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QJEEM14Ma8d56: JEE Main 2014 Mathematics - Class 12 JEEM14Ma8d56 Solution: The angle between the lines whose direction cosines satisfy the …

The angle between the lines whose direction cosines satisfy the equations l+m+n=0l+m+n=0 and l2=m2+n2{l^2} = {m^2} + {n^2} is :

Multiple Choice Options

  • A π6{\pi \over 6}
  • B π2{\pi \over 2}
  • C π3{\pi \over 3}
  • D π4{\pi \over 4}

Step-by-Step Verified Solution

Given

l+m+n=0l + m + n = 0 and l2=m2+n2{l^2} = {m^2} + {n^2}

Now, (−m−n)2=m2+n2{\left( { - m - n} \right)^2} = {m^2} + {n^2}

⇒mn=0⇒m=0  \Rightarrow mn = 0 \Rightarrow m = 0\,\, or   n=0\,\,n = 0

If m=0m=0 then l=−nl=-n

We know

l2+m2+n2=1⇒n=±12{l^2} + {m^2} + {n^2} = 1 \Rightarrow n = \pm {1 \over {\sqrt 2 }}

i.e. (l1,m1,n1)=(−12,0,12)\left( {{l_1},{m_1},{n_1}} \right) = \left( { - {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }}} \right)

If n=0n=0 then l=−ml=-m

l2+m2+n2=1   ⇒2m2=1{l^2} + {m^2} + {n^2} = 1\,\,\, \Rightarrow 2{m^2} = 1

⇒m=±12\Rightarrow m = \pm {1 \over {\sqrt 2 }}

Let m=12⇒l=−12m = {1 \over {\sqrt 2 }} \Rightarrow l = - {1 \over {\sqrt 2 }}

and n=0n=0

(l2,m2,n2)=(−12,12,0)\left( {{l_2},{m_2},{n_2}} \right) = \left( { - {1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }},0} \right)

∴\therefore cos⁡θ=12⇒θ=π3\cos \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 3}

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