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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2024 · Mathematics · Three Dimensional Geometry

If a direction cosines of a line are connected by relation l+2...

If a direction cosines <l,m,n>< l, m, n > of a line are connected by relation l+2m+n=0,2l−2m+3n=0l+2 m+n=0,2 l-2 m+3 n=0, then what is the value of l2+m2−n2l^2+m^2-n^2 ?

Options

  1. A1101\frac{1}{101}
  2. B29101\frac{29}{101}
  3. C41101\frac{41}{101}
  4. D92101\frac{92}{101}

Correct answer

B. 29101\frac{29}{101}

Step-by-step solution

l+2m+n=0l+2 m+n=0 ....(i)and2l−2m+3n=0and \quad 2 l-2 m+3 n=0 ....(ii) From (i) we have l=−2m−nl=-2 m-n ....(iii) Putting the value of ll in (ii) 2(−2m−n)−2m+3n=0∴−4m−2n−2m+3n=0∴n=6m Now, l=−2m−n=−2m−6m=−8ml−8m=mn=n6m∴l−8=m1=n6=l2+m2+n2(−8)2+(1)2+(6)2=1101l=−8101,m=1101n=6101\begin{aligned} & 2(-2 m-n)-2 m+3 n=0 \\ & \therefore-4 m-2 n-2 m+3 n=0 \\ & \therefore \quad n=6 m \\ & \text{ Now, } \\ & l=-2 m-n \\ & =-2 m-6 m=-8 m \\ & \frac{l}{-8 m}=\frac{m}{n}=\frac{n}{6 m} \\ & \therefore \quad \frac{l}{-8}=\frac{m}{1}=\frac{n}{6} \\ & =\sqrt{\frac{l^2+m^2+n^2}{(-8)^2+(1)^2+(6)^2}}=\frac{1}{\sqrt{101}} \\ & l=\frac{-8}{\sqrt{101}}, m=\frac{1}{\sqrt{101}} \\ & n=\frac{6}{\sqrt{101}} \end{aligned}\thereforeDirectionconsinesofalineDirection consines of a line\begin{aligned} & \left(\frac{-8}{\sqrt{101}}, \frac{1}{\sqrt{101}}, \frac{6}{\sqrt{101}}\right) \\ & \begin{aligned} \therefore \quad l^2+m^2-n^2 & =\frac{64}{101}+\frac{1}{101}-\frac{36}{101} \\ & =\frac{29}{101} \end{aligned}\end{aligned}
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