Question Number 52012 by adilmalik78623@gmail.com last updated on 02/Jan/19
Commented by mr W last updated on 02/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Jan/19
![pointB(3r+2,4r+3,5r+4) be the foot of perpendicular direction ratio between point A(1,2,3)and (3r+2,4r+3,5r+4) is (3r+1,4r+1,5r+1) 3(3r+1)+4(4r+1)+5(5r+1)=0 9r+16r+25r=−3−4−5 50r=−12 r=((−12)/(50)) d.r(3r+1,4r+1,5r+1) =((−36)/(50))+1,((−48)/(50))+1,((−60)/(50))+1 =((14)/(50)),(2/(50)),((−10)/(50)) =7,1,−5 foot of perpendicular (((−36)/(50))+2,((−48)/(50))+3,((−60)/(50))+4) (((64)/(50)),((102)/(50)),((140)/(50))) length(√((x_2 −x_1 )^2 +(y_2 −y_1 )^2 +(z_2 −z_1 )^2 )) (√((((64)/(50))−1)^2 +(((102)/(50))−2)^2 +(((140)/(50))−3)^2 )) [((14^2 +2^2 +(−10)^2 )/(50^2 ))]^(1/2) =(√((196+4+100)/(2500))) =(√((300)/(2500))) =((√3)/5) eqn ((x−1)/7)=((y−2)/1)=((z−4)/5) sir ls check...](https://www.tinkutara.com/question/Q52057.png)
$${pointB}\left(\mathrm{3}{r}+\mathrm{2},\mathrm{4}{r}+\mathrm{3},\mathrm{5}{r}+\mathrm{4}\right)\:{be}\:{the}\:{foot}\:{of}\:{perpendicular} \\ $$$${direction}\:{ratio}\:{between}\:{point}\:{A}\left(\mathrm{1},\mathrm{2},\mathrm{3}\right){and}\:\left(\mathrm{3}{r}+\mathrm{2},\mathrm{4}{r}+\mathrm{3},\mathrm{5}{r}+\mathrm{4}\right) \\ $$$${is}\:\left(\mathrm{3}{r}+\mathrm{1},\mathrm{4}{r}+\mathrm{1},\mathrm{5}{r}+\mathrm{1}\right) \\ $$$$\mathrm{3}\left(\mathrm{3}{r}+\mathrm{1}\right)+\mathrm{4}\left(\mathrm{4}{r}+\mathrm{1}\right)+\mathrm{5}\left(\mathrm{5}{r}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{9}{r}+\mathrm{16}{r}+\mathrm{25}{r}=−\mathrm{3}−\mathrm{4}−\mathrm{5} \\ $$$$\mathrm{50}{r}=−\mathrm{12} \\ $$$${r}=\frac{−\mathrm{12}}{\mathrm{50}}\: \\ $$$${d}.{r}\left(\mathrm{3}{r}+\mathrm{1},\mathrm{4}{r}+\mathrm{1},\mathrm{5}{r}+\mathrm{1}\right) \\ $$$$=\frac{−\mathrm{36}}{\mathrm{50}}+\mathrm{1},\frac{−\mathrm{48}}{\mathrm{50}}+\mathrm{1},\frac{−\mathrm{60}}{\mathrm{50}}+\mathrm{1} \\ $$$$=\frac{\mathrm{14}}{\mathrm{50}},\frac{\mathrm{2}}{\mathrm{50}},\frac{−\mathrm{10}}{\mathrm{50}} \\ $$$$=\mathrm{7},\mathrm{1},−\mathrm{5} \\ $$$${foot}\:{of}\:{perpendicular} \\ $$$$\left(\frac{−\mathrm{36}}{\mathrm{50}}+\mathrm{2},\frac{−\mathrm{48}}{\mathrm{50}}+\mathrm{3},\frac{−\mathrm{60}}{\mathrm{50}}+\mathrm{4}\right) \\ $$$$\left(\frac{\mathrm{64}}{\mathrm{50}},\frac{\mathrm{102}}{\mathrm{50}},\frac{\mathrm{140}}{\mathrm{50}}\right) \\ $$$${length}\sqrt{\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({y}_{\mathrm{2}} −{y}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)^{\mathrm{2}} } \\ $$$$\sqrt{\left(\frac{\mathrm{64}}{\mathrm{50}}−\mathrm{1}\right)^{\mathrm{2}} +\left(\frac{\mathrm{102}}{\mathrm{50}}−\mathrm{2}\right)^{\mathrm{2}} +\left(\frac{\mathrm{140}}{\mathrm{50}}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\left[\frac{\mathrm{14}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{10}\right)^{\mathrm{2}} }{\mathrm{50}^{\mathrm{2}} }\right]^{\frac{\mathrm{1}}{\mathrm{2}}} =\sqrt{\frac{\mathrm{196}+\mathrm{4}+\mathrm{100}}{\mathrm{2500}}}\:=\sqrt{\frac{\mathrm{300}}{\mathrm{2500}}} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{5}} \\ $$$${eqn}\:\frac{{x}−\mathrm{1}}{\mathrm{7}}=\frac{{y}−\mathrm{2}}{\mathrm{1}}=\frac{{z}−\mathrm{4}}{\mathrm{5}}\: \\ $$$$ \\ $$$${sir}\:{ls}\:{check}… \\ $$$$ \\ $$$$ \\ $$