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Solved Subjective Question on Vector Set 1

Posted on - 04-01-2017

Math

IIT JEE

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, Band 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.

 
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