AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

Given,
l+m+n=0 …… (I)
2lm+2ln−mn=0 …… (II)
From equation (I) and (II),
2l(−l)−mn=0
2l2+mn=0 .......(III)
Square both sides of equation (I),
l2+m2+n2+2lm+2mn+2nl=0
As l2+m2+n2=1,
So, 2lm+2mn+2nl=−1 .......(IV)
From equation (II) and (IV),
mn=−

1

3

From equation (III),
6l2−1=0
So, l1l2=−

1

6

From equation (I) and (II),
2lm+(2l−m)(−l−m)=0
2l2−lm−m2=0
2(

l

m

)2−(

l

m

)−1=0

l1l2

m1m2

=−

1

2

l1l2

(− 1 6 )

=

m1m2

( 2 6 )

Similarly,

l1l2

(− 1 6 )

=

n1n2

( 2 6 )

So,
cos‌θ|l1l2+m1m2+n1n2|
cos‌θ=

3

6

cos‌θ=

1

2

θ=

π

3