color{green} ✍️ Let veca and vecb be any two vectors, and k and m be any scalars. Then

color{blue} {"(i)" k veca + mveca = (k + m)veca}
color{blue} {"(ii)" k(mveca) = (km)veca}
color{blue} {"(iii)" k(veca + vecb) = kveca + kvecb}

color{blue} "Remarks"

(i) We can observe that whatever be the value of λ, the vector λveca is always collinear to the vector veca.
=> If the vectors veca and vec b are given in the component form, i.e. veca = a_1 hat i + a_2 hat j + a_3 hatk and vec b = b_1 hati + b_2 hatj + b_3 k, then the two vectors are collinear if and only if

b_1 hati + b_2 hat j + b_3 hat k = λ(a_1 hat i + a_2 hatj + a_3 hat k)

=> b_1 hat i + b_2 hat j + b_3 hat k = ( lamda a_1 ) hat i + ( lamda a_2 ) ha j + ( lamda a_3 ) hat k

=>  b_1 = λa_1 , b_2 = λa_2 , b_3 = λa_3

=> b_1/a_1 = b_2/a_2 = b_3/a_3 = lamda

(ii) If veca = a_1 hati + a_2 hatj + a_3 hat k , then a_1, a_2, a_3 are also called direction ratios of vec a.

(iii) In case if it is given that l, m, n are direction cosines of a vector, then lhat i + mhatj + n hat k = (cosα)hat i + (cosβ) hatj + (cos γ)hat 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.
Q 3118767609

Find the values of x, y and z so that the vectors vec a = x hat i + 2 hat j + z hat k and
vec b = 2 hat i + y hat j + hat k are equal.
Class 12 Chapter 10 Example 4
Solution:

Note that two vectors are equal if and only if their corresponding components
are equal. Thus, the given vectors vec a and vec b will be equal if and only if
x = 2, y = 2, z = 1
Q 3118867700

Let vec a = vec i + 2 hat j and vec b = 2 hat i + hat j Is | vec a | = | vec b | ? Are the vectors and vec a  and vec b
equal?
Class 12 Chapter 10 Example 5
Solution:

We have | vec a | = sqrt (1^2 + 2^2) = sqrt 5 and | b | = sqrt(2^2 + 1^2) = sqrt 5

So, | vec a | = | vec b |
. But, the two vectors are not equal since their corresponding components
are distinct.
Q 3128867701

Find unit vector in the direction of vector vec a = 2 hat i + 3 hat j + hat k
Class 12 Chapter 10 Example 6
Solution:

The unit vector in the direction of vector vec a is given by vec a = 1/(|vec a|) vec a

Now  | vec a | = sqrt( 2^2 + 3^2 + 1^2) = sqrt(14)

Therefore  hat a = 1/sqrt(14) ( 2 hat i + 3 hat j + hat k) = 2/sqrt(14) hat i + 3/sqrt(14) hat j + 1/sqrt(14) hat k
Q 3138867702

Find a vector in the direction of vector vec a = hat i − 2 hat j that has magnitude
7 units.
Class 12 Chapter 10 Example 7
Solution:

The unit vector in the direction of the given vector vec a is

hat a = 1/(|vec a|) vec a = 1/sqrt5 ( hat i - 2 hat j) = 1/sqrt5 hat i - 2/sqrt5 hat j

Therefore, the vector having magnitude equal to 7 and in the direction of vec a is

 7 hat a = 7 ( 1/sqrt5 hat i - 2/sqrt5 hat j) = 7/sqrt5 hat i - (14)/sqrt5 hat j
Q 3148867703

Find the unit vector in the direction of the sum of the vectors,
vec a = 2 hat i + 2 hat j – 5 hat k and vec b = 2 hat i + hat j + 3 hat k

Class 12 Chapter 10 Example 8
Solution:

The sum of the given vectors is

vec a + vec b ( = vec c  , say ) =  4 hat i + 3 hat j - 2 hat k

and  | vec c | = sqrt( 4^2 + 3^2 + (-2)^2) = sqrt(29)

Thus, the required unit vector is

hat c = 1/(|vec c|) vec c = 1/sqrt(29) ( 4 hat i + 3 hat j - 2 hat k) = 4/sqrt(29) hat i + 3/sqrt(29) hat j - 2/sqrt(29) hat k
Q 3158867704

Write the direction ratio’s of the vector vec a = hat i + hat j − 2 hat k and hence calculate
its direction cosines.
Class 12 Chapter 10 Example 9
Solution:

Note that the direction ratio’s a, b, c of a vector vec r = x hat i + y hat j + z hat k are just
the respective components x, y and z of the vector. So, for the given vector, we have
a = 1, b = 1 and c = –2. Further, if l, m and n are the direction cosines of the given
vector, then

l = a/(|vec r|) = 1/sqrt6 , m= b/(|vec r|) = 1/sqrt6 , n = c/(|vec r|) = (-2)/sqrt6 as  | vec r| = sqrt 6

Thus, the direction cosines are  ( 1/sqrt6 , 1/sqrt6 , 2/sqrt6)

### Vector joining two points

=> If P_1(x_1, y_1, z_1) and P_2(x_2, y_2, z_2) are any two points, then the vector joining P_1 and P_2 is the vector P_1P_2 (Fig).

=> Joining the points P_1 and P_2 with the origin O, and applying triangle law, from the triangle OP_1P_2, we have

vec(OP_1) + vec(P_1P_2) = vec(OP_2)

=> Using the properties of vector addition, the above equation becomes

vec(P_1P_2) = vec(OP_2) − vec (OP_1)

i.e vec(P_1 P_2) = ( x_2 hat i + y_2 hatj + z_2 hat k ) - ( x_1 hat i + y_1 hatj + z_1 hatk )

= ( x_2 - x_1 ) hat i + ( y_2 - y_1 ) hat j + ( z_2 - z_1 ) hat k

=> The magnitude of vector vec(P_1P_2) is given by

color{orange}{vec(P_1P_2) = sqrt ( ( x_2 - x_1)^2 + (y_2 - y_1)^2 + ( z_2 - z_1 )^2 )}
Q 3168867705

Find the vector joining the points P(2, 3, 0) and Q(– 1, – 2, – 4) directed
from P to Q.
Class 12 Chapter 10 Example 10
Solution:

Since the vector is to be directed from P to Q, clearly P is the initial point
and Q is the terminal point. So, the required vector joining P and Q is the vector vec(PQ)
given by

 vec(PQ) = (−1 − 2) hat i + (−2 − 3) hat j + (−4 − 0)hat k
i.e.  vec(PQ) = −3 hat i − 5 hat j − 4 hat k.

### Section formula

=> Let P and Q be two points represented by the position vectors vec(OP) and vec(OQ), respectively, with respect to the origin O. => Then the line segment joining the points P and Q may be divided by a third point, say R, in two ways internally (Fig) and externally (Fig ).

color{blue} "Case" I When R divides PQ internally (Fig).

=> If R divides vec(PQ) such that m vec (RQ) = n vec(PR),

=> where m and n are positive scalars, we say that the point R divides vec(PQ) internally in the ratio of m : n. Now from triangles ORQ and OPR, we have

vec(RQ) = vec (OQ) - vec (OR) = vec b - vecr

and  vec (PR) = vec (OR) - vec (OP) = vec r - vec a,

=> Therefore, we have m (vec b − vec r ) = n (vec r − vec a )

or  vec r = ( m vec b + n vec a)/( m + n ) (on simplification)

=> Hence, the position vector of the point R which divides P and Q internally in the ratio of m : n is given by

vec (OR) = ( m vec b + n vec a ) / ( m + n )

color{blue} "Case II" When R divides PQ externally (Fig 10.17). We leave it to the reader as an exercise to verify that the position vector of the point R which divides the line segment vec (PQ) externally in the ratio

m : n ("i.e" (PR)/(QR) = m /n ) is given by

vec ( OR) = ( m vec b - n vec a ) / ( m - n )

"Concept : " If R is the midpoint of PQ , then m = n. And therefore, from Case I, the midpoint R of vec(PQ), will have its position vector as

vec (OR) = ( vec a + vec b ) /2
Q 3178867706

Consider two points P and Q with position vectors vec( OP) = 3 vec a − 2 vec b
and vec (OQ) = vec a + vec b . Find the position vector of a point R which divides the line joining P and Q
in the ratio 2:1, (i) internally, and (ii) externally.
Class 12 Chapter 10 Example 11
Solution:

(i) The position vector of the point R dividing the join of P and Q internally in the
ratio 2:1 is

 vec(QR) = ( 2 ( veca + vec b) + ( 3 vec a - 2 vec b))/(2 +1) = (5 veca)/3

(ii) The position vector of the point R dividing the join of P and Q externally in the
ratio 2:1 is

 vec(QR) = ( 2 ( veca + vec b) - ( 3 vec a - 2 vec b))/(2 - 1) = 4 vec b - vec a
Q 3108867708

Show that the points A (2 hat i − hat j + hat k), B ( hat i − 3 hat j − 5 hat k), C(3 hat i − 4 j − 4 hat k) are
the vertices of a right angled triangle.
Class 12 Chapter 10 Example 12
Solution:

We have
vec(AB) = (1− 2) hat i + (−3 +1) hat j + (−5 −1) hat k = − hat i − 2 hat j − 6 hat k

 vec (BC) = (3 −1) hat i + (−4 + 3) hat j + (−4 + 5) hat k = 2 hat i − hat j + hat k

and vec (CA) = (2 − 3) hat i + (−1+ 4) hat j + (1+ 4) hat k = − hat i + 3 hat j + 5 hat k

Further, note that

| vec (AB) | ^2 = 41= 6 + 35 = | vec(BC) |^2 + | vec ( CA)|^2

Hence, the triangle is a right angled triangle.