Example.1
Two system of forces P, Q, R and P¢, Q¢, R¢
act along the side BC, CA, AB of a DABC. Prove that the
resultant will be parallel if
= 0.
Solution
Unit vector along
=cos (p –
C)
+sin(p–
C)
= – cosC
+ sinC
Unit vector along 
=cos(p + B)
+
sin(p +
B)
= – cos B
– sinB
If S and S¢ be resultant in the two
cases, then

= P
+ Q (– cos C
+ sin C
) + R( – cosB
– sinB
) ….(1)
Similarly
S¢ = (P¢ – Q¢
cosC – R¢
cosB)
+ (Q¢
sinC – R¢
sinB)
….(2)
If Q and Q¢ be the angles made by
the resultant with x–axis, then
tanq =
and
tanq¢ = 
if the resultant are parallel , then q = q¢ \
tanq =
tanq¢

upon solving we get
&⇒ (PQ¢ – P¢Q) sinC + (RP¢ –
R¢P)
sinB + (QR¢ – Q¢R) sinA = 0
= 0.
Example.2
Line L1 is parallel to a vector
and passes through a point A
(7,6,2) and the line L2 is parallel to a vector
and passes through a point B (5,
3,4). Now a line L3 parallel to a vector
intersects
the lines L1 and L2 at points C and D respectively. Find
.
Solution
P.V.
of C.
a∈R
P.V. of D,
b
∈ R
and we know that 

Hence
by comparing both
we get3a + 2b – 2c = 2-2a
+ b+ 2c = 3
&⇒ -4a
+ 3b + c = - 2
&⇒ a = 2, b = 1, c = 3
&⇒ 
Example.3
A circle is inscribed in an n-sided regular polygon a1
, a2., . . . ., an having each side a
unit. For any arbitrary point p on the circle, prove that
.
Solution
Let the centre of the incircle be the reference point.
Then 


[
= nR2 +nr2
– 2
= 
now
R =

R2 + r2
=
= 
&⇒
.
Example.4
In
a triangle PQR, S and T are points on QR and PR respectively, such that QS =3SR
and PT = 4TR. Let M be the point of intersection of PS and QT. Determine the
ratio QM:MT using vector methods. .
Solution
Let QM : MT = l :
1 and PM : MS = m :1 and
.
Now 
&⇒
…(1) 
Also
….(2)
From (1) and (2),
=
On comparing, we get
…..(3)
and
…..(4)
&⇒ 
&⇒ m =
16/3 and l = 15/4
So QM :MT = 15 :4. .
Example.5
Let
be non-coplanar
unit vectors, equally inclined to one another at an angle q.
If
, find the scalars p, q and r in
terms of q.
Solution

Taking dot product with
, we get
p cosq +
r cosq +
q = 0 .....(1).
p cosq +
q cosq +
r =
..... (2)
p + q cosq +
r cosq =
..... (3)
1 - cosq =
p
......(4)
We get (p, q, r) = 
Or, (p, q, r) =
.