Scalar (or Dot) Product of two vectors:
The scalar product of 
and 
,
written as 
, is defined to be the 
cosq
where q is the angle
between the vectors 
and 
i.e.
= ab cos q
Geometrical Interpretation
is
the product of length of one vector and length of the projection
of the other vector in the direction of former.
.
Projection of b in direction of =
.Projection of a in direction of
So projection of on = 
.
Properties
&⇒ 
(acosq)b
= (projection of 
) b = (projection of 
) a
(Distributive law)
Û 
are perpendicular to each other ( if )
= 
- If
then
- If
are non-zero, the
angle between them is given by 
- =
Example.1
If 
are
mutually perpendicular vectors of equal magnitude show that 
are equally inclined to 
.
Solution
We have 
= 
×
= 
= 3||2
&⇒
= 
Now × = ||2 = 
cosq1
where q1 is the angle between 
and
.
Hence ||2
= 
cosq1
&⇒
cosq1 = 
Similarly, if q2, q3 are the angles between and
, then cosq2 = , cosq3= .
Hence q1 = q2 = q3.
Example.2
Find the unit vector which makes angle of
45° with the vector 2i + 2j – k and an angle of 60° with the vector 
.
Solution
Let the unit vector be 
so
that 
&⇒
and 
&⇒c2 – c3 = 
&⇒
and 
Also 
&⇒ 
=
1
&⇒ c3 =
Hence the required vectors are 
.
Example.3
Find the component of 
in the direction of the vector 
Solution
A unit vector along 
is
Hence the component of 
along the given vector
= Projection of on
= 
=
Example.4
Show that any two opposite edges of a regular
tetrahedron are perpendicular.
Note: (A tetrahedron whose edges are all
equal in length is called a regular tetrahedron)..
Solution
Take
one of the vertices of the tetrahedron as the initial point. Let it be O. Then
the position vectors of the points A, B, C of the tetrahedron OABC are 
, where || =
|| = |
|.
Since all the edges are of equal length, all the four faces of the tetrahedron
are equilateral triangles
&⇒
&⇒
= 0. Hence the opposite edges are
perpendicular to each other.
Example.5
are three coplanar
vectors. If is not parallel to , show that.
Solution
Since 
are
coplanar, we may write 
&⇒
= l1
+ l2 
…(1)
and 
= l1
+ l2
…(2)
Solving (1) and (2), by Cramer’s rule, we
find that
and 
Substituting for l1 and l2, we get the
required expression. .
