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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let the direction cosines of two lines satisfy the equations:
$4\ell + m - n = 0$ and $2mn + 10n\ell + 3\ell m = 0$.
Then the cosine of the acute angle between these lines is:
Solution
Direction cosines of two lines satisfy the equation
$4\ell + m - n = 0 \quad ...(1)$
$2mn + 10n\ell + 3\ell m = 0 \quad ...(2)$
And we know
$\ell^2 + m^2 + n^2 = 1 \quad ...(3)$
From (1):
$n = 4\ell + m$
Putting in (2):
$(4\ell + m)(2m + 10\ell) + 3\ell m = 0$
$8\ell m + 40\ell^2 + 2m^2 + 10\ell m + 3\ell m = 0$
$40\ell^2 + 21\ell m + 2m^2 = 0$
$(8\ell + m)(5\ell + 2m) = 0$
Case 1: $8\ell + m = 0 \Rightarrow m = -8\ell$
Case 2: $5\ell + 2m = 0 \Rightarrow m = -\dfrac{5}{2}\ell$
So direction ratios
$L_1 = (\ell, -8\ell, -4\ell)$
$L_2 = \left(\ell, -\dfrac{5}{2}\ell, \dfrac{3}{2}\ell\right)$
Cosine of angle
$\cos\theta = \dfrac{\ell^2 + 20\ell^2 - 6\ell^2}{\sqrt{\ell^2 + 64\ell^2 + 16\ell^2};\sqrt{\ell^2 + \dfrac{25}{4}\ell^2 + \dfrac{9}{4}\ell^2}}$
$= \dfrac{15\ell^2}{\sqrt{81\ell^2};\sqrt{\dfrac{38}{4}\ell^2}}$
$= \dfrac{15\ell^2}{(9\ell)\left(\dfrac{\sqrt{38}}{2}\ell\right)}$
$= \dfrac{10}{3\sqrt{38}}$
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