vector tutorials science 2 - ncert board level
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Vector geometry problems for ncert cbse studentsTRANSCRIPT
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GROUP (E): CLASS WORK PROBLEMS Q-1) By vectors method, prove that the diagonals of rhombus are perpendicular to
each other.
Ans. Let ABCD be a rhombus
Let AB a= and AD b=
∴ABCD is a rhombus
( ) ( ) ( ) ( )l AB l BC l CD l AD∴ = = =
(By triangle law of addition of vectors)
BD BA AD a b= + = − +
Now AC AB BC a b= + = +
( ) ( ). .AC BD a b a b∴ = + − + . . . .a a a b b a b b= − + − + 2 2a b= − + 2 2 0b b= − =
AC BD∴ ⊥ ⇒ AC BD⊥
∴ the diagonals AC and BD of rhombus are right angles.
Q-2) Using vectors prove that, if the diagonals of a parallelogram are at right
angles then it is rhombus.
Ans. Let ABCD be a parallelogram
Let AC AB BC a b= + = +
And BD BA AD a b= + = − +
∴ the diagonals AC and BD are perpendicular
. 0AC BD∴ = ⇒ ( ) ( ) 0a b a b+ − + =
. . . . 0a a a b b a b b− + − + =
2 2b a= ⇒ ab = ⇒ ( ) ( ) ( ) ( )l AB l BC l CD l AD∴ = = =
∴ the parallelogram ABCD is a rhombus.
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Q-3) Show by vector method. If the diagonal of a parallelogram are congruent then
it is a rectangle.
Ans. Let ABCD be parallelogram
Let ,AB a AD b= =
From ∆ABC, by triangle law,
AC AB BC a b= + = +
From ∆ABC, by triangle law,
BD BC CD b a= + = −
We have to prove that ABCD is a rectangle i.e. to prove that a b⊥
Since the diagonals AC and BD are congruent
AC BD= ⇒ 2 2
AC BD=
( ) ( ) ( ) ( ). .a b a b b a b a+ + = − −
. . . . . . . .a a a b b a b b b b b a a b a a+ + + = − − +
22 2 22 . 2 .a b a b b a b a+ + = − +
2 . 2 . 0a b a b+ = ⇒ ( )4 . 0a b = ⇒ . 0a b =
a b⊥ ⇒ AB AD⊥ ⇒ AB AD⊥
∴The parallelogram ABCD is a rectangle
Q-4) Show by vector method that the sum of the squares of the diagonals of a
parallelogram is equal to the sum of the squares of its sides.
Ans. Let ABCD be a parallelogram and ,AB a AD b= = .
Then AB DC a= = and AD BC b= =
Now, OC a b= + and DB a b= − .
( ) ( )2 . .OC OC OC a b a b∴ = = + + . . . .a a a b b a b b= + + + 2 22 .a a b b= + + ( . . )a b b a=
( ) ( )2 . .DB DB DB a b a b= = − − . . . .a a a b b a b b= − − + 2 22 .a a b b= − + ………. (ii)
Adding (i) and (ii),we get
2 2 2 2 2 2OC DB a b a b+ = + + + 2 2 2 2AB BC C AD= + + +
This proves the result.
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Q-5) Using vectors, prove that altitudes of a triangle are concurrent.
Ans. In ∆ABC, let the altitudes AD and BE
intersect in H.
Let the line HC meet the line AB in F.
We have to prove that CH is also an
altitude of the ∆ABC.
i.e. To prove that HC AB∴ ⊥
we take H as the origin, let , ,a b c be the p.v. of A, B, C respectively w.r.t. H
, ,HA a HB b HC c∴ = = =
Since AD is an altitude AD BC∴ ⊥
i.e. 0HA BC∴ ⊥ =
( ) 0a c b∴ − = ⇒ . . 0a c a b− = … (i)
Also BE is an altitude
BE CA∴ ⊥ ⇒ . 0HB CA = ⇒ ( ) 0b a c− =
. . 0b a b c∴ − = … (ii)
Adding (i) and (ii)
. . . . 0a c a b b a b c− + − =
. . . . 0c a b a b a b c∴ − + − =
. . 0c a c b− = ⇒ . . 0c b c a− = ⇒ ( ). 0c b a− =
HC AB∴ ⊥
Q-6) Using vectors prove perpendicular bisector of sides of a triangle are
concurrent.
In ∆ABC, let the perpendicular bisectors
of BC and AB meet at 0.
Let N be the mid-point of AC
Let , , , ,a b c l m and n be p.v. of point A,B,
C, L, M, N w.r.t. origin O and L,M,N are
midpoint of BC, AB and AC respectively
Ans. , ,
2 2 2
b c a b a cl m and n
+ + +∴ = = =
Now OL BC⊥ OL BC⊥
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( ). 0l c b∴ − = ⇒ ( ) 02
b cc b
+∴ − =
. . 0c c b b∴ − = ⇒ . .c c b b∴ = …………. (i)
Similarly
OM AB⊥ . .a a b b∴ = ………… (ii)
. . .a a b b c c∴ = =
Consider ( ). .ON AC n c a= − ( )2
a cc a
+= −
( )1. .
2c c a a= − ( )
10 0
2= =
ON AC∴ ⊥
∴The perpendicular bisectors of the sides of triangle are concurrent.
Q-7) Using vectors prove that median of triangle are concurrent
Ans. Let ABC be a triangle and ED, andF are
the mid points of the sides ,BC CA and
AB respectively. Let , , , , ,p a b c d e and
f are the position vectors of , , , ,A B C D E
and F respectively with respect
to some origin O .By mid point formula
. 2
2
b cd d b c
+= ∴ = + … (i)
22
c ae e c a
+= ∴ = + … (ii)
22
a bf f a b
+= ∴ = + … (iii)
Adding , ,a b and c on both sides of (i), (ii) and (iii) respectively, we get
2d a a b c+ = + + ⇒ 2e b a b c+ = + + ⇒ 2f c a b c+ = + +
Dividing by 3
2
3 3
d a a b c+ + += … (iv)
2
3 3
e b a b c+ + += … (v)
2
3 3
f c a b c+ + += … (vi)
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∴from equations (iv), (v) and (vi), we can write
( )2 2 2
2 1 2 1 2 1 3
d a e b f c a b cg say
+ + + + += = = =
+ + + … (vi)
Let G be a point whose p.v. is g .
Therefore by section formula, G divides AD , BE and CF internally in ratio 1:2 .
Here ,AD BE and CF are the medians of a triangle ABC and all are passing
through same point G, (which is centroid of a triangle.) Hence medians of a triangle
are concurrent.
Q-8) Prove that the bisectors of the angles of a triangle are concurrent.
Ans. Let , ,a b c be the position vectors of the
vertices , ,A B C of ABC∆ and let the
lengths of the sides ,BC CA and AB be
, ,x y z respectively. If segments
, ,AD BE CF are the bisectors of the angles , ,A B C respectively, thenD divides the
sides BC in the ratio :AB AC i.e. yz : , E divides the side AC in the ratio :BA BC
i.e. xz : and F divides the side AB in the ratio :AC BC i.e. xy : .
Hence by section formula, the position vectors of the points D, E and F are
( ) ( ) ( ); ;y z d yb zc z x e zc xa x y f xa yb∴ + = + + = + + = +
( ) ( ) ( )y z d xa z x e yb x y f zc xa yb zc∴ + + = + + = + + = + +
( )( )
( )( )
( )( )
y z d xa z x e yb x y f zc xa yb zcp
y z x z x y x y z x y z
+ + + + + + + += = = =
+ + + + + + + + …(say)
This show that the point P whose position vector is p , lies on the three bisectors
respectively.
Hence three bisector segments are concurrent in the point whose position vectors
is xa yb zc
x y z
+ +
+ +. This point of concurrence of bisectors is called Incentre of ABC∆ .
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Q-9) Using vectors, prove angle subtended on a semi circle is right angle.
Ans. Let ACB be a semi-circle whose diameter
is AB , O be the centre of the circle. We
have to prove that ACB is a right angle
i.e. BCAC ⊥ We take O as origin, let
, ,a b c be the p.v.s. of the points , ,A B C w. r. t. origin
OA OB OC∴ = = ⇒ a b c= =
∴ O is the mid point of AB
AO OB∴ = ⇒ AO OB− = ⇒ a bor a b− = = −
Now ( ) ( ). .AC BC c a c b= − − ( ) ( )c b c b= + − . . . .c c c b b c b b= − + −22c b−= −
2 2 0c c= − =
AC BC AC BC∴ ⊥ ∴ ⊥
90ACB∴∠ = �⇒ ACB∠ is right angle
GROUP (F): CLASS WORK PROBLEMS
Q-1) Find two unit vectors perpendicular to vectors 2 6 3a i j k= − + & 4 3b i j k= + −
Ans. Unit vector perpendicular to both
a ba and b
a b
×= ±
×………. (i)
2 6 3
4 3 1
i j k
a b∴ × = −
−
( ) ( ) ( )6 9 2 12 6 24i j k= − − − − + +
3 14 30a b i j k∴ × = − + + ……….. (ii)
Now, ( ) ( ) ( )2 2 2
3 14 30a b× = − + + 9 196 900 1105= + + = ……. (iii)
Put (ii) and (iii) in (i)
∴ Unit vector perpendicular to both ˆa and b n=( )3 14 30
1105
i j k− + += ±
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Q-2) If 0a b c+ + = , Prove that a b b c c a× = × = ×
Ans. 0a b c+ + =
[ ] 0a a b c a∴ × + + = ×
( ) ( ) ( ) 0a a a b a c× + × + × =
( ) ( )0 0a b a c+ × + × =
a b a c∴ × = − ×
a b c a∴ × = × …………… (i)
0b a b c b × + + = ×
( ) ( ) ( ) 0b a b b b c× + × + × =
b a b c∴ × = − ×
a b b c∴ − × = − × …………… (ii)
From equations (i) and (ii),
a b b c c a× = × = ×
Q-3) In any ABC∆ by vector method, prove that a b c
SinA SinB SinC= =
Ans. Let ∆ABC
Let �(BC) = a, �(CA) = b and �(AB) = c
0BC CA AB+ + = ⇒ 0a b c+ + =
Angle between a and b is π - C
Angle between b and c is π - A
Angle between c and a is π - B
( )sin sina b ab C ab Cπ× = − =
( )sin sinb c bc A bc Aπ× = − =
( )sin sinc a ca B ca Bπ× = − =
Now from (i)
( ) 0a a b c a× + + = ×
0a a a b a c× + × + × =
0 0a b a c+ × + × =
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a b a c c a× = − × = ×
a b c a× = ×
c a a b∴ × = × ……………… (ii)
Similarly (ii) and (iii), we get
a b b c c a× = × = ×
∴ sin sin sinab C bc A ca B= =
Dividing by abc, we get
SinC SinA SinB
c a b∴ = = ⇒
a b c
SinA SinB SinC= =
Q-4) Prove that ( )sin sin cos cos sinα β α β α β− = −
Ans. Consider a unit circle in the xoy plane with the origin as centre. ∠AOB = α, ∠AOC =
β, where A is on the x = axis.
∴A (1, 0, 0), B (cos α, sinα, 0), c(cosβ, sinβ,0)
. . cos sin 1pv of B OB i j and OBα α∴ = = + = ,
. . cos sin 1pv of c OC i j and OCβ β∴ = = + =
Consider cos sin 0
cos sin 0
i j k
OC OB β β
α α
× =
( ) ( ) ( )0 0 sin cos cos sini j k α β α β= − + − ( )sin cos cos sink α β α β= − … (i)
Also ∠COB = α - β
( )[ . .sin ]OC OB OC OB kα β∴ × = − sin( )kα β= − … (ii)
From (i) and (ii), ( )sin sin .cos cos .sinα β α β α β− = −
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Q-5) If a−
and -
b represent adjacent sides of a parallelogram prove that Area a b− −
= ×
Ans. Let � OACB be a parallelogram,
and OA a OB b= = be the vectors a long
the adjacent sides Draw seg AM ⊥ side
OB and let AOM θ∠ =
In ∆ AOM
AMsin
OAθ = sinAM OA θ=
Area of parallelogram
( ) ( )OACB base height=
( ) ( )OB OA sinθ=
sinOA OB θ=
OA OB= ×
a b= ×
Q-6) If , , ,a b c d are four distinct vectors such that a b c d× = × and a c b d× = ×
prove that a d− is parallel b c−
Ans. We know that, two non-zero vectors are parallel is their cross product is zero
vector , , ,a b c d are distinct vectors.
To show a d− and b c− are parallel vectors, we have to show
( ) ( ) 0a d b c× − × =
( ) ( ) ( ) ( )a d b c a b c d b c× − × = × × − × × (distributive law)
( ) ( ) ( ) ( )a b a c d b d c= × − × − × + ×
( ) ( ) ( ) ( )a b a c b d c d= × − × + × − × ( ) ( ) ( ) ( )c d b d b d c d= × − × + × − × 0=
( )a d∴ − is parallel to ( )b c−
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Q-7) If , , ,a b c d are three vectors such that 0a ≠ and a b a c× = × and . .a b a c= .
Show that cb =
Ans. Given, a b a c× = ×
0a b a c× − × =
( ) 0a b c∴ × − =
Given that 0a ≠ , 0b c− = or a is
Perpendicular to ( )b c− …………. (i)
Also, given
. .a b a c=
( ). 0a b c∴ − =
Given that 0a ≠ , 0b c− = or a to b c− …………….. (ii)
From (i) and (ii), Vectors a and b c− are simultaneously parallel and perpendicular,
but it is impossible
0b c− = ( )0a ≠∵
b c=
Hence proved.
Q-8) If a b c× = and c a b× = , then prove that . 1a a = .
Ans. a b c× =
and c a c b⇒ ⊥ ⊥ …………(i)
c a b× =
and b c b a⇒ ⊥ ⊥ …………(ii)
∴ from (i) and (ii)
, ,a b c are mutually perpendicular ……(iii)
a b c× =
a b c× =
( ) ( )sin90a b c=
a b (i) = c ,a a b b = =
c = ab ……………(iv)
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c a b× =
sin90c a b=
ca (1) =b ,c c a a = =
(ab) a = b [uning iv]
2 1a =
. 1a a =
Q-9) Show that ( )12 3 6
7i j k+ + , ( )
13 6 2
7i j k− + and ( )
16 2 3
7i j k+ − form a right handed
system of mutually perpendicular unit vectors.
Ans. Let ( )1
2 3 67
a i j k= + +
( )13 6 2
7b i j k= − +
( )16 2 3
7c i j k= + −
( ) ( )1 12 3 6 3 6 2
7 7a b i j k i j k× = + + × − +
1
2 3 649
3 6 2
i j k
=
−
( ) ( ) ( )1
6 36 4 18 12 949
i j k = + − − + − −
( )142 14 21
49i j k= + −
( )76 2 3
49i j k= + −
( )16 2 3
7i j k c= + − =
i.e a b c× =
, ,a b c⇒ form right handed set ….(i)
Also and c a c b⊥ ⊥ ……….(ii)
Now
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( ) ( )1 1. 2 3 6 . 3 6 2
7 7a b i j k i j k= + + − +
( ) ( ) ( )1
2 3 3 6 6 249
= + − +
[ ]1
6 18 1249
= − +
( )1
049
=
0=
a b⇒ ⊥ …….(iii)
From (i) (ii) And (iii)
, ,a b c form right handed system of mutually ⊥ or vectors ………(iv)
( ) ( ) ( )2 3 21 1
2 3 6 4 9 367 7
a = + + = + +
17
7a = × 1a∴ =
a∴ is unit vector
Similarly 1b c= = ……..(v)
, ,a b c∴ from right handed system of mutually perpendicular unit vector
GROUP-(G): CLASS WORK EXAMPLES
Q-1) Find [ ]abc if
2 3a i j k= + + , 2 4b i j k= + + , 2c i j k= + +
Ans.
2 1 3
1 2 4
1 1 2
abc =
= 2(4 – 4) – 1(2 – 4) + 3(1 – 2)
[ ] 1abc∴ = −
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Q-2) a i j pk= − + , 4b i j k= + − , 2c i j k= + + find p, if [ ] 0abc =
Ans. 0abc =
1 1
1 1 4 0
2 1 1
p−
∴ − =
∴1(1 + 4) + 1(1 +8) + p(1 – 2) = 0
14 – p = 0 ⇒∴p = 14
Q-3) Find the volume of the parallelepiped if 2a i= 3b j= and 4c k= are the
coterminus edges of the parallelepiped.
Ans. Volume of parallelepiped = [ ]abc
2 0 0
0 3 0
0 0 4
= = 2 (12) -0 + 0 = 24 cubic units.
Q-4) Show that vectors a i j k= + + , b i j k= − + & 2 3 2c i j k= + + are coplanar.
Ans. If [ ] 0abc = then vectors ,a b and c are coplanar.
L.H.S.=
1 1 1
1 1 1
2 3 2
abc = − = 1( – 2 – 3) – 1(2 – 2) + 1(3 + 2)= – 5 + 5 = 0
= R. H. S. , , ,a b c∴ are coplanar vectors.
Q-5) Find the value of λλλλ if4 3i k+ , 2i j kλ+ + . 8 2 7i j k+ + are coplanar.
Ans. Let kia 34 += ; 2b i j kλ= + + ; 8 2 7c i j k= + +
If the vectors , ,a b c are coplanar then [ ] 0abc =
4 0 3
2 1 0
8 2 7
λ∴ = ⇒ ( ) ( )4 7 2 3 4 8 0λ λ∴ − + − = ⇒ 28 8 12 24 0λ λ∴ − + − =
4 4λ∴ = − ⇒ 1λ∴ = −
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Q-6) Show that A, B, C, D are coplanar, if A ≡≡≡≡ (2, 1, -3), B ≡≡≡≡ (3, 3,, 0),
C ≡≡≡≡ (7, -1, 4) and D ≡≡≡≡ (2, -5, -7).
Ans. We know that, four points A, B, C, and D are coplanar if [ ] 0AB AC AD =
( ) ( ) ( )3 2 3 1 0 3AB b a i j k= − = − + − + +
2 3AB i j k∴ = + +
( ) ( ) ( )7 2 1 1 4 3AC c a i j k= − = − + − − + +
5 2 7AC i j k∴ = − +
( ) ( ) ( )2 2 5 1 7 3AD d a i j k= − = − + − − + − + ⇒ 6 4j k= − +
. . .L H S AB AC AD ∴ =
1 2 3
5 2 7
0 6 4
= −
− −
= 1(8 + 42) – 2(- 20) + 3 (-30)
= 50 + 40 – 90 = 0 = R. H. S.
∴ , ,AB AC AD are coplanar vectors.
∴The points A, B, C and D are coplanar.
Q-7) Find value of m if points (2, -1, 1), (4, 0, 3), (m, 1, 1), (2, 4, 3) are coplanar.
Ans. Let A ≡ (2, -1, 1), B ≡ (4, 0, 3), C ≡ (m, 1, 1) and D ≡ (2, 4, 3)
Given that the points A, B, C, D are coplanar 0AB AC AD ∴ = ……… (i)
( ) ( ) ( )4 2 1 1 3 1AB b a i j k= − = − + + + −
2 2AB i j k= + +
( ) ( ) ( )2 1 1 1 1AC c a m i j k= − = − + + + −
( )2 2AC m i j= − +
( ) ( ) ( )2 2 4 1 3 1AD d a i j k= − = − + + + − 5 2j k= +
0AB AC AD ∴ =
2 1 2
2 2 0 0
0 5 2
m − =
2(4) – 1(2m – 4) + 2(5m – 10) = 0⇒ ∴8 – 2m + 4 + 10m – 20 = 0
∴8m = 8⇒ ∴m = 1
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Q-8) If O is the origin, A (1,2,3) B(2,3,4) and P(x,y,z) are coplanar points prove that x
2 0x y z− + − = , using vector method.
Ans. The points O, A, B and P are coplanar. (given)
[ ] 0OA OB OP∴ =
2 3OA a i j k∴ = = + +
2 3 4OB b i j k= = + +
OP p xi y j zk= = + +
[ ] 0OA OB OC∴ =
1 2 3
2 3 4 0
x y z
∴ =
( ) ( ) ( )1 3 4 2 2 4 3 2 3 0z y z x y x∴ − − − + − =
3 4 4 8 6 9 0z y z x y x∴ − − + + − =
2 0x y z∴ − + − =
2 0x y z∴ − − =
Q-9) If A ≡≡≡≡ (1, 1, 1), B ≡≡≡≡ (2, -1, 3), C ≡≡≡≡ (3, -2, -2) and D ≡≡≡≡ (3, 3, 4), find the volume of
parallelepiped with segments AB, AC and AD as concurrent edges.
Ans. We know that, the volume of parallelepiped with concurrent edges AB, AC
and AD = , ,AB AC AD
( ) ( ) ( )2 1 1 1 3 1AB b a i j k= − = − + − − + −
2 2AB i j k∴ = − +
( ) ( ) ( )3 1 2 1 2 1AC c a i j k= − = − + − − + − −
2 3 3AC i j k∴ = − −
( ) ( ) ( )3 1 3 1 4 1AD d a i j k= − = − + − + −
2 2 3AD i j k∴ = + +
∴volume of parallelepiped
1 2 2
2 3 3
2 2 3
−
= − −
= 1 (-9 + 6) + 2 (6 + 6) + 2 (4 + 6)
= -3 + 24 + 20 = 41 cubic units.
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Q-10) Find the volume of a tetrahedron whose vertices are A ≡≡≡≡ (3,7,4), B ≡≡≡≡ (5,-2, 3),
C ≡≡≡≡ (-4,5,6) and D ≡≡≡≡ (1,2,3)
Ans. We know that, the volume of tetrahedron ABCD1
6AB AC AD =
( ) ( ) ( )5 3 2 7 3 4AB b a i j k= − = − + − − + −
2 9AB i j k∴ = − −
( ) ( ) ( )4 3 5 7 6 4AC c a i j k= − = − − + − + − 7 2 2i j k= − − +
( ) ( ) ( )1 3 2 7 3 4AD d a i j k= − = − + − + − 2 5i j k= − − −
∴Volume of tetrahedron1
6AB AC AD =
2 9 11
7 2 26
2 5 1
− − = − − − − −
( ) ( ) ( )1
2 2 10 9 7 4 1 35 46 = + + + − −
[ ]1 1 4624 24 99 31 92
6 6 3= + − = = cubic units.
Q-11) Prove that [ ] 2[ ]a b b c c a abc+ + + =
Ans. L.H.S = [ ]a b b c c a+ + +
( ) ( ) ( )[ ]a b b c c a= + + × +
( ) [ ]a b b c b a c c c a= + × + × + × + ×
. [ ] . [ ], [ 0]a b c b a c a b b c b a c a c c= × + × + × + × + × + × × =∵
( ) ( ) ( ) ( ) ( ) ( ). . . . . .a b c a b a a c a b b c b b a b c a= × + × + × + × + × + ×
( ) ( ). .a b c b c a= × + ×
[ ] [ ]a b c b c a= +
[ ] [ ]a b c a b c= +
2[ ] . . .a b c R H S= =
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Q-12) If 2 5c a b= + , show that [ ] 0a b c = if the angle between a and b is 3
π,
Also show that 2 2 24 10 25c a ab b= + + , where a a= and b b= .
Ans. . . [ ]L H S a b c= = [ 2 5 ] 2 5a b a b c a b+ = =∵
[ 2 ] [ 5 ]ab a ab b= + 2[ ] 5[ ]ab a ab b= + = 2.0 + 5.0 = 0 = R.H.S.
Now, 2 5c a b= +
( ) ( ). 2 5 . 2 5c c a b a b∴ = + +
2 4 . 10 . 10 . 25 .c a a a b b a b b= + + + ( )224 20 . 25 . .a a b b a b b a= + + =∵
( )2 2 24 20 cos 25 ,3 3
c a ab bπ πθ= + + =
2 2 214 20 25
2c a ab b∴ = + + 2 24 10 25a ab b= + +
Q-13) Show that the statement ( ).[( ) ( )] 2 .a b b c c a a b c= − − × − = × is true only when
,a b and c are coplanar.
Ans. If ,a b , c are coplanar then [ ] 0a b c =
i.e. ( ). 0a b c× =
( ) ( ) ( ). . . .[ ]L H S a b b c c a= − − × −
( ).[ ]a b b c b a c c c a= − × − × − × + ×
( ).[ ]a b b c a b c a= − × + × + ×
.[ ] .[ ]a b c a b c a b b c a b c a= × + × + × − × + × + ×
( ) ( ) ( ) ( ) ( ) ( ). . . . . .a b c a a b a c a b b c b a b b c a= × + × + × − × + × + ×
( ) ( ). .a b c b c a= × − × [ ] [ ]a b c b c a= − [ ] [ ]a b c a b c= − = 0.
2 .a b c× 2[ ]a b c= = 2.0 = 0 [ , , ]a b c are coplanar∵ .
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Q-14) If 2 3a i j k= − − , 2b i j k= − + and 2 3c i j k= + + . Find [ ]a b c . State with
reasons whether [ ]a b c is a right handed or left handed triplet.
Ans. 2 1 3
2 1 1
1 2 3
a b c
− −
= −
=2(-3 – 2) + 196 – 1) – 3(4 + 1)
=2(-5) + (5) – 3(5)
= -10 + 5 – 15 = -20
Since [ ]a b c is negative
∴a b c is left handed triplet.
Q-15) If 2u i j k= − − + , 3v i k= + and w j k= − then find the value of
( ) ( ) ( ).u w u v v w + × × ×
.
Ans. let ( ) ( )2u w i j k j k i j+ = − − + + − = − − ………… (i)
Let 1 2 1
3 0 1
i j k
b u v= × = − − = 2 4 6i j k− + + …………. (ii)
Let 3 0 1
0 1 1
i j k
c v w= × =
−
3 3i j k= − + + …………. (iii)
Now ( ) ( ) ( ).[ ]u w u v v w+ × × × ( )1 1 0
2 4 6
1 3 3
a b c a bc
− −
= × = = − −
= (-1) (12 – 18) + 1 (-6 + 6) + 0 (-6 + 4) = (-1) (-6) + 0 + 0 = 6
( ) ( ) ( ). 6u w u v v w ∴ + × × × =
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Q-16) If a b c are three non coplanar vectors then find the value of
. .
. .
a b c a b c
c a b c a b
× ×+
× ×
Ans. . . [ ] [ ]
. . [ ] [ ]
a b c a b c abc bac
c a b c a b c ab c ab
× ×+ = +
× ×
[ ] [ ]
[ ] [ ]
abc abc
c ab c ab
−= + ⇒
[ ] [ ]1 1 0
[ ] [ ]
abc abc
abc abc= − = − =
Q-17) a b c are three non-coplanar vectors. If , ,b c c a a b
p q rabc abc abc
× × ×= = =
then prove that
(i) . . . 3a p b q c r+ + =
(ii) ( ) ( ) ( ). . . 3a b p b c q c a r+ + + + + =
Ans. i) L.H.S. . . .a p b q c r= + +
( ) ( ) ( ). . .
[ ] [ ] [ ]
a b c b c a c a b
abc abc abc
× × ×= + +
[ ] [ ] [ ]
[ ]
abc bc a c ab
abc
+ +=
[ ] [ ] [ ] 3[ ]3
[ ] [ ]
abc abc abc abc
abc abc
+ += = =
ii) L.H.S. = ( ) ( ) ( ). . .a b p b c q c a r+ + + + +
( )( ) ( ) ( )
( )( ).
[ ] [ ] [ ]
b c b c c a a ba b c a
abc abc abc
× + × ×= + + + +
Since abc is a scalar. Let abc
= k
( ) ( )( ) 1
. .b c
a b p a b abc abck k
+ + = + = + { }1
0a b ck
= + 1 1
0 1a b c kk k = = × =
Similarly ( ). 1b c q+ =
And ( ). 1c a r+ =
( ) ( ) ( ). . . 1 1 1 3a b p b c q c a r+ + + + + = + + =
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Q-18) Prove that [ ] 0a b b c c a− − − =
Ans. L.H.S.= ( ) ( ) ( ).a b b c c a − − × −
( ) [ ]a b b c b a c c c a= − × − × − × + ×
( ) 0a b b c b a c a = − × − × − + ×
( ) ( ) ( ). . 0 0 0 0 .a b c a b a b c a× − × + − + − × [ ] [ ]a b c b c a= −
[ ] [ ]a b c a b c= − = 0.
Q-19) If the vectors a i + j + k, I + b j + k, I + j + ck (a ≠≠≠≠ b ≠≠≠≠ c ≠≠≠≠1) are coplanar then
show that 1 1 1
1.1 1 1a b c
+ + =− − −
Ans. Let p ai j k= + + , q i bj k= + + , r i j ck= + + and , ,p q r are coplanar.
[ ] 0pqr∴ =
1 1
1 1 0
1 1
a
pqr b
c
∴ = =
Applying C1 – C2 and C2 – C3
1 0 1
0 1 1 0
1 1
a
b
c c c
−
− =
− −
∴ by taking (a – 1), (b – 1), (1 – c) common
( ) ( ) ( )
11 0
1
11 1 1 0 1 0
1
1 11
a
a b cb
c
c
−
∴ − − − = =−
−
⇒ ∴a ≠ b ≠ c ≠1
∴(a -1) (b – 1) (c – 1) ≠ 0⇒
11 0
1
10 1 0
1
1 11
a
b
c
c
−
=−
−
( )1 1
1 01 1 1
c
c b a
− + − =
− − −
1 10
1 1 1
c
c b a− − =
− − − ⇒
1 10
1 1 1
c
c b a+ + =
− − −
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1 11 1
1 1 1
c
c b a
+ + + =
− − − ⇒
1 1 11
1 1 1
c c
c b a
− ++ + =
− − −
1 11
1 1 1
c
c b a+ + =
− − −
Q-20) If ( ) ( ) ( )d a b u b c c aλ δ= × + × + × . Find ( ).d a b c+ + Also if 1
8abc = .
Show that λλλλ + u + d = ( )8 .d a b c + +
Ans. ( ) ( ) ( ) ( ) ( ). .d a b c a b u b c c a a b cλ δ + + = × + × + × + +
[ . . . ] [ . . . ] [ . . . ]a b a a b b a b c b c a b c b b c c c a a c a b c a cλ µ δ= × + × + × + × + × + × + × + × + ×
0 0 . . 0 0 0 . 0a b c u b c a c a bλ δ = + + × + × + + + + × +
abc u bca cabλ δ = + + ( ) abcλ µ δ = + +
If 1
8abc = ⇒ i.e. ( ) ( )
1.
8d a b c uλ δ+ + = + + = , ∴λ + u + d = ( )8 .d a b c + +
Q-21) Show that if four points A, B, C D with position vectors , ,a b c ,d are
coplanar then bcd cad abd abc + + =
Ans. ∵four points A, B, C, D are coplanar
AB AC AD∴ are also coplanar
0AB AC AD ∴ =
( ). 0AB AC AD∴ × =
( ) ( ) ( ). 0b a c a d a − − × − =
( ). 0b a c d c a a d a a − × − × − × + × =
( ) ( ) ( ) ( ) ( ) ( ). . . . . . 0b c d b c a b a d a c d a c a a a d× − × − × − × + × + × = 0a a × =
0bc d bc a bad ac d ac a aad − − − + + =
0 0 0bc d abc abd c ad − − − + + =
bc d abd cad abc + + =
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Q-22) If 3c a b= + , show that, 0abc = Also, if 2 2 23 3c a ab ab= + + , prove
that the angle between a and b is 6
π .
Ans. 3c a b= +
i.e c is linear combination of and a b
, ,a b c∴ are coplanar
0abc ∴ =
3c a b= +
3c a b∴ = +
Squaring both sides.
2 23c a b= +
( ) ( )2
3 . 3c a b a b= + +
2. 3 . 3 . .c a a a b b a ab b= + + +
2 2 26 . 9c a a b b= + +
( )22 2
6 6 cos 96
c a a bπ
= + +
2 2 236 . 9
2c a a b b= + +
2 2 236 . 9
2c a ab b= + +
2 2 23 3 9c a ab b= + +