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# Multiplication of Vectors by a Real Number

## Multiplication of Vectors by real numbers

Multiplying a vector A with a positive number λ gives a vector whose magnitude is changed by the factor λ but the direction is the same as that of A: |λA| = λ|A|     if λ > 0
For example, if A is multiplied by 2, the resultant vector 2A is in the same direction as A and has a magnitude twice of |A| as shown in Fig. (a) below.
Multiplying a vector A by a negative number −λ gives another vector whose direction is opposite to the direction of A and whose magnitude is λ times |A|.
Multiplying a given vector A by negative numbers, say –1 and –1.5, gives vectors as shown in Fig (b) below.
The factor λ by which a vector A is multiplied could be a scalar having its own physical dimension. Then, the dimension of λ A is the product of the dimensions of λ and A. For example, if we multiply a constant velocity vector by duration (of time), we get a displacement vector.

Dot product or scalar product:-
The dot product of two vectors A and B, is a scalar, which is equal to the product of the magnitudes of A and B and the Cosine of the smaller angle between them. If Θ is the smaller angle between A and B, then AB = AB
bar A.bar B=ABcosθ
(i) hat i.hat i=hat j. hat j=hat k.hat k=1
(ii) hat i.hat j=hat j.hat k=hat k.hat i=0
(iii) If barA=A_x hat i +A_y hat j +A_z hat k and barB=B_x hat i +B_y hat j +B_z hat k
then barA .bar B=A_x B_x +A_y B_x +A_z B_z

Properties of Scalar product:-
1. It obeys commutative law.
barA .bar B=barB .bar A

2. It obeys distributive law.
barA .(bar B + bar C)= barA .bar B+ barA .bar C

3. Scalar (Dot) product of two mutually perpendicular vectors is zero i.e.
(barA .bar B) = AB cos 90^o=0

4. Scalar (Dot) product will be maximum when θ = 0o i.e., vectors are parallel to each other.
(barA .bar B)_(max)= |A||B|

5. If bar a  and bar b are unit vectors then |bar a|=|bar b|=1and bar a.bar b=1.1 cos θ= cos θ

6. Dot product of unit vectors hat i, hat j, hat k
hat i. hat i =hat j. hat j =hat k. hat k = 1
hat i. hat j =hat j. hat k =hat k. hat i = 0

7. Square of a vector bar a . bar a =|a||a| cos 0=a^2

8. If the two vectors bar A and bar B, in terms of their rectangular components, are barA=A_x hat i +A_y hat j +A_z hat k and barB=B_x hat i +B_y hat j +B_z hat k

then barA .bar B=(A_x hat i +A_y hat j +A_z hat k)(B_x hat i +B_y hat j +B_z hat k)
barA .bar B=A_x B_x +A_y B_x +A_z B_z

Vector product(Cross product):
The cross product of two vectors and, represented by x is a vector, which is equal to the product of the magnitudes of A and B and the sine of the smaller angle between them. If Θ is the smaller angle between A and B, then = AB Sin θ
where hat n is a unit vector perpendicular to the plane containing bar A and bar B.
(i) hat i × hat i =hat j × hat j =hat k × hat k = 0
(ii) hat i × hat j =hat k,  hat j × hat k =hat i,    hat k × hat i =hat j
hat j × hat i =-hat k, hat k × hat j =-hat i, hat i × hat k =-hat j
(iii) If barA=A_x hat i +A_y hat j +A_z hat k and barB=B_x hat i +B_y hat j +B_z hat k
bar A × bar B = (A_x B_z- A_z B_y) hat i +(A_z B_x- A_x B_z) hat j +(A_x B_y- A_y B_x) hat k

Properties of Cross Product
(i) Cross product of two vectors is not commutative
bar a × bar b ≠ bar b × bar a
bar a × bar b =-bar b × bar a

(ii) Cross product is not assosiative
bar a × (bar b × bar c) ≠ (bar a × bar b) × bar c

(iii) Cross product obeys distributive law
bar a × (bar b + bar c) = bar a × bar b + bar a × bar c

(iv) If θ = 0 or pi it means the two vectors are collinear.
bar a × bar b = bar 0
and conversely, if bar a × bar b = bar 0 then the vector bar a and bar b are parallel provided bar a and bar b are non-zero vectors.

(v) If  θ = 90o, and hat n is the unit vector perpendicular to both bar a and bar b
bar a × bar b=|a||b|  sin 90^o hat n =|a||b|  hat n

(vi) The vector product of any vector with itself is bar 0
bar a × bar a = bar 0

(vii) If bar a × bar b = bar 0,then
bar a=0 or bar b =0 or bar a|| bar b

(viii) If bar a  and bar b are unit vectors, then bar a × bar b=1.1 sin θ  hat n = sin θ  hat n

(ix) Cross product of unit vectors hat i, hat j and hat k
hat i × hat i =hat j × hat j =hat k × hat k = 0

hat i × hat j =hat k = - hat j × hat j
hat j × hat k =hat i = -hat k × hat j
hat k × hat i =hat j = -hat i × hat k

(x) If the two vectors bar A and bar B in terms of their rectangular components are
bar A = a_1 hat i + b_1 hat j + c_1 hat k
bar B = a_2 hat i + b_2 hat j + c_2 hat k
bar A × bar B  = (a_1 hat i + b_1 hat j + c_1 hat k)×( a_2 hat i + b_2 hat j + c_2 hat k)
It can be found by the determinant method
i.e.,    bar A × bar B =[(hat i,hat j,hat k),("a"_1 ,"b"_1,"c"_1),("a"_2,"b"_2,"c"_2)]
= hat i(b_1 c_2 - b_2 c_1) - hat j(a_1 c_2 - a_2 c_1) + hat k(a_1 b_2 - a_2 b_1)

• For motion in a plane, velocity is defined as:
bar v =(bar r_2 - bar r_1)/(t_2 - t_1)= ((x_2 hat i + y_2 hat j) - (x_1 hat i + y_1 hat j))/((t_2 - t_1)) = (x_2 - x_1)/(t_2 - t_1) hat i + (y_2 - y_1)/(t_2 - t_1) hat j = v_x hat i and v = sqrt(a_x^2 + a_y^2)
• For motion in a plane, acceleration is defined as:
bar a =(bar v_2 - bar v_1)/(t_2 - t_1)= ((v_(x_2) hat i + v_(y_2) hat j) - (v_(x_1) hat i + v_(y_1) hat j))/((t_2 - t_1)) = ((v_(x_2) - v_(x_1))/(t_2 - t_1)) hat i + ((v_(y_2) - v_(y_1))/(t_2 - t_1)) hat j and v = sqrt(a_x^2 + a_y^2)
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