Example.1
Prove by vector method: sin(q – f) =
sinq.cosf –
cosq
sinf.
Solution
Let r1, r2
be the position vectors of the points A(x1, y1) and B(x2,
y2) in xy-plane making angles q and f
respectively with x-axis ( q > f) Hence x1
= r1 cosq , y1 = r1 sinq
x2 = r2
cos f, y2
= r2 sinf ....(i).
Now 
= r2r1sin(q - f)
....(ii) 
But 
=
and 
=
&⇒ = 
´ 
= (x2 y1 –x1
y2) . .
.(iii)
(as i´j = - j´i =
and i´i =
j´j
= 0 )
From (ii) and (iii),
r1 r2 sin( q - f)= ( x2 y1
–x1 y2)
so r1 r2 sin(q -f) =
x2y1 –x1 y2 = r1 r2
[ sinq
cos f-
sinf
cosq ]
&⇒ sin(q - f)=
sinq
cos f -
sinf
cos q
Example.2
Two
given points P and Q in the rectangular cartesian coordinates lie on y = 2x
+ 2 such that 
and 
where 
is
a unit vector along the x – axis . Find the magnitude of 
Solution
Let
P(x1, y1), Q(x2, y2) be the two
points on y = 2x+2
= projection of 
on x – axis, so x1 = -1
&⇒ y1
= 2
= projection of 
on x – axis, so x2 = 2
&⇒ y2
= 16
If 
is a unit
vector along y– axis, then
= - 
+ 2 
,
= 2+
16
&⇒ - 4 
= 6+
8
&⇒ |- 4 |
= 
Example.3
Show, by vector method, that the angular bisectors of a
triangle are concurrent and find an expression for the position vector of the
point of concurrency in terms of the position vectors of the vertices. .
Solution
Let
be the position vectors of the
vertices of the triangle ABC. Let the bisectors of the angles A, B and C meet
the opposite sides at D, E and F respectively.
Let AB = c, BC = a and CA = b . .
Let
I be the point of intersection of AD and BE. .
Hence
position vector of 
.
Also
Hence position vector of 
.
The symmetry of the result shows that the point I also
lies on the intersection of the bisectors of the angles B and C. .
Hence the three bisectors are concurrent at I. .
Example.4
In DABC,
using vector method show that the distance between the circumcentre and the
orthocentre is 
where R is the
circumradius of the DABC.
Solution
In DABC,
S is the circumcentre, H is the orthocente Here S is the origin of the
system and position vectors of A
, B
and C
= 
( D is the
mid–point of BC )
= 
Now, 
= a2 + b2
+ c2 + 2
= a2 + b2
+ c2 + 2ab cos2C + 2bc cos2A + 2ca cos2B
= 3R2 + 2R2
(cos2A + cos 2B + cos2C)
= 3R2 + 2R2
(–1 – 4 cosA cos B cos C)
= 3R2 – 2R2
– 8R2 cos A cos B cos C
= R2 – 8R2
cosA cosB cosC
SH = R
.
Example.5
Let
ABC and PQR be any two triangles in the same plane. Assume that the
perpendiculars from the points A, B, C to the sides QR, RP, PQ respectively are
concurrent. Using vector method or otherwise, prove that the perpendiculars
from P, Q, R to BC, CA, AB respectively are also concurrent. .
Solution
ABC and PQR are the given triangles. Let the
perpendiculars from A, B, C to the sides QR, PR and PQ intersect at O. Take O
as the initial point. Let 
, 
be the position vector of A, B,
C, P, Q and R respectively. Since OA, OB and OC are perpendicular to QR,
RP & PQ.
,
= 0 and 
= 0
Let the perpendiculars from P and Q on BC and CA
respectively intersect at the point X whose position vector is taken as 
. It implies
and 
= 0
&⇒ 
= 
and 
Adding, we have
=
- 
= 
=-
&⇒
=0
&⇒XR
is perpendicular to AB.
Hence perpendicular from R to AB passes through X.
Alternative solution:
Let the vertices of triangle ABC be A(x1, y1),
B(x2, y2) and C(x3, y3). Let the
vertices of triangle PQR be P(p1, q1), Q(p2, q2)
and R(p3, q3). Now the slope of QR is (q3 – q2)/
(p3 – p2), so the slope of the line perpendicular to it
is – (p3 – p2) / (q3 – q2).
Therefore the equation of the perpendicular from A to QR is y – y1 =
&⇒ (p2 – p3)x + (q2 – q3)y
= (p2 – p3)x1 + (q2 – q3)y1. …(1).
Similarly, the equations of the perpendiculars from B to
RP and C to PQ are(p3 – p1)x + (q3 – q1)y
= (p3 – p1)x2 + (q3 – q1)y2 …(2)
and (p1 – p2)x + (q1 –
q2)y = (p1 – p2)x3 + (q1
– q2)y3 …(3)
The lines (1), (2)
and (3) are stated to be concurrent, i.e., they meet at a point. The x- and
y-values on the LHS will then be the same at this point. The left hand sides of
(1), (2) and (3) will add up
to zero, giving .
(p2–p3)x1+(p3–p1)x2+(p1
–p2)x3+(q2-q3)y1+(q3
–q1)y2+(q1 –q2)y3=0
…(4)
In other words, equation (4) is equivalent to saying
that the perpendiculars from the vertices of ABC to the sides of PQR meet at a
point. But (4) can be rewritten in the form.
(x2
–x3)p1+(x3 –x1)p2+(x1
–x2)p3+(y2 –y3)q1+(y3
–y1)q2+(y1 –y2)q3 =0 …(5)
Equation (5) is the mirror image of (4),
which is equivalent to saying that the perpendiculars from the vertices of PQR
to the sides of ABC meet at a point.
