Question

$\huge \underline \mathfrak \pink{Hey :) \: Questions \: }$



ASAP !SOLVE THIS PLOX :(​

Answer 1

Solution: Given that the direction cosines of the lines are given by the equations

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

and

From (1), we have n= -l-m

Putting value of n in (2), we get

$⇒ {l}^{2} + {m}^{2} - ( - m) {}^{2} = 0$

$⇒ - 2lm = 0$

$⇒l = 0 \: or \: m = 0$

$When \: l=0, we \: have n = -m$

$When \: m =0, we \: have n= -l$

∴Direction cosines of the first line are '0,m,-m' and direction cosines of the second line are ':.' .Then, direction ratios of the first line are 0,1,-1and direction ratios of the second line are 1,0,-1

$Let \: a _{1} = 0, \: b _{1} \: 1,c _{1} = - 1 \: and \: a _{2} = 1, b _{2} = 0, c _{2} = - 1$

$Then \: \cos( \alpha ) = | \frac{ a_1 a_2 + b_1b_2 + c_1c_2 }{ \sqrt{a\frac{2}{1} + b \frac{2}{1}c \frac{2}{1} } \sqrt{a \frac{2}{2} + b \frac{2}{2} + c \times \frac{2}{2} } }|$

$= | \frac{(0)(1)+(1)(0)+(-1)(-1)}{ \sqrt{(0) {}^{2} + (1) {}^{2} + ( - 1) {}^{2} } \sqrt{(1) {}^{2} + (0) {}^{2} + ( - 1) {}^{2} } } | = \frac{1}{( \sqrt{2})( \sqrt{2)} } = \frac{1}{2} .$

Hence α=60° , which is the required angle.

Answer 2

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Hence α=60° , which is the required angle.