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Question Number 57669 by rahul 19 last updated on 09/Apr/19

Commented by mr W last updated on 09/Apr/19

x=3k−2 y=1 z=4k D=d^2 =(3k−2−1)^2 +(1+1)^2 +(4k−2)^2 =25k^2 −34k+17 (dD/dk)=50k−34=0 ⇒k=((17)/(25)) d^2 =((17^2 −34×17+17×25)/(25))=((17×8)/(25)) ⇒d=((2(√(34)))/5) A=((ld)/2)=((5×2(√(34)))/(2×5))=(√(34)) ⇒(1) is correct.

$${x}=\mathrm{3}{k}−\mathrm{2} \\ $$$${y}=\mathrm{1} \\ $$$${z}=\mathrm{4}{k} \\ $$$${D}={d}^{\mathrm{2}} =\left(\mathrm{3}{k}−\mathrm{2}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{4}{k}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$=\mathrm{25}{k}^{\mathrm{2}} −\mathrm{34}{k}+\mathrm{17} \\ $$$$\frac{{dD}}{{dk}}=\mathrm{50}{k}−\mathrm{34}=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{17}}{\mathrm{25}} \\ $$$${d}^{\mathrm{2}} =\frac{\mathrm{17}^{\mathrm{2}} −\mathrm{34}×\mathrm{17}+\mathrm{17}×\mathrm{25}}{\mathrm{25}}=\frac{\mathrm{17}×\mathrm{8}}{\mathrm{25}} \\ $$$$\Rightarrow{d}=\frac{\mathrm{2}\sqrt{\mathrm{34}}}{\mathrm{5}} \\ $$$${A}=\frac{{ld}}{\mathrm{2}}=\frac{\mathrm{5}×\mathrm{2}\sqrt{\mathrm{34}}}{\mathrm{2}×\mathrm{5}}=\sqrt{\mathrm{34}} \\ $$$$ \\ $$$$\Rightarrow\left(\mathrm{1}\right)\:{is}\:{correct}. \\ $$

Commented by rahul 19 last updated on 09/Apr/19

thank you sir!

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

((x+2)/3)=((y−1)/0)=(z/4)=r pointD{(3r−2),1,4r}lies on given st line direction ratio betweenA (1,−1,2) andD{(3r−2),1,4r} is {3r−3,2,4r−2} AD⊥given[st line (3r−3)×3+2×0+(4r−2)×4=0 9r−9+16r−8=0 r=((17)/(25)) point D{3r−2,1,4r}→{((51)/(25))−2,1,((68)/(25))} PointD={(1/(25)),1,((68)/(25))} distance between AD=(√((1−(1/(25)))^2 +(−1−1)^2 +(2−((68)/(25)))^2 )) =(√(((576)/(625))+4+((324)/(625)))) =(√((576+2500+324)/(625))) =(√((3400)/(625))) =(√(((34×4)/(25)) )) =(2/5)(√(34)) area of triangle=(1/2)×(2/5)(√(34)) ×5=(√(34)) unit

$$\frac{{x}+\mathrm{2}}{\mathrm{3}}=\frac{{y}−\mathrm{1}}{\mathrm{0}}=\frac{{z}}{\mathrm{4}}={r} \\ $$$${pointD}\left\{\left(\mathrm{3}{r}−\mathrm{2}\right),\mathrm{1},\mathrm{4}{r}\right\}{lies}\:{on}\:{given}\:{st}\:{line} \\ $$$${direction}\:{ratio}\:{betweenA}\:\left(\mathrm{1},−\mathrm{1},\mathrm{2}\right)\:{andD}\left\{\left(\mathrm{3}{r}−\mathrm{2}\right),\mathrm{1},\mathrm{4}{r}\right\} \\ $$$${is}\:\left\{\mathrm{3}{r}−\mathrm{3},\mathrm{2},\mathrm{4}{r}−\mathrm{2}\right\} \\ $$$${AD}\bot{given}\left[{st}\:{line}\right. \\ $$$$\left(\mathrm{3}{r}−\mathrm{3}\right)×\mathrm{3}+\mathrm{2}×\mathrm{0}+\left(\mathrm{4}{r}−\mathrm{2}\right)×\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{9}{r}−\mathrm{9}+\mathrm{16}{r}−\mathrm{8}=\mathrm{0} \\ $$$${r}=\frac{\mathrm{17}}{\mathrm{25}} \\ $$$${point}\:{D}\left\{\mathrm{3}{r}−\mathrm{2},\mathrm{1},\mathrm{4}{r}\right\}\rightarrow\left\{\frac{\mathrm{51}}{\mathrm{25}}−\mathrm{2},\mathrm{1},\frac{\mathrm{68}}{\mathrm{25}}\right\} \\ $$$${PointD}=\left\{\frac{\mathrm{1}}{\mathrm{25}},\mathrm{1},\frac{\mathrm{68}}{\mathrm{25}}\right\} \\ $$$${distance}\:{between}\:{AD}=\sqrt{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{25}}\right)^{\mathrm{2}} +\left(−\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}−\frac{\mathrm{68}}{\mathrm{25}}\right)^{\mathrm{2}} }\: \\ $$$$=\sqrt{\frac{\mathrm{576}}{\mathrm{625}}+\mathrm{4}+\frac{\mathrm{324}}{\mathrm{625}}} \\ $$$$=\sqrt{\frac{\mathrm{576}+\mathrm{2500}+\mathrm{324}}{\mathrm{625}}}\: \\ $$$$=\sqrt{\frac{\mathrm{3400}}{\mathrm{625}}}\:=\sqrt{\frac{\mathrm{34}×\mathrm{4}}{\mathrm{25}}\:}\:=\frac{\mathrm{2}}{\mathrm{5}}\sqrt{\mathrm{34}}\: \\ $$$${area}\:{of}\:{triangle}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{5}}\sqrt{\mathrm{34}}\:×\mathrm{5}=\sqrt{\mathrm{34}}\:{unit} \\ $$

Commented by rahul 19 last updated on 09/Apr/19

thank you sir!

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