Find the angle between the lines l_1 :y − x −4 = 0 and l_2 :2x − y = 7 and hence the perpendicular distance from one point on l_1 to l


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Question Number 36290 by Rio Mike last updated on 31/May/18

$Find the angle between the lines$ $l_{1} : y - x - 4 = 0 and l_{2} : 2 x - y = 7$ $and hence the perpendicular$ $distance from one point on l_{1}$ $to l_{2} .$
Commented by prof Abdo imad last updated on 01/Jun/18

$w e h a v e L_{1} : x - y + 4 = 0 s o u (1, 1) i s a v e c t o r$ $d i r e c t o r o f L_{1}$ $L_{2} : 2 x - y - 7 = 0 s o v (1, 2) i s a v e c t o r d i r e c t o r o f$ $L_{2} \Rightarrow (L_{1}, L_{2}) \equiv (u, v) [2 π]$ $c o s (u, v) = \frac{u . v}{∣∣ u ∣∣ . ∣∣ v ∣∣} = \frac{1 \times 1 + (1 \times 2)}{\sqrt{2} \sqrt{5}} = \frac{3}{\sqrt{2} \sqrt{5}} = \frac{3}{\sqrt{10}}$ $s i n (u, v) = \frac{d e t (u, v)}{∣∣ u ∣∣ . ∣∣ v ∣∣} = \frac{∣_{1 2}^{1 1} ∣}{\sqrt{10}} = \frac{1}{\sqrt{10}} \Rightarrow$ $t a n (u, v) = \frac{1}{3} \Rightarrow θ = (u, v) = a r c t a n (\frac{1}{3}) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 31/May/18

	$y = x + 4$ $m_{1} = t a n α = 1$ $y = 2 x - 7$ $m_{2} = t a n β = 2$ $l e t a n g l e b e t w e e n t w o l i n e i s θ$ $θ = β - α$ $t a n θ = t a n (β - α)$ $t a n θ = \frac{t a n β - t a n α}{1 + t a n β . t a n α}$ $t a n θ = \frac{2 - 1}{1 + 2.1}$ $t a n θ = \frac{1}{3}$ $θ = {t a n}^{- 1} (\frac{1}{3})$ $θ = 18^{0} 20^{'} a c u t e a n g l e$ $(0, 4) i s a p o i n t l i e s o n s t l i n e y = x + 4$ $p e r p e n d i c u l a r d i s t a n c e f r o m (0, 4) 2 x - y - 7 = 0$ $i s =∣ \frac{2 \times 0 - 4 - 7}{\sqrt{2^{2} + {(- 1)}^{2}}} ∣$ $= \frac{11}{\sqrt{5}}$
	Answered by MJS last updated on 31/May/18

	$l_{1} : - x + y - 4 = 0 \Rightarrow n_{1} = (\begin{matrix} - 1 \\ 1 \end{matrix})$ $l_{2} : 2 x - y - 7 = 0 \Rightarrow n_{2} = (\begin{matrix} 2 \\ - 1 \end{matrix})$ $\cos α = \frac{∣ n_{1} * n_{2} ∣}{∣ n_{1} ∣ \times ∣ n_{2} ∣} = \frac{∣ (- 1) \times 2 + 1 \times (-) 1 ∣}{\sqrt{{(- 1)}^{2} + 1^{2}} \times \sqrt{2^{2} + {(- 1)}^{2}}} =$ $= \frac{3 \sqrt{10}}{10}$ $α \approx 18.43 °$
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