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STD 12 - 11. three dimension geometry — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 11. three dimension geometry4 Marks
MCQ
An angle between the lines whose direction cosines are given by the equations, $l+ 3m + 5n\, = 0$ and $5lm -2mn + 6nl = 0$ , is
  • A
    ${\cos ^{ - 1}}\left( {\frac{1}{8}} \right)$
  • ✓
    ${\cos ^{ - 1}}\left( {\frac{1}{6}} \right)$
  • C
    ${\cos ^{ - 1}}\left( {\frac{1}{3}} \right)$
  • D
    ${\cos ^{ - 1}}\left( {\frac{1}{4}} \right)$
✓

Answer

Correct option: B.
${\cos ^{ - 1}}\left( {\frac{1}{6}} \right)$
b
Given

$l+3 m+5 n=0$      ....$(1)$

and $5 l m-2 m n+6 n l=0$      .....$(2)$

From eq. $( 1 )$ we have $l=-3 m-5 n$

Put the value of $l$ in eq. $(2),$ we get;

$5(-3 m-5 n) m-2 m n+6 n(-3 m-5 n)=0$

$\Rightarrow 15 m^{2}+45 m n+30 n^{2}=0$

$\Rightarrow m^{2}+3 m n+2 n^{2}=0$

$\Rightarrow m^{2}+2 m n+m n+2 n^{2}=0$

$\Rightarrow(m+n)(m+2 n)=0$

$\therefore m=-n$ or $m=-2 n$

For $m=-n, l=-2 n$

And for $m=-2 n, l=n$

$\therefore(l, m, n)=(-2 n,-n, n)$ Or $(l, m, n)$

$=(n,-2 n, n)$

$\Rightarrow(l, m, n)=(-2,-1,1)$ Or $(l, m, n)$

$=(1,-2,1)$

Therefore, angle between the lines is given as:

$\cos (\theta)=\frac{(-2)(1)+(-1) \cdot(-2)+(1)(1)}{\sqrt{6} \cdot \sqrt{6}}$

$\Rightarrow \cos (\theta)=\frac{1}{6} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{6}\right)$

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