vector tutorials science 2 - ncert board level

Mahesh Tutorials Science

Vectors

137

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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140

( ). 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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141

∴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− = −


Vectors

145

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−


Vectors

146

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


Vectors

148

( ) ( )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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149

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∴ = −


Vectors

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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.


Vectors

152

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= =


Vectors

153

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∵ .


Vectors

154

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 ∴ + × × × =


Vectors

155

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+ + + + + = + + =


Vectors

156

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+ + =

− − −


Vectors

157

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 + + =


Vectors

158

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= + +

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