Vector
Vectors and their representation :
Vector quantities are specified by definite magnitude and definite direction. A vector is generally represented by a directed line segment, say AB . A is called the initial point and B is called the terminal point. The magnitude of vector AB is expressed by AB .
Zero vector:
A vector of zero magnitude i.e. which has the same initial and terminal point, is called a zero vector. It is denoted by O. The direction of zero vector is indeterminate.
Unit vector:
A vector of unit magnitude in the direction of a vector a is called unit vector along a and is denoted by
aˆ, symbolically | a | a aˆ .
Example # 1 : Find unit vector of iˆ2jˆ3kˆ Solution : a = iˆ2jˆ3kˆ if a = axiˆ + ayjˆ + azkˆ then |a| = ax2ay2az2 |a| = 14 aˆ = |aa| = 14 1 iˆ – 14 2 jˆ + 14 3 kˆ Equal vectors:
Two vectors are said to be equal if they have the same magnitude, direction and represent the same physical quantity.
Collinear vectors:
Two vectors are said to be collinear if their directed line segments are parallel irrespective of their directions. Collinear vectors are also called parallel vectors. If they have the same direction they are named as like vectors otherwise unlike vectors.
Symbolically, two non-zero vectors
a
andb
are collinear if and only if, ab, where R b 1 1 b a = 2 2 b a = 3 3 b a ( =)
Vectors a = a1iˆ + a2jˆ + a3kˆ and b = b1iˆ + b2jˆ + b3kˆ are collinear if
1 1 b a = 2 2 b a = 3 3 b a
Example # 2 : Find values of x & y for which the vectors
a = (x + 2) iˆ – (x – y) jˆ + kˆ
b= (x – 1) iˆ + (2x + y) jˆ + 2 kˆ are parallel. Solution : a and b are parallel if
1 x 2 x = y x 2 x y = 2 1 x = – 5, y = – 20 Coplanar vectors:
A given number of vectors are called coplanar if their line segments are all parallel to the same plane. Note that “two vectors are always coplanar”.
Multiplication of a vector by a scalar :
If a is a vector and m is a scalar, then ma is a vector parallel to a whose magnitude ism times that of a. This multiplication is called scalar multiplication. If a and bare vectors and m, n are scalars, then : a m m ) a ( ) a ( m , m(na) n(ma) (mn)a a n a m a ) n m ( , m(ab) ma mb
Self Practice Problems :
(1) Given a regular hexagon ABCDEF with centre O, show that
(i) OB – OA = OC – OD (ii) OD + OA = 2 OB + OF
(iii) AD + EB + PC = 4 AB
(2) The vector iˆ jˆkˆ bisects the angle between the vectors c and 3iˆ4jˆ. Determine the unit vector along c.
(3) The sum of the two unit vectors is a unit vector. Show that the magnitude of the their difference is 3 . Answers : (2) kˆ 15 14 jˆ 15 2 iˆ 3 1
Addition of vectors :
(i) If two vectors a and b are represented by OA and OB, then their sum ab is a vector represented byOC, where OC is the diagonal of the parallelogram OACB.
(ii)
a
b
b
a
(commutative) (iii)(
a
b
)
c
a
(
b
c
)
(associative) (iv)a
0
a
0
a
(v)a
(
a
)
0
(
a
)
a
(vi) |ab||a||b| (vii) |ab|||a||b||
Example # 3 : If aiˆ2jˆ3kˆ andb 2iˆ4jˆ5kˆ represent two adjacent sides of a parallelogram, find unit vectors parallel to the diagonals of the parallelogram.
Solution : Let ABCD be a parallelogram such that AB = a and BC = b.
Then, AB + BC = AC AC = ab = 3iˆ6jˆ2kˆ | AC | = 9364 = 7 AB + BD = AD BD = ADAB = ba = iˆ2jˆ8kˆ | BD | = 1464 = 69
Unit vector along AC =
| AC | AC = 7 1
kˆ 2 jˆ 6 iˆ 3 and Unit vector along BD =
| BD | BD = 69 1
kˆ 8 jˆ 2 iˆ Example # 4 : ABCDE is a pentagon. Prove that the resultant of the forces AB , AE ,BC, DC, ED and AC
is 3AC.
Solution : Let R be the resultant force
R = AB + AE + BC + DC + ED + AC
R = ( AB + BC) + ( AE + ED + DC) + AC
R = AC + AC + AC
R = 3AC. Hence proved.
Position vector of a point:
Let O be a fixed origin, then the position vector of a point P is the vector OP. If a and b are position vectors of two points A and B, then
AB = ba = position vector (p.v.) of B position vector (p.v.).) of A. DISTANCE FORMULA
Distance between the two points A(a) and B(b) is AB = ab SECTION FORMULA
a point which divides AB in the ratio m: n is given by n m b m a n r .
Note : Position vector of mid point of AB =
2 b a
.
Example # 5 : ABCD is a parallelogram. If L, M be the middle point of BC and CD, express AL and AM in terms of AB and AD . Also show that AL + AM =
2 3
AC.
Solution : Let the position vectors of points B and D be respectively b and d referred to A as origin of reference.
Then AC = AD + DC = AD + AB [ DC = AB ]
AC = d
+ b AB = b, AD = d i.e. position vector of C referred to A is d + b AL = p.v. of L, the mid point of BC.
= 2 1 [p.v. of B + p.v. of C] = 2 1
b d b = AB + 2 1 AD Similarly AM = 21
ddb
= AD + 2 1 AB AL + AM = b + 2 1 d + d+ 2 1 b = 2 3 b + 2 3 d = 2 3 ( b + d ) = 2 3 AC.Example # 6 : If ABCD is a parallelogram and E is the mid point of AB. Show by vector method that DE trisect AC and is trisected by AC.
Solution : Let AB = a and AD = b
Then BC = AD = b and AC = AB + AD = a + b Also let K be a point on AC, such that AK : AC = 1 : 3
AK = 3 1 AC AK = 3 1 (a + b) ...(i)
Again E being the mid point of AB, we have AE = 2 1
a
Let M be the point on DE such that DM : ME = 2 : 1
AM = 2 1 AE 2 AD = 3 a b ...(ii) From (i) and (ii) we find that
AK = 31 (a + b) = AM , and so we conclude that K and M coincide. i.e. DE trisect AC and is trisected by AC. Hence proved.
Self Practice Problems
(4) Express vectors BC, CA and AB in terms of the vectors OA, OB and OC
(5) If a,b are position vectors of the points (1, –1), (–2, m), find the value of m for which a and b are collinear.
(6) The position vectors of the points A, B, C, D are iˆ jˆkˆ, 2iˆ5jˆ, 3iˆ2jˆ3kˆ, iˆ6jˆkˆ respectively. Show that the lines AB and CD are parallel and find the ratio of their lengths. (7) The vertices P, Q and S of aPQS have position vectorsp,q and s respectively..
(i) If M is the mid point of PQ, then find position vector of M in terms of pandq
(ii) Find t, the position vector of T on SM such that ST : TM = 2 : 1, in terms of p,q ands.
(iii) If the parallelogram PQRS is now completed. Express r, the position vector of the point R in terms of p,q and s
(8) D, E, F are the mid-points of the sides BC, CA, AB respectively of a triangle. Show FE =
2 1
BC and that the sum of the vectors AD , BE , CF is zero.
(9) The median AD of a ABC is bisected at E and BE is produced to meet the side AC in F. Show that AF = 3 1 AC and EF = 4 1 BF..
(10) Point L, M, N divide the sides BC, CA, AB ofABC in the ratios 1 : 4, 3 : 2, 3 : 7 respectively. Prove that AL + BM + CN is a vector parallel to CK, when K divides AB in the ratio 1 : 3. Answers : (4) BCOCOB, CAOAOC, ABOBOA (5) m = 2 (6) 1 : 2 (7) m = 2 1 ) q p ( , t = 2 1 ) s q p ( , r = 2 1 ) s p q (
Angle between two vectors :
It is the smaller angle formed when the initial points or the terminal points of the two vectors are brought together. Note that 0º 180º .
Vector equation of a line :
Parametric vector equation of a line passing through two pointA(a) andB(b) is given by r = at(ba), where 't' is a parameter. If the line passes through the point
A a
( )
and is parallel to the vector b, then its equation is r a tb. Note : r is the p.v. of the point on the line.A vector in the direction of the bisector of the angle between the two vectors aandb is b b a a .
Hence bisector of the angle between the two vectors a and b is
a
b
, where R+. Bisector of the exterior angle between a and b is
a
b
, R+.Note that the equations of the bisectors of the angles between the lines
r
=a
+
b
and
r
=a
+
c
are :
r
=
a
+ t
b c
and
r
=a
+ p
c b
.Scalar product (Dot Product) of two vectors :
Geometrical interpretation of scalar product :
Let a and b be vectors represented by OA and OB respectively. Let be the angle between OA and
OB. Draw BL OA and AM OB.
FromOBL and OAM, we have OL = OB cos and OM = OA cos .
(b)
(a)
Here OL and OM are known as projections of b on a
anda on b respectively.. Now, a.b
= |a| | b| cos = |a|(| b| cos ) = |a| (OB cos ) = |a| (OL)
= (Magnitude of a) (Projection of b on a) ...(i) Again a.b = |a| | b| cos = | b| (|a| cos )
= | b| (OA cos) = | b| (OM)
= (magnitude of b) (Projection of a on b) ...(ii)
Thus geometrically interpreted, the scalar product of two vectors is the product of modulus of either vector and the projection of the other in its direction.
(i) iˆ . iˆ = jˆ . jˆ = kˆ . kˆ = 1; iˆ . jˆ = jˆ . kˆ = kˆ . iˆ = 0 (ii) Projection of | b | b . a b on a
(iii) If a = a1iˆ + a2jˆ + a3kˆ and b = b1iˆ + b2jˆ + b3kˆ , then a.b = a1b1+ a2b2+ a3b3 2 3 2 2 2 1 a a a a , 32 2 2 2 1 b b b b
(iv) The angle between a and b is given by
| b | | a | b . a cos , 0 (v) a.b a b cos,(0)
note that if is acute, then a.b > 0 and if is obtuse, then a.b < 0 (vi) ab = |a|2|b|22|a||b|cos
, where is the angle between the vectors (vii) a.a a2a2 (viii) a.bb.a (commutative)
(ix) a .(bc)a.ba.c (distributive) (x) a.b 0a b (a0,b 0)
(xi) (ma). b = a.(mb) = m(a.b) (associative), where m is a scalar.. Note:
(a) Maximum value of
a
.
b
isa
b
(b) Minimum value of
a
.
b
is– a
b
(c) Any vector
a
can be written asa
=
a i i. a. j j a k k. .Example # 7 : Find the value of p for which the vectors a 3iˆ2jˆ9kˆ and biˆpjˆ3kˆ
are (i) perpendicular (ii) parallel
Solution : (i) a b a.b = 0
3iˆ2jˆ9kˆ
.
iˆpjˆ3kˆ
= 0 3 + 2p + 27 = 0 p = – 15 (ii) vectors a = 3iˆ2jˆ9kˆ and b = iˆpjˆ3kˆ are parallel iff
1 3 = p 2 = 3 9 3 = p 2 p = 3 2
Example # 8 : If a + b + c = 0, |a| = 3, |b| = 5 and |c| = 7, find the angle between a and b. Solution : We have, abc 0 ab = –c
ab .
ab =
c .
c ab 2 = |c|2 a 2 + b 2 + 2a.b = c 2 2 a + b 2 + 2 a b cos = c 2 9 + 25 + 2 (3) (5) cos = 49 cos = 2 1 = 3 .Example # 9 : Find the values of x for which the angle between the vectors a = 2x2 iˆ + 4x jˆ + kˆ and b = 7 iˆ – 2 jˆ + x kˆ is obtuse.
Solution : The angle between vectors a and b is given by cos =
| b | | a | b . a
Now, is obtuse cos < 0
| b | | a | b . a < 0 a.b < 0 [ |a|,|b|0] 14x2– 8x + x < 0 7x (2x – 1) < 0 x(2x – 1) < 0 0 < x < 2 1
Example # 10 : D is the mid point of the side BC of aABC, show that AB2+ AC2= 2 (AD2+ BD2) Solution : We have AB = AD + DB AB2= 2 ) DB AD ( AB2= AD2+ DB2+ 2AD . DB ...(i) Also we have AC = AD + DC AC2= (ADDC)2 AC2= AD2+ DC2+ AD 2 . DC ...(ii)
Adding (i) and (ii), we get AB2+ AC2= 2AD2+ 2BD2+ 2 AD . (DBDC) AB2+ AC2= 2(AD2+ BD2) DB +
DC = 0
Example # 11 : If a = iˆ + jˆ + kˆ and b = 2 iˆ – jˆ + 3 kˆ , then find
(i) Component of b along a. (ii) Component of b in plane of a &b but to a.
Solution : (i) Component of b along a is
2 | a | b . a a Here a . b = 2 – 1 + 3 = 4 2 | a | = 3 Hence 2 | a | b . a a = 3 4 a = 3 4 (iˆ + jˆ + kˆ)
(ii) Component of b in plane of a &b but to a is b –
2 | a | b . a a. = 3 1
kˆ 5 jˆ 7 iˆ 2 Self Practice Problems :
(11) If a and b are unit vectors and is angle between them, prove that tan 2 = | b a | | b a | . (12) Find the values of x and y if the vectors a = 3iˆxjˆkˆ and b = 2iˆ jˆykˆ are mutually
perpendicular vectors of equal magnitude.
(13) Let a = x2iˆ2jˆ2kˆ, b= iˆjˆkˆ andc = x2iˆ5jˆ4kˆ be three vectors. Find the values of x for which the angle between a and b is acute and the angle between b and c is obtuse.
(14) The points O, A, B, C, D are such that OAa, OB b , OC 2a 3b , OD a 2b .
Given that the length of OA is three times the length of OB. Show that BD and AC are perpendicular.
(15) ABCD is a tetrahedron and G is the centroid of the base BCD. Prove that AB2+ AC2+ AD2= GB2+ GC2+ GD2+ 3GA2
Answers : (12) x = – 12 31 , y = 12 41 (13) (– 3, –2) (2, 3)
Vector product (Cross Product) of two vectors:
(i) If a,b are two vectors and is the angle between them, then a xb a b sinnˆ, where nˆ is the unit vector perpendicular to both aandb such that a,b andnˆ forms a right handed screw system. (ii) Geometrically a xb = area of the parallelogram whose two adjacent sides are represented by
b and a .
(iii) iˆiˆ jˆjˆkˆkˆ0 ; iˆjˆkˆ,jˆkˆ iˆ,kˆiˆ jˆ
(iv) If a = a1iˆ +a2jˆ + a3kˆ and
b
= b1iˆ + b2jˆ + b3kˆ , then3 2 1 3 2 1 b b b a a a kˆ jˆ iˆ b a (v)
a x b
b x a
(not commutative) (vi)(
m a
)
b
=a
(
m b
)
= m(
)
a
b
(associative), where m is a scalar.. (vii)a x b c
(
)
(
a x b
)
(
a x c
)
(distributive)(viii) ab0a andb are parallel (collinear)
(
a
0
,
b
0
)
i.e.
a
K b
, where K is a scalar.. (ix) Unit vector perpendicular to the plane of a and b is nˆ =| b a | b a
(x) A vector of magnitude ‘r’ and perpendicular to the plane of a and b is
| b a | ) b a ( r
(xi) If is the angle between
b a b x a sin then , b and a
(xii) If a,b andc are the position vectors of 3 points A, B and C respectively, then the vector area of
ABC =
1
2
a x b
b x c
cx a
. The points A, B and C are collinear if a xb b xc c xa 0 (xiii) Area of any quadrilateral whose diagonal vectors are d1andd2 is given by d1xd22
1
(xiv) Lagrange's Identity : For any two vectors
b . b b . a b . a a . a ) b . a ( b a ) b x a ( ; b and a 2 2 2 2
Example # 12 : Find a vector of magnitude 9, which is perpendicular to both the vectors 4iˆ– jˆ3kˆ and 2iˆjˆ2kˆ.
Solution : Let a = 4iˆjˆ3kˆ and b = 2iˆjˆ2kˆ. Then b a = 2 1 2 3 1 4 kˆ jˆ iˆ = (2 – 3) iˆ – (–8 + 6) jˆ + (4 – 2) kˆ = iˆ2jˆ2kˆ |ab| = 2 2 2 2 2 ) 1 ( = 3 Required vector = 9 | b a | b a = 3 9 ) kˆ 2 jˆ 2 iˆ ( = ± (3iˆ6jˆ6kˆ)
Example # 13 : For any three vectors a,b,c, show that a(bc)b(ca)c(ab)0. Solution : We have, a × (bc) + b × (ca) + c × (ab)
= abacbcbacacb [Using distributive law] = abacbcabacbc= 0 [ ba–ab etc] Example # 14 : For any vector a, prove that |aiˆ|2 + |ajˆ|2 + |akˆ|2 = 2 2
| a |
Solution : Let a = a1iˆa2jˆa3kˆ. Then iˆ
a = (a1iˆa2jˆa3kˆ) × iˆ = a1(iˆ + aiˆ) 2(jˆiˆ) + a3 (kˆiˆ) = –a2kˆa3jˆ |aiˆ|2 = a 2 2+ a 3 2 jˆ a = (a1iˆa2jˆa3kˆ) × jˆ = a1kˆa3iˆ 2 | jˆ a | = a2 1+ a3 2 kˆ a (aiiˆa2jˆa3kˆ)× k = – aijˆa2iˆ |akˆ|2 = a 1 2+ a 2 2 2 | iˆ a | + 2 | jˆ a | + 2 | kˆ a | = a22+ a 3 3+ a 1 2+ a 3 2+ a 1 2+ a 2 2 = 2 (a12+ a 2 2+ a 3 2) = 2 |a|2
Example # 15 : Let OA = a, OB = 10a + 2b and OC = b where O is origin. Let p denote the area of the quadrilateral OABC and q denote the area of the parallelogram with OA and OC as adjacent sides. Prove that p = 6q.
Solution : We have,
p = Area of the quadrilateral OABC
p = 2 1 | AC OB | = 2 1 | ) OA OC ( OB | p = 2 1 | ) a b ( ) b 2 a 10 ( | = 2 1 | ) a b ( 2 ) b b ( 2 ) a a ( 10 ) b a ( 10 | p = 2 1 | ) b a ( 2 0 0 ) b a ( 10 | = 6 |ab| ...(i)
and q = Area of the parallelogram with OA and OC as adjacent sides q = |OAOC| = |ab| ...(ii)
Self Practice Problems :
(16) If p and q are unit vectors forming an angle of 30º. Find the area of the parallelogram having q
2 p
a and b2pq as its diagonals.
(17) Prove that the normal to the plane containing the three points whose position vectors are c
, b ,
a lies in the direction bccaab
(18) ABC is a triangle and EF is any straight line parallel to BC meeting AC, AB in E, F respectively. If BR and CQ be drawn parallel to AC, AB respectively to meet EF in R and Q respectively, prove that ARB = ACQ.
Answers : (16) 3/4 sq. units
Shortest distance between two lines :
If two lines in space intersect at a point, then obviously the shortest distance between them is zero. Lines which do not intersect and are also not parallel are called skew line. For Skew lines the direction of the shortest distance would be perpendicular to both the lines.
Let LM be the shortest distance vector between the lines L1and L2. Then LM is perpendicular to both p and q i.e. LM is parallel to p × q. Therefore the magnitude of the shortest distance vector (i.e. | LM |) would be equal to that of the projection of AB along the direction of the line of shortest distance. |LM| Projection of ABonLM = Projection of ABonp xq = q x p ) q x p ( . ) a b ( q x p ) q x p ( . AB
(i) The two lines directed alongp andq will intersect only if shortest distance = 0
i.e. (ba).(p xq)0 i.e.
b a lies in the plane containing p andq.
ba · pq0.(ii) If two parallel lines are given by r1a1Kb andr2 a2Kb, then distance (d) between them is given by b ) a a ( x b d 2 1
Scalar triple product (Box Product) (S.T.P.) :
(i) The scalar triple product of three vectors a,b andc is defined as: a xb.c a b c sin cos. , where is the angle between a,b (i.e. a bθ) and is the angle between a xb andc (i.e.(a b) c
=) . It is (i.e. ab.c) also written as
a b c
and spelled as box product. (ii) Scalar triple product geometrically represents the volumeof the parallelopiped whose three coterminous edges are represented by a,b andc .ie.V = [a b c]
(iii) In a scalar triple product the position of dot and cross can be interchanged i.e. ] b a c [ ] a c b [ ] c b a [ c . ) b x a ( ) c x b ( . a (iv) a.(bxc)a.(cxb) .ie. [abc][acb] (v) If a = a1iˆ + a2jˆ + a3kˆ ; b = b1iˆ +b2jˆ +b3kˆ and c = c1iˆ + c2jˆ + c3kˆ , then 3 2 1 3 2 1 3 2 1 c c c b b b a a a ] c b a [ . In general, ifaa1a2m a3n;bb1b2m b3n and c c1 c2m c3n then
mn
c c c b b b a a a c b a 3 2 1 3 2 1 3 2 1 , where ,m andn are non-coplanar vectors.
(vi) If a,b andc are coplanar [a b c]0.
(vii) Scalar product of three vectors, two of which are equal or parallel is 0 [a b c]0,
(viii) If
a b c
,
,
are non-coplanar, then[a b c] 0 for right handed system and [a b c] 0 for left handed system.(ix) [iˆ jˆkˆ] = 1 (x)[Ka b c]K[a b c] (xi)[(ab) c d ] [a c d] [b c d] (xii)
a
b
b c
c a
= 0 and
a
b
b c
c a
= 2
a b c
. (xiii)
a b c
2= c . c b . c a . c c . b b . b a . b c . a b . a a . a Tetrahedron and its properties :
(a) The volume of the tetrahedron OABC with O as origin and the position vectors of A, B and C being c and b , a respectively is given by 6 1 V
a b c
(b) If the position vectors of the vertices of tetrahedron are a,b,c andd, then the position vector of its centroid is given by (a b c d)
4
1 .
note that this is also the point of concurrency of the lines joining the vertices to the centroids of the opposite faces and is also called the centre of the tetrahedron. In case the tetrahedron is regular it is equidistant from the vertices and the four faces of the tetrahedron.
Example # 16 : Find the volume of a parallelopiped whose sides are given by 3iˆ7jˆ5kˆ, 5iˆ7jˆ3kˆ and7iˆ5jˆ3kˆ
Solution : Let a 3iˆ7jˆ5kˆ, b5iˆ7jˆ–3kˆ
and c7iˆ5jˆ3kˆ.
We know that the volume of a parallelopiped whose three adjacent edges are a,b,c is
] c b a [ .
Now [abc] = 3 5 7 3 7 5 5 7 3 = –3 (–21 – 15) – 7 (15 + 21) + 5 (25 – 49) = 108 – 252 – 120 = –264
So required volume of the parallelopiped = [a b c] = | – 264 | = 264 cubic units
Example # 17 : Simplify [ab bc ca] Solution : We have : ] a c c b b a [ = {(ab)(bc)} . (ca) [By definition] = (abacbbbc) . (ca) [By distribution law] = (abcabc) . (ca) [ bb0]
= (ab).c – (ab).a + (ca).c – (ca).a + (bc).c – (bc).a [By distribution law] = [a b c] – [a b a] + [c a c] – [c a a] + [b c c] – [b c a]
= [a b c] – [b c a] [ When any two vectors are equal, scalar triple product is zero ] = [a b c] – [a b c] = 0 [ [b c a] = [a b c]]
Example # 18 : Find the volume of the tetrahedron whose four vertices have position vectors a, b, c and d. Solution : Let four vertices be A, B, C, D with position vectors a, b, c and d respectively..
DA = (a – d) DB = ( b – d) DC = (c – d) Hence volume V = 6 1 [a – d b – d c – d] = 6 1 (a – d) . [( b – d) × (c – d)] = 6 1 (a – d) . [ b × c – b × d + c × d] = 6 1 {[a b c] – [a b d] + [a c d] – [ d b c]} = 6 1 {[a b c] – [a b d] + [a c d] – [ b c d]}
Example # 19 : Show that the vectors a2iˆ4jˆ2kˆ, b4iˆ2jˆ2kˆ andc 2iˆ2jˆ4kˆ are coplanar..
Solution : [a b c] = 4 2 2 2 2 4 2 4 2 = – 2(–8 – 4) – 4(16 – 4) – 2(–8 – 4) = 24 – 48 + 24 = 0 So vectors a, b, c are coplanar
Self Practice Problems :
(19) Show that {(a + b + c) × (c – b)} . a = 2
a b c
.(20) Show that a.(bc)(abc)0
(21) One vertex of a parallelopiped is at the point A (1, –1, –2) in the rectangular cartesian co-ordinate. If three adjacent vertices are at B(–1, 0, 2), C(2, –2, 3) and D(4, 2, 1), then find the volume of the parallelopiped.
(22) Find the value of m such that the vectors 2iˆjˆkˆ , iˆ2jˆ3kˆ and 3iˆmjˆ5kˆ are coplanar.. (23) Show that the vector a,b,c are coplanar if and only if b ,c ca, ab are coplanar..
Answers : (21) 72 (22) – 4
Vector triple product :
Let a,b andc be any three vectors, then the expression a x(b x c) is a vector & is called a vector triple product.
Geometrical interpretation of a x(b x c)
Consider the expression
a x b x c
(
)
which itself is a vector, since it is a cross product of two vectors ) c x b ( anda . Now
a x b x c
(
)
is a vector perpendicular to the plane containing a and(b xc) butb x c
is a vector perpendicular to the plane containing b andc, thereforea x b x c
(
)
is a vector which lies in the plane of bandc and perpendicular toa
. Hence we can express a x (b x c) in terms of bandc i.e. a x (b x c) =xb yc, where x , y are scalars.
a x b x c
(
)
=( . )
a c b
( . )
a b c
(
a x b x c
)
=( . )
a c b
( . )
b c a
In general(
a x b x c
)
a x b x c
(
)
Example # 20 : For any vector a, prove that iˆ(aiˆ) jˆ(ajˆ)kˆ(ak) = 2a
Solution : Let a a1iˆa2jˆa3kˆ.
Then iˆ(aiˆ)jˆ(aˆjˆ)kˆ(akˆ)
= {(iˆ.iˆ)a(iˆ.a)iˆ} + {(jˆ. jˆ)a(jˆ.a)jˆ} + {(kˆ.kˆ)a(kˆ.a)kˆ} = {(a(iˆ.a)iˆ}{a(jˆ.a)jˆ} + {a(kˆ.a)kˆ}
= 3a{(iˆ.a)iˆ(jˆ.a)jˆ(kˆ.a)kˆ = 3a(a1iˆa2jˆa3kˆ) = 3aa 2a
Example # 21 :
Prove that a{b(cd)} = (b.d)(ac) – (b.c) (ad) Solution : We have, a{b(cd)} = a{(b.d)c(b.c)d}
= a{(b.d)c}a{(b.c)d} [by dist. law] = (b.d)(a c) (b.c)(a d)
Example # 22 : Let a = + jˆiˆ 2 – kˆ3 , b = iˆ + 2 – kˆjˆ 2 and c = 2 –iˆ + kˆ . Find the value(s) of , if any,,jˆ such that
ab
bc
×
ca
= 0. Solution :
ab
bc
×
ca
=
a b c
b ×
c a
=
a b c
a.b c b.c a
which vanishes if (i)
a.b c =
b.c a (ii)
a b c
= 0(i)
a.b c =
b.c a leads to the equation 23+ 10 + 12 = 0, 2+ 6 = 0 and 6 – 6 = 0, which do not have a common solution.(ii)
a b c
= 0 1 2 2 2 1 3 2 = 0 3 = 2 = 3 2Example # 23 : If A B = a, A .a = 1 and AB = b, then prove that A = 2
| a | a b a and B =
2
2 | a | 1 | a | a a b .Solution : Given A B a ...(i)
a.
A B = a.a a.A . Ba = a.a 1 + a.B = |a|2 a.B = |a|2 – 1 ...(ii) Given AB b a
AB = a × b
a.B A –
a.A Bab
|a|2 1
A B = ab ...(iii) [Using equation (ii)] solving equation (i) and (iii) simultaneously, we getA = 2 | a | a b a and B =
2
2 | a | 1 | a | a a b Example # 24 : Solve for rsatisfying the simultaneous equations rb cb, r.a0 provided a is not perpendicular to b.
Solution : (rc) × b = 0 rc and b are collinear rc kb
r = ckb ...(i) r.a = 0
(ckb) . a = 0 k = – b . a c . a
putting in (i) we get
b . a c . a c r b
Example # 25 : If xakxb, where k is a scalar and a,b are any two vectors, then determine x in terms of a,b and k.
Solution : xakxb ...(i)
Premultiply the given equation vectorially by a ) a x ( a + k (ax) = ab (a.a)x (a.x)ak(ax)ab ...(ii) Premultiply (i) scalarly by a
] a x a [ + k(a.x) = a.b b . a ) x . a ( k ...(iii)
Substituting xa from (i) and a.x from (iii) in (ii) we get
x = 2 2 k a 1 a k ) b . a ( ) b a ( b k
Self Practice Problems :
(24) Prove that a(bc)b(ca)c(ab)0.
(25) Find the unit vector coplanar with iˆ + jˆ + 2 kˆ and iˆ + 2 jˆ + kˆ and perpendicular to iˆ + jˆ + kˆ . (26) Prove that a{a(ab)}(a.a)(ba). (27) Given that 2 p 1 x (p.x) p , show thatq p.q 2 1 x .
p and find x in terms of p and q.
(28) If x.a = 0, x.b = 0 and x.c = 0 for some non-zero vector x, then show that [a b c] = 0
(29) Prove that r = ] c b a [ ) c b ( ) a . r ( + ] c b a [ ) a c ( ) b . r ( + ] c b a [ ) b a ( ) c . r (
where a,b,c are three non-coplanar vectors Answers : (25) ± 2 1 ( – jˆ + kˆ ) and x = q – p | p | 2 q . p 2
Linear combinations :
Given a finite set of vectors
a b c
, , ,...
, then the vectorr
xa
yb
zc
...
is called a linear combination ofa b c
, , ,...
for any x, y, z... R. We have the following results:(b) Fundamental Theorem in plane : Let
a b
,
be non zero, non collinear vectors, then any vector
r
coplanar
with
a b
,
can be expressed uniquely as a linear combination of a andb i.e.there exist some unique x, y
R such thatxa
yb
r
. (c) Ifa b c
, ,
are nonzero, noncoplanar vectors, thenxa yb zc x a y b z c
'
'
'
x
x
' ,
y
y
' ,
z
z
'
(d) Fundamental theorem in space: Let
a b c
, ,
be nonzero, noncoplanar vectors in space.Then any vector
r
can be uniquely expressed as a linear combination ofa b c
, ,
i.e. there exist some unique x,y, z R such thatxa
yb
zc
r
.(e) Ifx1,x2,..., xn are n non zero vectors and k1,k2,...,kn are n scalars and if the linear co m bin atio nk1x1k2x2...knxn0 k10,k20,...,kn 0, the n we say th at ve cto rs
n 2
1,x ,...,x
x
are linearly independent vectors.
(f) Ifx1,x2,..., xn are not linearly independent then they are said to be linearly dependent vectors. i.e. if k1x1k2x2k3x3...krxr ...knxn 0 and if there exists at least one kr 0, then
n 2
1,x ,...,x
x
are said to be linearly dependent vectors.
Note 1: If kr 0; k1x1k2x2k3x3...krxr ...knxn0
k x
r r
k x
k x
k
rx
r
k
rx
r
k x
n n
1 1 2 2...
1.
1 1.
1...
k
k
x
k
k
x
k
k
x
k
k
x
k
k
x
r r r r r r r r n r n1
1
1
1
1
1 1 2 2 1 1
...
.
...
xr c1x1c2x2...cr–1xr–1+ cr 1xr 1 ... cnxn i.e.
x
r is expressed as a linear combination of vectors x1,x2,...,xr1,xr1,...,xn Hence xr withx1,x2,...,xr1,xr1,...,xnforms a linearly dependent set of vectors. Note 2: If
a
= 3 iˆ + 2 jˆ + 5 kˆ thena
is expressed as a Linear Combination of vectors iˆ , jˆ , kˆ . AlsoAlso
a
, iˆ , jˆ , kˆ form a linearly dependent set of vectors. In general, in 3 dimensional space every set of four vectors is a linearly dependent system. iˆ , jˆ , kˆ are Linearly Independent set of vectors. For K1iˆ + K2jˆ + K3kˆ = 0 K1= K2= K3= 0
Two vectorsa and b are linearly dependent
a
is parallel tob
i.e.ab0linear dependence of aand b. Conversely if ab0 then a and b are linearly independent.
If three vectors
a b c
, ,
are linearly dependent, then they are coplanar i.e.[a b c]= 0. Conversely if ]c b a
Example # 26 : Given that position vectors of points A, B, C are respectively
a – 2 b + 3c, 2a + 3 b – 4c, – 7 b + 10c then prove that vectors AB and AC are linearly dependent.
Solution : Let A, B, C be the given points and O be the point of reference then
OA = a – 2 b + 3c, OB = 2a + 3 b – 4c and OC= – 7 b + 10c
Now AB = p.v. of B – p.v. of AA
= OB – OA = (a + 5 b – 7c) and AC = p.v. of C – p.v of AA
= OCOA = – (a5b7c) = – AB
AC = AB where = – 1. Hence AB and AC are linearly dependent
Example # 27 : Prove that the vectors 5a + 6 b + 7c, 7a – 8 b + 9c and 3a + 20 b + 5c are linearly dependent, where a, b,c being linearly independent vectors.
Solution : We know that if these vectors are linearly dependent , then we can express one of them as a linear combination of the other two.
Now let us assume that the given vector are coplanar, then we can write 5a + 6 b + 7c =( 7a – 8 b + 9c) + m (3a + 20 b + 5c)
where, m are scalars
Comparing the coefficients of a, b and c on both sides of the equation 5 = 7 + 3m ...(i)
6 = – 8 + 20 m ...(ii) 7 = 9 + 5m ...(iii) From (i) and (iii) we get
4 = 8 =
2 1
= m which evidently satisfies (ii) equation too. Hence the given vectors are linearly dependent .
Self Practice Problems :
(30) Does there exist scalars u, v, w such that ue1ve2we3 iˆ where e1kˆ , e2 jˆkˆ, kˆ
2 jˆ e3 ?
(31) Consider a base a,b,c and a vector 2a3bc. Compute the co-ordinates of this vector relatively to the base p,q,r where p 2a3b, qa2bc, r3ab2c.
(32) If a and b are non-collinear vectors and A =(x + 4y)a + (2x + y + 1) b and B = (y – 2x + 2)
a+ (2x – 3y – 1) b, find x and y such that 3A 2B.
(33) If vectors a,b,c be linearly independent, then show that
(i) a2b3c, 2a3b4c, b2c are linearly dependent (ii) a3b2c, 2a4bc, 3a2bc are linearly independent.
(34) Given that iˆ ,jˆ iˆ2jˆ are two vectors. Find a unit vector coplanar with these vectors and perpendicular to the first vector iˆ . Find also the unit vector which is perpendicular to thejˆ plane of the two given vectors.
(35) If with reference to a right handed system of mutually perpendicular unit vectors iˆ,ˆ,jkˆ, jˆ
iˆ 3
, 2iˆjˆ3kˆ. Express in the form 12 where 1 is parallel to and
2
is perpendicular to .
(36) Prove that a vector r in space can be expressed linearly in terms of three non-coplanar,, non-zero vectors a,b,c in the form
] c b a [ c ] b a r [ b ] a c r [ a ] c b r [ r Answers : (30) No (31) (0, – 7/5, 1/5) (32) x = 2, y = –1 (34) ± 2 1 ) jˆ iˆ ( ; kˆ (35) jˆ 2 1 iˆ 2 3 1 , jˆ 3kˆ 2 3 iˆ 2 1 2
Reciprocal system of vectors :
If
a b c
, ,
&
a' , b' , c'
are two sets of non-coplanar vectors such that
a. a' = b. b' = c. c' = 1
, then the two systems are called Reciprocal System of vectors.Note : ] c b a [ b x a c and ] c b a [ a x c b , ] c b a [ c x b = a
Example # 28 : If a,b,c and a,b,c be the reciprocal system of vectors, prove that (i) a.ab.bc.c3 (ii) aabbcc0 Solution : (i) We have : a.a = b.b = c.c = 1
a .
a + b.b + c.c = 1 + 1 + 1 = 3
(ii) We have : a = (bc), b = (ca) and c = (ab), where = ] c b a [ 1 } c ) b . a ( b ) c . a {( )} c b ( a { ) c b ( a a a } a ) c . b ( c ) a . b {( )} a c ( b { ) a c ( b b b and ccc(ab){c(ab)}{(c.b)a(c.a)b} aabbcc = {(a.c)b(a.b)c}{(b.a)c(b.c)a}{(c.b)a(c.a)b} =[(a.c)b(a.b)c(b.a)c(b.c)a(c.b)a(c.a)b] =[(a.c)b(a.b)c(a.b)c(b.c)a(b.c)a(a.c)b] = 0 0
Equation of a plane :
(i) The equation(rr0).n 0 represents a plane containing the point with position vector n
where ,
r0
The above equation can also be written as r.nd, where d = r0.n
(ii) Angle between two planes is the angle between two normals drawn to the planes and the angle between a line and a plane is the compliment of the angle between the line and the normal to the plane.
(iii) The length of perpendicular (p) from a point having position vector a to the plane r.nd is given by p = | n | | d n . a |
(iv) If (ra).n10 and (ra).n20 are the equations of two planes, then the equation of line of intersection of these planes is given by ra(n1n2).
Test of collinearity :
Three points A,B,C with position vectors
a b c
, ,
respectively are collinear, if & only if there exist scalars x, y, z not all zero simultaneously such thatxa
yb zc
= 0, where x + y + z = 0.Test of coplanarity :
Four points A, B, C, D with position vectors
a b c d
, , ,
respectively are coplanar if and only if there exist scalars x, y, z, w not all zero simultaneously such thatxa + yb + zc + wd
= 0, wherex + y + z + w = 0.
Example # 29 : Show that the vectors 2ab3c, ab2c and ab3c are non-coplanar vectors. Solution : Let, the given vectors be coplanar.
Then one of the given vectors is expressible in terms of the other two. Let 2ab3c = x
ab2c
+ y
ab3c
, for some scalars x and y.. 2ab3c = (x + y) a + (x + y) b + (–2x – 3y) c 2 = x + y, –1 = x + y and 3 = – 2x – 3y.
Solving first and third of these equations, we get x = 9 and y = –7. Clearly these values do not satisfy the second equation.
Hence the given vectors are not coplanar.
Example # 30 : Prove that four points 2a3bc, a2b3c, 3a4b2c and a6b6c are coplanar.. Solution : Let the given four points be P, Q, R and S respectively. These points are coplanar if the vectors
PQ, PR andPS are coplanar. These vectors are coplanar iff one of them can be expressed as a linear combination of other two. So let PQ = x PR + y PS
a5b4c = x
abc
+ y
a9b7c
a5b4c = (x – y) a + (x – 9y) b + (–x + 7y) c
x – y = –1, x – 9y = –5, –x + 7y = 4 [Equating coeff. of a,b,c on both sides] Solving the first two equations of these three equations, we get x = –
2 1 , y = 2 1 . These values also satisfy the third equation. Hence the given four points are coplanar. Self Practice Problems :
that 3a2bc2d0 , show that the terminal A, B, C, D of these vectors are coplanar. Find the point (P) at which AC and BD meet. Also find the ratio in which P divides AC and BD. (38) Show that the vector abc, bca and 2a3b4c are non-coplanar, where a,b,c are
any non-coplanar vectors.
(39) Find the value of for which the four points with position vectors jˆkˆ, 4iˆ5jˆkˆ,3iˆ9jˆ4kˆ and 4iˆ4jˆ4kˆ are coplanar..
Answers : (37) 4 c a 3 p
P divides AC in 1 : 3 and BD in 1 : 1 ratio (39) = 1
Application of vectors :
(a) Work done against a constant force
F
over a displacement
s
is defined as W F.s(b) The tangential velocity
V
of a body moving in a circle is given byV x r, where
r
is the position vector of the point P.(c) The moment of
F
about ’O’ is defined as M r xF,where r is the position vector of P w.r.t. ’O’. The direction ofM
is along the normal to the plane OPN such that
r F
,
&
M
form a right handed system.(d) Moment of the couple =
(
r
1
r
2)
x F
, where r1andr2 are position vectors of the point of the application of the forces F andF.Example # 31 : Forces of magnitudes 5 and 3 units acting in the directions 6iˆ2jˆ3kˆ and 3iˆ–2jˆ6kˆ respectively act on a particle which is displaced from the point (2, 2, –1) to (4, 3, 1). Find the work done by the forces.
Solution : Let F be the resultant force and d be the displacement vector. Then, F = 9 4 36 ) kˆ 3 jˆ 2 iˆ 6 ( 5 + 3 36 4 9 ) kˆ 6 jˆ 2 – iˆ 3 ( = 7 1 ) kˆ 33 jˆ 4 iˆ 39 (
and d = (4iˆ3jˆkˆ) – (2iˆ2jˆkˆ) = 2iˆjˆ2kˆ Total work done = F. d =
7 1 ) kˆ 33 jˆ 4 iˆ 39 ( .(2iˆ jˆ2kˆ) = 7 1 (78 + 4 + 66) = 7 148 units. Self Practice Problems :
(40) A point describes a circle uniformly in the iˆ , jˆ plane taking 12 seconds to complete one revolution. If its initial position vector relative to the centre is iˆ and the rotation is from iˆ to jˆ , find the position vector at the end of 7 seconds. Also find the velocity vector.
(41) The force represented by 3iˆ2kˆ is acting through the point 5iˆ4jˆ3kˆ. Find its moment about the point iˆ3jˆkˆ.
(42) Find the moment of the couple formed by the forces 5iˆkˆ and 5iˆkˆ acting at the points (9, –1, 2) and (3, –2, 1) respectively Answers : (40) – 2 1
jˆ iˆ 3 , 12
jˆ 3iˆ (41) 2iˆ20jˆ3kˆ (42) iˆjˆ5kˆ
Miscellaneous solved examples
Example # 32 : Show that the points A, B, C with position vectors 2iˆ jˆkˆ , iˆ3jˆ5kˆ and 3iˆ4jˆ4kˆ respectively are the vertices of a right angled triangle. Also find the remaining angles of the triangle.
Solution : We have,
AB = Position vector of B – Position vector of A = (iˆ3jˆ5kˆ) – (2iˆjˆkˆ) = iˆ2jˆ6kˆ
BC = Position vector of C – Position vector of B = (3iˆ4jˆ4kˆ) – (iˆ3jˆ5kˆ) = 2iˆjˆkˆ and, CA = Position vector of A – Position vector of C
= (2iˆjˆkˆ) – (3iˆ4jˆ4kˆ) = iˆ3jˆ5kˆ
Since AB + BC + CA = (iˆ2jˆ6kˆ) + (2iˆ jˆkˆ) + (iˆ3jˆ5kˆ) = 0 So A, B and C are the vertices of a triangle.
Now, BC . CA = (2iˆjˆkˆ) . (iˆ3jˆ5kˆ) = –2 – 3 + 5 = 0
BC CA BCA =
2 Hence ABC is a right angled triangle.
Since A is the angle between the vectors AB and AC. Therefore
cos A = | AC | | AB | AC . AB = 2 2 2 2 2 2 ) 5 ( ) 3 ( 1 ) 6 ( ) 2 ( ) 1 ( ) kˆ 5 jˆ 3 iˆ ( . ) kˆ 6 jˆ 2 iˆ ( = 25 9 1 36 4 1 30 6 1 = 35 41 35 = 41 35 A = cos–1 41 35 cos B = | BC | | BA | BC . BA = 12 22 62 22 ( 1)2 (1)2 ) kˆ jˆ iˆ 2 ( . ) kˆ 6 jˆ 2 iˆ ( 6 2 2 6 6
Example # 33 : If a,b,c are three mutually perpendicular vectors of equal magnitude, prove that abc is equally inclined with vectors a,b and c.
Solution : Let |a| = |b| = |c| = (say). Since a,b,c are mutually
perpendicular vectors, therefore a.b = b.c = c.a = 0 ...(i) Now, abc 2 = a.a + b.b + c.c + 2a.b + 2b.c + 2c.a = |a|2 | + |b|2 + 2 | c | [Using (i) ] = 32 [ |a| = | b | = |c| =] |a b c| = 3 ...(ii)
Suppose abc makes angles1,2,3 with a,b and c respectively. Then,
cos1= | c b a | | a | ) c b a ( . a = | c b a | | a | c . a b . a a . a = | c b a | | a | | a | 2 = |a b c| | a | = 3 = 3 1 [Using (ii)] 1 = cos–1 3 1 Similarly,2= cos–1 3 1 and3= cos–1 3 1 1=2=3.
Hence, abc is equally inclineded with a,b and c
Example # 34 : Prove using vectors : If two medians of a triangle are equal, then it is isosceles.
Solution : Let ABC be a triangle and let BE and CF be two equal medians. Taking A as the origin, let the position vectors of B and C be b and c respectively. Then,
P.V. of E = 2 1 c and P.V. of F = 2 1 b BE = 21 (c 2b) CF = 2 1 ) c 2 b ( Now, BE = CF |BE| = |CF| |BE|2 = |CF|2 2 ) b 2 c ( 2 1 = 2 ) c 2 b ( 2 1 4 1 2 | b 2 c | = 4 1 2 | c 2 b | |c2b|2 = |b2c|2 (c2b) . (c2b) = (b2c) . (b2c) c.c – 4b.c + 4b.b = b.b – 4b.c + 4c.c |c|2 – c . b 4 + 4 2 | b | = 2 | b | – 4b.c + 4|c|2
3 |b|2 = 3 |c|2 |b|2 = |c|2
AB = AC Hence triangle ABC is an isosceles triangle. Example # 35 : Using vectors : Prove that cos (A + B) = cos A cos B – sin A sin B
Solution : Let OX and OY be the coordinate axes and let iˆ and jˆ be unit vectors along OX and OY respectively. LetXOP = A and XOQ = B. Drawn PL OX and QM OX.
Clearly angle between OP and OQ is A + B
InOLP, OL = OP cos A and LP = OP sin A. Therefore OL = (OP cos A) iˆ and LP = (OP sin A)
jˆNow, OL + LP = OP
OP = OP [(cos A ) iˆ – (sin A)A) jˆ ] ...(i) InOMQ, OM = OQ cos B and MQ = OQ sin B. Therefore, OM = (OQ cos B) iˆ , MQ = (OQ sin B) jˆ Now, OM + MQ= OQ
OQ = OQ[(cosB)iˆ(sinB)jˆ] ...(ii) From (i) and (ii), we get
OP . OQ = OP [(cos A) iˆ – (sin A)A) jˆ ] . OQ [(cos B) iˆ + (sin B) jˆ ]
= OP . OQ [cos A cos B – sin A sin B]
But, OP . OQ = |OP| |OQ| cos (A + B) = OP . OQ cos (A + B) OP . OQ cos (A + B) = OP . OQ [cos A cos B – sin A sin B] cos (A + B) = cos A cos B – sin A sin B
Example # 36 : Prove that in any triangle ABC
(i) c2= a2+ b2– 2ab cos C (ii) c = bcosA + acosB. Solution : (i) InABC, AB + BC + CA = 0
BC + CA = – AB ...(i)
Squaring both sides (BC)2+ ( CA)2+ 2 ( BC). CA = ( AB )2 a2+ b2+ 2 ( BC.CA) = c2 c2 = a2+ b2+ 2 ab cos ( – C) c2= a2+ b2– 2ab cosC (ii) (BC+ CA). AB = – AB . AB BC . AB + CA. AB = – c2 – ac cosB – bc cos A = – c2 acosB + bcosA = c.
Example # 37 : If D, E, F are the mid-points of the sides of a triangle ABC, prove by vector method that area of
DEF = 4 1
(area ofABC)
Solution : Taking A as the origin, let the position vectors of B and C be b and c respectively. Then the position vectors of D, E and F are
2 1 ) c b ( , 2 1 c and 21 b respectively..
Now, DE = 2 1 c – 2 1 ) c b ( = 2 b and DF = 2 1 b – 2 1 ((bc) = 2 c
Vector area ofDEF = =
= = = (vector area ofABC)
Hence area ofDEF = area ofABC.
Example # 38 : P, Q are the mid-points of the non-parallel sides BC and AD of a trapezium ABCD. Show that APD = CQB.
Solution : Let = and =
Now DC is parallel to AB there exists a scalar t such that = t = t
= + =
The position vectors of P and Q are and respectively..
Now 2 = × = × = (1 + t) Also 2 = × = × = = = = = Hence Prove.
Example # 39 : Let and are unit vectors and is a vector such that = and then find the value of .
Solution : Given = and
×
× = (as )
(using . = 1 and = 0, since unit vector)
= 0 (as 0) ...(i)
Now
(given + u)
– .
– 0 (as = 0)
1 (as = = 1)
= 1
Example # 40 : In any triangle, show that the perpendicular bisectors of the sides are concurrent.
Solution : Let ABC be the triangle and D, E and F are respectively middle points of sides BC, CA and AB. Let the perpendicular bisectors of BC and CA meet at O. Join OF. We are required to prove that OF is to AB. Let the position vectors of A, B, C with O as origin of reference be , and respectively.. = ( + ), = ( + ) and = ( + ) Also = – , = – and = – Since OD BC ( + ) . ( – ) = 0 b2= c2 ...(i) Similarly OE CA ( + ) . ( – ) = 0 a2= c2 ...(ii)
from (i) and (ii) we have a2– b2= 0
( + ) . ( – ) = 0 ( + ) . ( – ) = 0
Example # 41 : A, B, C, D are four points in space. using vector methods, prove that
AC2+ BD2+ AD2+ BC2 AB2+ CD2 what is the implication of the sign of equality. Solution : Let the position vector of A, B, C, D be and respectively then
AC2+ BD2+ AD2+ BC2= . + . + . + . = + – 2 + + – 2 + + – 2 + + – = + – 2 + + – 2 + + + + + 2 + – – – – = . + . + = AB2+ CD2+ . AB2+ CD2 AC2+ BD2+ AD2+ BC2 AB2+ CD2
for the sign of equality to hold, = 0
