Menu Close

Angle-between-the-lines-x-1-1-y-1-1-z-1-2-and-x-1-3-1-y-1-3-1-z-1-4-is-

Tinku Tara 0 Comments
Pin




Question Number 53034 by gopikrishnan005@gmail.com last updated on 16/Jan/19
Angle between the lines ((x−1)/1)=((y−1)/1)=((z−1)/2)and ((x−1)/(−(√(3−1))))=((y−1)/( (√(3−1))))=((z−1)/4) is
$$\mathrm{Angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{lines}\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{1}}=\frac{\mathrm{y}−\mathrm{1}}{\mathrm{1}}=\frac{\mathrm{z}−\mathrm{1}}{\mathrm{2}}\mathrm{and}\:\frac{\mathrm{x}−\mathrm{1}}{−\sqrt{\mathrm{3}−\mathrm{1}}}=\frac{\mathrm{y}−\mathrm{1}}{\:\sqrt{\mathrm{3}−\mathrm{1}}}=\frac{\mathrm{z}−\mathrm{1}}{\mathrm{4}}\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19
cosθ=((a_1 a_2 +b_1 b_2 +c_1 c_2 )/( (√(a_1 ^2 +b_1 ^2 +c_1 ^2 )) ×(√(a_2 ^2 +b_2 ^2 +c_2 ^2 )))) use this formula ((x−x_1 )/a_1 )=((y−y_1 )/b_1 )=((z−z_1 )/c_1 ) ((x−x_2 )/a_2 )=((y−y_2 )/b_2 )=((z−z_2 )/c_2 )
$${cos}\theta=\frac{{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{b}_{\mathrm{1}} {b}_{\mathrm{2}} +{c}_{\mathrm{1}} {c}_{\mathrm{2}} }{\:\sqrt{{a}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} +{c}_{\mathrm{1}} ^{\mathrm{2}} }\:×\sqrt{{a}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +{c}_{\mathrm{2}} ^{\mathrm{2}} }} \\ $$$${use}\:{this}\:{formula} \\ $$$$\frac{{x}−{x}_{\mathrm{1}} }{{a}_{\mathrm{1}} }=\frac{{y}−{y}_{\mathrm{1}} }{{b}_{\mathrm{1}} }=\frac{{z}−{z}_{\mathrm{1}} }{{c}_{\mathrm{1}} } \\ $$$$\frac{{x}−{x}_{\mathrm{2}} }{{a}_{\mathrm{2}} }=\frac{{y}−{y}_{\mathrm{2}} }{{b}_{\mathrm{2}} }=\frac{{z}−{z}_{\mathrm{2}} }{{c}_{\mathrm{2}} }\:\:\: \\ $$
Answered by ajfour last updated on 16/Jan/19
b_1 ^� = i^� +j^� +2k^� b_2 ^� = −((√3)+1)i^� +((√3)−1)j^� +4k^� cos θ = ((−(√3)−1+(√3)−1+8)/( (√6)(√(((√3)+1)^2 +((√3)−1)^2 +16)))) = ((√6)/( (√(24)))) = (1/2) ⇒ θ = (π/3) .
$$\bar {{b}}_{\mathrm{1}} =\:\hat {{i}}+\hat {{j}}+\mathrm{2}\hat {{k}} \\ $$$$\bar {{b}}_{\mathrm{2}} =\:−\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)\hat {{i}}+\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\hat {{j}}+\mathrm{4}\hat {{k}} \\ $$$$\mathrm{cos}\:\theta\:=\:\frac{−\sqrt{\mathrm{3}}−\mathrm{1}+\sqrt{\mathrm{3}}−\mathrm{1}+\mathrm{8}}{\:\sqrt{\mathrm{6}}\sqrt{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{16}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{24}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\Rightarrow\:\:\theta\:=\:\frac{\pi}{\mathrm{3}}\:. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *