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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 :2x − y = 7 and hence the perpendicular distance from one point on l_1 to l_2 .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{lines} \\ $$$$\:\:\:{l}_{\mathrm{1}} :\mathrm{y}\:−\:\mathrm{x}\:−\mathrm{4}\:=\:\mathrm{0}\:\mathrm{and}\:{l}_{\mathrm{2}} :\mathrm{2}{x}\:−\:{y}\:=\:\mathrm{7}\: \\ $$$$\mathrm{and}\:\mathrm{hence}\:\mathrm{the}\:\mathrm{perpendicular}\: \\ $$$$\mathrm{distance}\:\mathrm{from}\:\mathrm{one}\:\mathrm{point}\:\mathrm{on}\:{l}_{\mathrm{1}} \\ $$$$\mathrm{to}\:{l}_{\mathrm{2}} . \\ $$$$ \\ $$

Commented by prof Abdo imad last updated on 01/Jun/18

we have L_1 : x−y+4=0 so u(1,1) is a vector director of L_1 L_2 : 2x−y −7=0 so v( 1,2) is a vector director of L_2 ⇒ (L_1 ,L_2 ) ≡ (u ,v)[2π] cos(u,v) = ((u.v)/(∣∣u∣∣.∣∣v∣∣)) = ((1×1 +(1×2))/((√2) (√5))) =(3/((√2)(√5)))=(3/(√(10))) sin(u,v) =((det(u,v))/(∣∣u∣∣.∣∣v∣∣)) = ((∣_(1 2) ^(1 1) ∣)/(√(10))) = (1/(√(10))) ⇒ tan(u,v) = (1/3) ⇒ θ=(u,v) =arctan((1/3)).

$${we}\:{have}\:{L}_{\mathrm{1}} \::\:\:{x}−{y}+\mathrm{4}=\mathrm{0}\:{so}\:{u}\left(\mathrm{1},\mathrm{1}\right)\:{is}\:{a}\:{vector} \\ $$$${director}\:{of}\:{L}_{\mathrm{1}} \\ $$$${L}_{\mathrm{2}} :\:\mathrm{2}{x}−{y}\:−\mathrm{7}=\mathrm{0}\:{so}\:{v}\left(\:\mathrm{1},\mathrm{2}\right)\:{is}\:{a}\:{vector}\:{director}\:{of} \\ $$$${L}_{\mathrm{2}} \:\Rightarrow\:\left({L}_{\mathrm{1}} ,{L}_{\mathrm{2}} \right)\:\equiv\:\left({u}\:,{v}\right)\left[\mathrm{2}\pi\right] \\ $$$${cos}\left({u},{v}\right)\:=\:\frac{{u}.{v}}{\mid\mid{u}\mid\mid.\mid\mid{v}\mid\mid}\:=\:\:\frac{\mathrm{1}×\mathrm{1}\:+\left(\mathrm{1}×\mathrm{2}\right)}{\sqrt{\mathrm{2}}\:\sqrt{\mathrm{5}}}\:=\frac{\mathrm{3}}{\sqrt{\mathrm{2}}\sqrt{\mathrm{5}}}=\frac{\mathrm{3}}{\sqrt{\mathrm{10}}} \\ $$$${sin}\left({u},{v}\right)\:=\frac{{det}\left({u},{v}\right)}{\mid\mid{u}\mid\mid.\mid\mid{v}\mid\mid}\:=\:\frac{\mid_{\mathrm{1}\:\:\:\:\:\:\mathrm{2}} ^{\mathrm{1}\:\:\:\:\:\mathrm{1}} \mid}{\sqrt{\mathrm{10}}}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{10}}}\:\Rightarrow \\ $$$${tan}\left({u},{v}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:\theta=\left({u},{v}\right)\:={arctan}\left(\frac{\mathrm{1}}{\mathrm{3}}\right). \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 31/May/18

y=x+4 m_1 =tanα=1 y=2x−7 m_2 =tanβ=2 let angle between two line isθ θ=β−α tanθ=tan(β−α) tanθ=((tanβ−tanα)/(1+tanβ.tanα)) tanθ=((2−1)/(1+2.1)) tanθ=(1/3) θ=tan^(−1) ((1/3)) θ=18^0 20′ acute angle (0,4) is a point lies on st line y=x+4 perpendicular distance from(0,4) 2x−y−7=0 is=∣((2×0−4−7)/(√(2^2 +(−1)^2 )))∣ =((11)/(√5))

$${y}={x}+\mathrm{4} \\ $$$${m}_{\mathrm{1}} ={tan}\alpha=\mathrm{1} \\ $$$${y}=\mathrm{2}{x}−\mathrm{7} \\ $$$${m}_{\mathrm{2}} ={tan}\beta=\mathrm{2} \\ $$$${let}\:{angle}\:{between}\:{two}\:{line}\:{is}\theta \\ $$$$\theta=\beta−\alpha \\ $$$${tan}\theta={tan}\left(\beta−\alpha\right) \\ $$$${tan}\theta=\frac{{tan}\beta−{tan}\alpha}{\mathrm{1}+{tan}\beta.{tan}\alpha} \\ $$$${tan}\theta=\frac{\mathrm{2}−\mathrm{1}}{\mathrm{1}+\mathrm{2}.\mathrm{1}} \\ $$$${tan}\theta=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\theta={tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\theta=\mathrm{18}^{\mathrm{0}} \mathrm{20}'\:\:{acute}\:{angle} \\ $$$$\left(\mathrm{0},\mathrm{4}\right)\:{is}\:{a}\:{point}\:{lies}\:{on}\:{st}\:{line}\:{y}={x}+\mathrm{4} \\ $$$${perpendicular}\:{distance}\:{from}\left(\mathrm{0},\mathrm{4}\right)\:\mathrm{2}{x}−{y}−\mathrm{7}=\mathrm{0} \\ $$$${is}=\mid\frac{\mathrm{2}×\mathrm{0}−\mathrm{4}−\mathrm{7}}{\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }}\mid \\ $$$$=\frac{\mathrm{11}}{\sqrt{\mathrm{5}}} \\ $$

Answered by MJS last updated on 31/May/18

l_1 : −x+y−4=0 ⇒ n_1 = (((−1)),(1) ) l_2 : 2x−y−7=0 ⇒ n_2 = ((2),((−1)) ) cos α=((∣n_1 ∗n_2 ∣)/(∣n_1 ∣×∣n_2 ∣))=((∣(−1)×2+1×(−)1∣)/((√((−1)^2 +1^2 ))×(√(2^2 +(−1)^2 ))))= =((3(√(10)))/(10)) α≈18.43°

$${l}_{\mathrm{1}} :\:−{x}+{y}−\mathrm{4}=\mathrm{0}\:\Rightarrow\:{n}_{\mathrm{1}} =\begin{pmatrix}{−\mathrm{1}}\\{\mathrm{1}}\end{pmatrix} \\ $$$${l}_{\mathrm{2}} :\:\mathrm{2}{x}−{y}−\mathrm{7}=\mathrm{0}\:\Rightarrow\:{n}_{\mathrm{2}} =\begin{pmatrix}{\mathrm{2}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mid{n}_{\mathrm{1}} \ast{n}_{\mathrm{2}} \mid}{\mid{n}_{\mathrm{1}} \mid×\mid{n}_{\mathrm{2}} \mid}=\frac{\mid\left(−\mathrm{1}\right)×\mathrm{2}+\mathrm{1}×\left(−\right)\mathrm{1}\mid}{\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }×\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }}= \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}} \\ $$$$\alpha\approx\mathrm{18}.\mathrm{43}° \\ $$

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