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 ∴ = + - + . . . . aa ab ba bb =- + - + 2 2 a b =- + 2 2 0 b 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 . 0 AC BD ∴ = ⇒ ( ) ( ) 0 a b a b + - + = . . . . 0 aa ab ba bb - + - + = 2 2 b a = ⇒ a b = ⇒ ( ) ( ) ( ) ( ) l AB l BC l CD l AD ∴ = = = ∴ the parallelogram ABCD is a rhombus.
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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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( ). 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= + −