TS EAMCET 5 May 2018 Shift 2 Solved Paper

The length of the projections of AB on the coordinate planes are d1,d2 and d3
Now,
AB=d
(cos‌α‌i+cos‌β‌j+cos‌γ‌k) =(d‌cos‌α‌i+d‌cos‌β‌j+d‌cos‌γ‌k)
Let θ be the angle between AB and the XY plane.
Then,
d=d‌cos‌θ
Now,
sin‌θ=

AB.k

|AB||k|

=

d‌cos‌γ

d.1

=cos‌γ
=sin(

π

2

−γ)
Hence,
θ=

π

2

−γ
Therefore,
d2=d‌sin‌α
d3=d‌sin‌β
Now,
cos2α+cos2β+cos2γ=1
1−sin2α+1−sin2β+1 −sin2γ=1
sin2α+sin2β+sin2γ=2
(

d2

d

)2+(

d3

d

)2+(

d1

d

)2=2
Therefore,
d12+d22+d32=2d2