Vector dot product
Vector dot product or scalar product:
The dot or scalar product of two vectors 
and 
is defined as the product of the magnitudes of and and the cosine of the angle between them.
If A and B creates angle θ then, . = AB
.
Properties of dot product:
(i) . = BA = AB = . [dot product is commutative]
(ii) 
.(
+
) = . + . [dot product is distributive]
(iii) If the angle between and is 
then, . = ABcos = AB
(iv) . = AAcos = 
(v) If the angle between and is 
then, . = ABcos = 0
(vi) 
. = 
. = 
. = 1.1cos = 1
(vii) . = . = . = 1.1cos = 0
Vector dot product or scalar product is used to calculate work done by a force, projection of a vector, angle between two vectors. We can show two vectors are mutually perpendicular using dot product of vectors.
Concepts on dot product:
1. Work done: A force 
= (2î + 3ĵ – k̂) N is applied on a body to move it from point P (1, 2, -1) m to point Q (2, 4, 1) m. What is the work done required?
Force = (2î + 3ĵ – k̂) N and displacement of the body 
= (2 – 1)î + (4 – 2)ĵ + (1 + 1)k̂ = (î + 2ĵ + 2k̂) m.
Work done W = . = (2î + 3ĵ – k̂).(î + 2ĵ + 2k̂) = 2 + 6 – 2 = 6J.
2. Projection: Projection of 
on 
or component of on . PQ represents the magnitude of 
. PR and QS are two perpendiculars drawn on and PT is parallel to RS. The component of on is RS = PT = PQcos
[where is the angle between and ] .
= A.1.cos = .
Example: What is the component of = 2î + 3ĵ + k̂ on = î + 2ĵ + 2k̂?
Component of on = . = 
= 
= 
.
Vector component of on is (.
) = (
) = 
.
3. Angle between two vectors: If is the angle between and then = 
.
Example: What is the angle between = 2î + 3ĵ + k̂ and = î + 2ĵ – 2k̂?
= 
= 
= 
.
4. Perpendicular vectors: If and are perpendicular to each other then, . = 0.
Example: What is the value of n so that = 2î – 3ĵ + k̂ and = î + nĵ + 2k̂ are perpendicular to each other?
Here (2î – 3ĵ + k̂).( î + nĵ + 2k̂) = 0
Or, 2 – 3n +2 = 0
n = 
.
5. Equation of a plane and distance of plane from origin:
= 2î+3ĵ+2k̂ is perpendicular to a plane. The terminal point of another vector 
= î+2ĵ+k̂ touches the plane. Find the equation of the plane.
Let us consider, Q is a point on the plane with coordinates (x, y, z). The position vector of point Q is xî + yĵ + zk̂.
= (1 – x)î + (2 – y)ĵ+(1 – z)k̂ is lying on the plane which is perpendicular to .
Therefore, . = 0
Or, (2î + 3ĵ + 2k̂).[(1 – x)î + (2 – y)ĵ + (1 – z)k̂] = 0
Or, 2(1 – x) + 3(2 – y) + 2(1 – z) = 0
Or, 2 – 2x + 6 – 3y + 2 – 2z = 0
2x + 3y + 2z = 10
The distance of plane from origin is projection of on = .
= 
= 
.
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