SSC CGL Tier II 1 Dec 2016 Paper 1

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Question : 72 of 100

Marks: +1, -0

If

a^{2}+b^{2}+c^{2}=16, x^{2}+y^{2}+z^{2} =25

and

a x+b y+c z=20,

then the value of

{a+b+c}/{x+y+z}

is

$3/5$
$5/ 3$
$4 /5$
$5/ 4$

Solution:

(3)

(a x+b y+c z)^{2}

= (a^{2}+b^{2}+c^{2} ) (x^{2}+y^{2}+z^{2} )

=400

⇒ a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}+2 a b x y

+2 b c y z+2 a c x z

=a^{2} x^{2}+a^{2} y^{2}+a^{2} z^{2}+b^{2} x^{2}+

b^{2} y^{2}+b^{2} z^{2}+c^{2} x^{2}+c^{2} y^{2}+c^{2} z^{2}

⇒ a^{2} y^{2}+a^{2} z^{2}+b^{2} x^{2}+b^{2} z^{2}+

c^{2} x^{2}+c^{2} y^{2}

=2 a b x y+2 b c y z+2 a c x z

⇒ a^{2} y^{2}-2 a b x y+b^{2} x^{2}+a^{2} z^{2}+

c^{2} x^{2}-2 a c x z+b^{2} z^{2}+c^{2} y^{2}+

2 b c y z=0

⇒(a y-b x)^{2}+(a z-c x)^{2}+(b z-c y)^{2}

=0

⇒ a y-b x=0⇒ a y=b x⇒ {a}/{b}= {x}/{y}

a z-c x=0 ⇒ a z=c x ⇒{a}/{c}= {x}/{z}

∴ a=k x ; b=k y ; c=k z

∴ a^{2}+b^{2}+c^{2}=16

⇒ k^{2} (x^{2}+y^{2}+z^{2} )=16

⇒ k^{2} × 25=16

⇒ k^{2}= {16}/{25} ⇒ k= {4}/{5}

∴{a+b+c}/{x+y+z}=k= {4}/{5}

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