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VITEEE 2009 Paper

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Question : 95 of 120

Marks: +1, -0

The angle between the lines whose direction cosines satisfy the equations

l

l^{2} + m^{2} - n^{2} = 0

is

$\frac{\pi}{6}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$

Solution:

l

+ m + n = 0, ⇒

l

= - m - n

and

l^{2} + m^{2} - n^{2} = 0

∴

(-m-n)^{2}+m^{2}-n^{2}=0\Rightarrow 2m^{2}+2mn=0

⇒ 2m(m + n) = 0 ⇒ m = 0 or m + n = 0

If m = 0, then

l

= -n

∴

\frac{l_{1}}{-1}=\frac{m_{1}}{0}=\frac{n}{1}

and if m + n = 0 ⇒ m = -n, then

l=0

∴

\frac{l_{2}}{0}=\frac{m_{2}}{-1}=\frac{n_{2}}{1}

i.e.,

(l_{1},m_{1},n_{1})=(-1,0,1)

and

(l_{2},m_{2},n_{2})=(0,-1,1)

∴

\cos\theta=\frac{0+0+1}{\sqrt{1+0+1}\sqrt{0+1+1}}=\frac{1}{2}\Rightarrow \theta=\frac{\pi}{3}

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