The projection of a line segment $OP$ through orig
Question:

The projection of a line segment OPOP through origin OO, on the coordinate axes are 8,5,68, 5, 6. Then, the length of the line segment OPOP is equal to

Updated On: Jun 14, 2022
  • 5 5
  • 55 5\sqrt{5}
  • 105 10\sqrt{5}
  • None of these
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The Correct Option is B

Solution and Explanation

Let l, l, m m and n n be the direction cosine's of the given line segment P
\therefore l=cosα,   m=cosβ,  n=cosγ l=\cos \,\alpha ,\,\,\,m=\cos \beta ,\,\,n=\cos \,\gamma
where α,β,γ \alpha ,\beta ,\gamma
are the angles which the line segment PQ makes with the axes. Suppose length of line segment
PQ=r PQ=r
This, projection of line segment PQ on x-axis
=PQ  cosα=rl =PQ\,\,\cos \,\alpha =rl
Also, the projection of line segment PQ on x-axis
=8 =8
\therefore lr=8 lr=8
Similarly mr=5,  nr=6 mr=5,\,\,nr=6
Now, on squaring adding these equations, we get
(lr)2+(mr)2+(nr)2=82+52+62 {{(lr)}^{2}}+{{(mr)}^{2}}+{{(nr)}^{2}}={{8}^{2}}+{{5}^{2}}+{{6}^{2}}
r2(l2+m2+n2)=64+25+36 {{r}^{2}}({{l}^{2}}+{{m}^{2}}+{{n}^{2}})=64+25+36
(    l2+m2+n2=1) (\because \,\,\,\,{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1)
\Rightarrow r2=125 {{r}^{2}}=125
\Rightarrow r=53 r=5\sqrt{3}
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