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
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 = .
= (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
mutually perpendicular vectors of equal magnitude show that are equally inclined to .
We have = ×
= = 3||2
Now × = ||2 = cosq1
where q1 is the angle between and
Similarly, if q2, q3 are the angles between and
, then cosq2 = , cosq3= .
Hence q1 = q2 = q3.
Find the unit vector which makes angle of
45° with the vector 2i + 2j – k and an angle of 60° with the vector .
Let the unit vector be so
&⇒c2 – c3 =
&⇒ c3 =
Hence the required vectors are .
Find the component of in the direction of the vector
A unit vector along is
Hence the component of along the given vector
= Projection of on
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)..
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.
are three coplanar
vectors. If is not parallel to , show that.
coplanar, we may write
= l1+ l2 …(1)
and = l1+ l2 …(2)
Solving (1) and (2), by Cramer’s rule, we
Substituting for l1 and l2, we get the
required expression. .