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Question Number 45259 by rahul 19 last updated on 11/Oct/18
Find distance of point (1,1,1) from the line passing through (2,3,4) & (−1,2,3) ?
$${Find}\:\:{distance}\:{of}\:{point}\:\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:{from} \\ $$$${the}\:{line}\:{passing}\:{through}\:\left(\mathrm{2},\mathrm{3},\mathrm{4}\right)\:\& \\ $$$$\left(−\mathrm{1},\mathrm{2},\mathrm{3}\right)\:? \\ $$
Commented by rahul 19 last updated on 11/Oct/18
Another method: Let P(1,1,1) , Q(2,3,4) , R(−1,2,3) and M be foot of perpendicular... Now in ΔPQR, area of Δ= (1/2)∣PQ×QR∣=(1/2)∣PM×QR∣ ⇒ ∣(i+2j+3k)×(3i+j+k)∣=PM×(√(11)) ⇒ PM= (√((90)/(11)))
$${Another}\:{method}: \\ $$$${Let}\:{P}\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:,\:{Q}\left(\mathrm{2},\mathrm{3},\mathrm{4}\right)\:,\:{R}\left(−\mathrm{1},\mathrm{2},\mathrm{3}\right) \\ $$$${and}\:{M}\:{be}\:{foot}\:{of}\:{perpendicular}… \\ $$$${Now}\:{in}\:\Delta{PQR}, \\ $$$${area}\:{of}\:\Delta=\:\frac{\mathrm{1}}{\mathrm{2}}\mid{PQ}×{QR}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mid{PM}×{QR}\mid \\ $$$$\Rightarrow\:\mid\left({i}+\mathrm{2}{j}+\mathrm{3}{k}\right)×\left(\mathrm{3}{i}+{j}+{k}\right)\mid={PM}×\sqrt{\mathrm{11}} \\ $$$$\Rightarrow\:{PM}=\:\sqrt{\frac{\mathrm{90}}{\mathrm{11}}} \\ $$
Answered by ajfour last updated on 11/Oct/18
let eq. of line be: r^� =a^� +λb^� b^� = 3i^� +j^� +k^� ; ∣b^� ∣=(√(11)) d^� = (1/( (√(11)))) determinant ((i^� ,j^� ,k^� ),(1,2,3),(3,1,1)) = (1/( (√(11))))(−i^� +8j^� −5k^� ) ∣d^� ∣ = (√((90)/(11))) .
$${let}\:{eq}.\:{of}\:{line}\:{be}:\:\:\:\bar {\boldsymbol{{r}}}=\bar {\boldsymbol{{a}}}+\lambda\bar {\boldsymbol{{b}}} \\ $$$$\bar {\boldsymbol{{b}}}\:=\:\mathrm{3}\hat {{i}}+\hat {{j}}+\hat {{k}}\:\:\:;\:\:\mid\bar {\boldsymbol{{b}}}\mid=\sqrt{\mathrm{11}} \\ $$$$\:\:\bar {{d}}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}\begin{vmatrix}{\hat {{i}}}&{\hat {{j}}}&{\hat {{k}}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\end{vmatrix} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}\left(−\hat {{i}}+\mathrm{8}\hat {{j}}−\mathrm{5}\hat {{k}}\right) \\ $$$$\:\mid\bar {{d}}\mid\:=\:\sqrt{\frac{\mathrm{90}}{\mathrm{11}}}\:\:. \\ $$
Commented by rahul 19 last updated on 11/Oct/18
Thank you sir ! ☺️��
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18
eqnof st line pass ing through (2,3,4) and (−1,2,3) is ((x−2)/(−1−2))=((y−3)/(2−3))=((z−4)/(3−4))=r (−3r+2,−r+3,−r+4) point lies on st line and assumed that it is the foot of perpendicular drswn from (1,1,1) on the st line. direction ratio between (1,1,1) and (−3r+2,−r+3,−r+4) is (−3r+1,−r+2,−r+3) now (−3r+1)×(−3)+(−r+2)×(−1)+(−r+3)×−1=0 9r−3+r−2+r−3=0 11r=8 r=(8/(11)) now distance bdtween (1,1,1) and(−3r+2,−r+3,−r+4) (√((−3r+2−1)^2 +(−r+3−1)^2 +(−r+4−1)^2 )) =(√(((−3r+1)^2 +(−r+2)^2 +(−r+3)^2 )) =(√((((−24+11)/(11)))^2 +(((−8+22)/(11)))^2 +(((−8+33)/(11)))^2 )) =(√(((169)/(121))+((196)/(121))+((625)/(121)))) =(√((990)/(121))) =(√((90)/(11)))
$${eqnof}\:{st}\:{line}\:{pass}\:{ing}\:{through}\:\left(\mathrm{2},\mathrm{3},\mathrm{4}\right)\:{and}\:\left(−\mathrm{1},\mathrm{2},\mathrm{3}\right) \\ $$$${is}\:\frac{{x}−\mathrm{2}}{−\mathrm{1}−\mathrm{2}}=\frac{{y}−\mathrm{3}}{\mathrm{2}−\mathrm{3}}=\frac{{z}−\mathrm{4}}{\mathrm{3}−\mathrm{4}}={r} \\ $$$$\left(−\mathrm{3}{r}+\mathrm{2},−{r}+\mathrm{3},−{r}+\mathrm{4}\right)\:{point}\:{lies}\:{on}\:{st}\:{line}\:{and} \\ $$$${assumed}\:{that}\:{it}\:{is}\:{the}\:{foot}\:{of}\:{perpendicular} \\ $$$${drswn}\:{from}\:\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:{on}\:{the}\:{st}\:{line}. \\ $$$${direction}\:{ratio}\:{between}\:\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:{and}\:\left(−\mathrm{3}{r}+\mathrm{2},−{r}+\mathrm{3},−{r}+\mathrm{4}\right) \\ $$$${is}\:\left(−\mathrm{3}{r}+\mathrm{1},−{r}+\mathrm{2},−{r}+\mathrm{3}\right) \\ $$$${now}\:\left(−\mathrm{3}{r}+\mathrm{1}\right)×\left(−\mathrm{3}\right)+\left(−{r}+\mathrm{2}\right)×\left(−\mathrm{1}\right)+\left(−{r}+\mathrm{3}\right)×−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{9}{r}−\mathrm{3}+{r}−\mathrm{2}+{r}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{11}{r}=\mathrm{8}\:\:\:{r}=\frac{\mathrm{8}}{\mathrm{11}} \\ $$$${now}\:{distance}\:{bdtween}\:\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:{and}\left(−\mathrm{3}{r}+\mathrm{2},−{r}+\mathrm{3},−{r}+\mathrm{4}\right) \\ $$$$\sqrt{\left(−\mathrm{3}{r}+\mathrm{2}−\mathrm{1}\right)^{\mathrm{2}} +\left(−{r}+\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} +\left(−{r}+\mathrm{4}−\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$=\sqrt{\left(\left(−\mathrm{3}{r}+\mathrm{1}\right)^{\mathrm{2}} +\left(−{r}+\mathrm{2}\right)^{\mathrm{2}} +\left(−{r}+\mathrm{3}\right)^{\mathrm{2}} \right.}\: \\ $$$$=\sqrt{\left(\frac{−\mathrm{24}+\mathrm{11}}{\mathrm{11}}\right)^{\mathrm{2}} +\left(\frac{−\mathrm{8}+\mathrm{22}}{\mathrm{11}}\right)^{\mathrm{2}} +\left(\frac{−\mathrm{8}+\mathrm{33}}{\mathrm{11}}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\frac{\mathrm{169}}{\mathrm{121}}+\frac{\mathrm{196}}{\mathrm{121}}+\frac{\mathrm{625}}{\mathrm{121}}} \\ $$$$=\sqrt{\frac{\mathrm{990}}{\mathrm{121}}}\:=\sqrt{\frac{\mathrm{90}}{\mathrm{11}}} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 11/Oct/18
thank you sir☺️��

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