Example.1
In a quadrilateral PQRS 
, 
, 
, M is the midpoint
of 
and X is a
point on SM such that SX = 
SM. Then prove that P, X and
R are collinear.
Solution
&⇒ 
also 
&⇒ 
,
hence P, X and R are collinear.
Example.2
Unit vectors 
are perpendicular to each other
and the unit vector 
is inclined at an angle
q to both . If 
are
real, prove that 
Solution
Example.3
Given that 
are
the position vectors of points P and Q respectively. Find the equation for the
plane passing through Q and perpendicular to the line PQ. What is the distance
from the point (-1, 1, 1) to the plane?
Solution
= 
= - 
Equation of plane
passing through Q and perpendicular to PQ is
&⇒ 
&⇒ 
&⇒ 
…(1)
Let 
Hence distance from 
to the plane (1) is
= 
=
=
5 units
Example.4
Find the vector 
which
makes equal angles with the vectors 
and is
perpendicular to the vector 
= (1,-1,2)
with 
and the angle between 
and the unit vector 
is obtuse.
Solution
….(1)
is perpendicular to
&⇒
&⇒ x–y+2z =0 …(2)
Also it is given
that , 
(
makes
equal angles with 
)
Since
We get 
&⇒ xy – 2yz +3xz = 2xz +3xy– yz
&⇒ 2xy+yz– zx = 0 ….(3).
As the angle between
is given to be obtuse.
….(4)
from (2) & (3)
we get, 2y (y – 2z) + yz – z(y – 2z) = 0
&⇒ (y – z)2 = 0,y= z and x
= –z
From (1) get z2
+ z2 + z2 = 12
&⇒
z2 = 4
\ z = –2 (z = y < 0)
Hence 
Example.5
Show that midpoints of the
three diagonals of a complete quadrilateral are collinear . .
Solution
A complete quadrilateral
is a figure made by four straight lines no three of which are
concurrent. .
Let ABCD be a quadrilateral and let P, Q and R be the
respective midpoints of the diagonals BD, CA and EF. .
Let 
be the position vectors of A,
B, C, D, E, F, P, Q & R respectively.
then 
, 
and 
since F lies on the two lines BC and AD
and E lies on the two lines BA and CD respectively, we have
. . . (1)
.
. . (2)
. . .
(3)
. . . (4)
where s, t , u and v
are scalars
adding (1) , (2), (3)
and (4) , we get
(2 – s – u)(
) + (1 – t + u)(
) - 2(
) = 0
setting 1 – t + u = 1 + s
– u and 2 – s – v =t + v
so that (2 – s – u) + (1
– t + u) –2 = 0
This shows that the points P, Q and
R are collinear since the sum (2 – s - u) + ( 1 – t + u) – 2= 0 . .
