Prove by vector method: sin(q – f) = sinq.cosf – cosq sinf.

Let r_{1}, r_{2}
be the position vectors of the points A(x_{1}, y_{1}) and B(x_{2},
y_{2}) in xy-plane making angles q and f
respectively with x-axis ( q > f) Hence x_{1}
= r_{1} cosq , y_{1} = r_{1} sinq

x_{2} = r_{2}
cos f, y_{2}
= r_{2} sinf ....(i).

Now = r_{2}r_{1}sin(q - f) ....(ii)

But = and =

&⇒ = ´

= (x_{2} y_{1} –x_{1}
y_{2}) . .
.(iii)

(as i´j = - j´i = and i´i = j´j = 0 )

From (ii) and (iii),

r_{1} r_{2} sin( q - f)= ( x_{2} y_{1}
–x_{1} y_{2})

so r_{1} r_{2} sin(q -f) =
x_{2}y_{1} –x_{1} y_{2} = r_{1} r_{2}
[ sinq
cos f-
sinf
cosq ]

&⇒ sin(q - f)=
sinq
cos f -
sinf
cos q

Two
given points P and Q in the rectangular cartesian coordinates lie on y = 2^{x
+ 2} such that and where is
a unit vector along the x – axis . Find the magnitude of

Let
P(x_{1}, y_{1}), Q(x_{2}, y_{2}) be the two
points on y = 2^{x+2}

= projection of on x – axis, so x_{1} = -1

&⇒ y_{1}
= 2

= projection of on x – axis, so x_{2} = 2

&⇒ y_{2}
= 16

If is a unit vector along y– axis, then

= - + 2 ,= 2+ 16

&⇒ - 4 = 6+
8

&⇒ |- 4 |
=

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. .

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. .

In DABC, using vector method show that the distance between the circumcentre and the orthocentre is where R is the circumradius of the DABC.

In DABC, S is the circumcentre, H is the orthocente Here S is the origin of the system and position vectors of A, Band C

= ( D is the mid–point of BC )

=

Now,

= a^{2} + b^{2}
+ c^{2} + 2

= a^{2} + b^{2}
+ c^{2} + 2ab cos2C + 2bc cos2A + 2ca cos2B

= 3R^{2} + 2R^{2}
(cos2A + cos 2B + cos2C)

= 3R^{2} + 2R^{2}
(–1 – 4 cosA cos B cos C)

= 3R^{2} – 2R^{2}
– 8R^{2} cos A cos B cos C

= R^{2} – 8R^{2}
cosA cosB cosC

SH = R.

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. .

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.

Let the vertices of triangle ABC be A(x_{1}, y_{1}),
B(x_{2}, y_{2}) and C(x_{3}, y_{3}). Let the
vertices of triangle PQR be P(p_{1}, q_{1}), Q(p_{2}, q_{2})
and R(p_{3}, q_{3}). Now the slope of QR is (q_{3} – q_{2})/
(p_{3} – p_{2}), so the slope of the line perpendicular to it
is – (p_{3} – p_{2}) / (q_{3} – q_{2}).
Therefore the equation of the perpendicular from A to QR is y – y_{1} =

&⇒ (p_{2} – p_{3})x + (q_{2} – q_{3})y
= (p_{2} – p_{3})x_{1} + (q_{2} – q_{3})y_{1}. …(1).

Similarly, the equations of the perpendiculars from B to
RP and C to PQ are(p_{3} – p_{1})x + (q_{3} – q_{1})y^{
}= (p_{3} – p_{1})x_{2} + (q_{3} – q_{1})y_{2} …(2)

and (p_{1} – p_{2})x + (q_{1} –
q_{2})y = (p_{1} – p_{2})x_{3} + (q_{1}
– q_{2})y_{3 }…(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 .

(p_{2}–p_{3})x_{1}+(p_{3}–p_{1})x_{2}+(p_{1}
–p_{2})x_{3}+(q_{2}-q_{3})y_{1}+(q_{3}
–q_{1})y_{2}+(q_{1} –q_{2})y_{3}=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.

(x_{2}
–x_{3})p_{1}+(x_{3} –x_{1})p_{2}+(x_{1}
–x_{2})p_{3}+(y_{2} –y_{3})q_{1}+(y_{3}
–y_{1})q_{2}+(y_{1} –y_{2})q_{3} =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.