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Question Number 78522 by jagoll last updated on 18/Jan/20
what is the line passing through (2,2,1) and parallel to 2i^� − j^� − k^� ?
$${what}\:{is}\:{the}\: \\ $$$${line}\:{passing}\:{through}\:\left(\mathrm{2},\mathrm{2},\mathrm{1}\right) \\ $$$${and}\:{parallel}\:{to}\:\mathrm{2}\hat {{i}}\:−\:\hat {{j}}\:−\:\hat {{k}}\:? \\ $$
Commented by mr W last updated on 18/Jan/20
((x−2)/2)=((y−2)/(−1))=((z−1)/(−1))
$$\frac{{x}−\mathrm{2}}{\mathrm{2}}=\frac{{y}−\mathrm{2}}{−\mathrm{1}}=\frac{{z}−\mathrm{1}}{−\mathrm{1}} \\ $$
Commented by jagoll last updated on 18/Jan/20
how about if line orthogonal sir?
$${how}\:{about}\:{if}\:{line}\:{orthogonal}\:{sir}? \\ $$
Commented by mr W last updated on 18/Jan/20
((x−2)/(10))=((y−2)/(13))=((z−1)/7)
$$\frac{{x}−\mathrm{2}}{\mathrm{10}}=\frac{{y}−\mathrm{2}}{\mathrm{13}}=\frac{{z}−\mathrm{1}}{\mathrm{7}} \\ $$
Commented by jagoll last updated on 19/Jan/20
how get (10,13,7) sir?
$$\mathrm{how}\:\mathrm{get}\:\left(\mathrm{10},\mathrm{13},\mathrm{7}\right)\:\mathrm{sir}? \\ $$
Commented by mr W last updated on 19/Jan/20
line 1 (given): vector 2i−j−k i.e. (x/2)=(y/(−1))=(z/(−1)) line 2 through point (2,2,1) and ⊥ line 1: say ((x−2)/a)=((y−2)/b)=((z−1)/c) line 2 ⊥ line 1: 2×a−1×b−1×c=0 ⇒a=((b+c)/2) a point should both on line 1 and on line 2: point on line 1: (x/2)=(y/(−1))=(z/(−1))=t ⇒x=2t ⇒y=−t ⇒z=−t point on line 2: ((x−2)/a)=((y−2)/b)=((z−1)/c)=s ⇒x=2+sa=2t ⇒y=2+sb=−t ⇒z=1+sc=−t ⇒sc=−t−1 ⇒sb=−t−2 ⇒sa=2t−2 a=((b+c)/2) ⇒sa=((sb+sc)/2)⇒2t−2=((−t−2−t−1)/2) ⇒4t−4=−2t−3 ⇒t=(1/6) ⇒sa=2×(1/6)−2=−((10)/6) ⇒sb=−(1/6)−2=−((13)/6) ⇒sc=−(1/6)−1=−(7/6) ⇒a:b:c=10:13:7 ⇒line 2: ((x−2)/(10))=((y−2)/(13))=((z−1)/7)
$${line}\:\mathrm{1}\:\left({given}\right): \\ $$$${vector}\:\mathrm{2}{i}−{j}−{k} \\ $$$${i}.{e}.\:\frac{{x}}{\mathrm{2}}=\frac{{y}}{−\mathrm{1}}=\frac{{z}}{−\mathrm{1}} \\ $$$$ \\ $$$${line}\:\mathrm{2}\:{through}\:{point}\:\left(\mathrm{2},\mathrm{2},\mathrm{1}\right)\:{and}\:\bot\:{line}\:\mathrm{1}: \\ $$$${say}\:\frac{{x}−\mathrm{2}}{{a}}=\frac{{y}−\mathrm{2}}{{b}}=\frac{{z}−\mathrm{1}}{{c}} \\ $$$${line}\:\mathrm{2}\:\bot\:{line}\:\mathrm{1}: \\ $$$$\mathrm{2}×{a}−\mathrm{1}×{b}−\mathrm{1}×{c}=\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{{b}+{c}}{\mathrm{2}} \\ $$$${a}\:{point}\:{should}\:{both}\:{on}\:{line}\:\mathrm{1}\:{and}\:{on} \\ $$$${line}\:\mathrm{2}: \\ $$$${point}\:{on}\:{line}\:\mathrm{1}:\:\frac{{x}}{\mathrm{2}}=\frac{{y}}{−\mathrm{1}}=\frac{{z}}{−\mathrm{1}}={t} \\ $$$$\Rightarrow{x}=\mathrm{2}{t} \\ $$$$\Rightarrow{y}=−{t} \\ $$$$\Rightarrow{z}=−{t} \\ $$$${point}\:{on}\:{line}\:\mathrm{2}:\:\frac{{x}−\mathrm{2}}{{a}}=\frac{{y}−\mathrm{2}}{{b}}=\frac{{z}−\mathrm{1}}{{c}}={s} \\ $$$$\Rightarrow{x}=\mathrm{2}+{sa}=\mathrm{2}{t} \\ $$$$\Rightarrow{y}=\mathrm{2}+{sb}=−{t} \\ $$$$\Rightarrow{z}=\mathrm{1}+{sc}=−{t} \\ $$$$ \\ $$$$\Rightarrow{sc}=−{t}−\mathrm{1} \\ $$$$\Rightarrow{sb}=−{t}−\mathrm{2} \\ $$$$\Rightarrow{sa}=\mathrm{2}{t}−\mathrm{2} \\ $$$${a}=\frac{{b}+{c}}{\mathrm{2}}\:\Rightarrow{sa}=\frac{{sb}+{sc}}{\mathrm{2}}\Rightarrow\mathrm{2}{t}−\mathrm{2}=\frac{−{t}−\mathrm{2}−{t}−\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{t}−\mathrm{4}=−\mathrm{2}{t}−\mathrm{3} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow{sa}=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{2}=−\frac{\mathrm{10}}{\mathrm{6}} \\ $$$$\Rightarrow{sb}=−\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{2}=−\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\Rightarrow{sc}=−\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}=−\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\Rightarrow{a}:{b}:{c}=\mathrm{10}:\mathrm{13}:\mathrm{7} \\ $$$$\Rightarrow{line}\:\mathrm{2}:\:\:\frac{{x}−\mathrm{2}}{\mathrm{10}}=\frac{{y}−\mathrm{2}}{\mathrm{13}}=\frac{{z}−\mathrm{1}}{\mathrm{7}} \\ $$
Commented by mr W last updated on 19/Jan/20
clear? maybe there are other easier ways.
$${clear}? \\ $$$${maybe}\:{there}\:{are}\:{other}\:{easier}\:{ways}. \\ $$
Commented by jagoll last updated on 19/Jan/20
line passing trought point (2,2,1) ⇒((x−2)/a)=((y−2)/b)=((z−1)/c) orthogonal to vector 2i^� − j^� −k^� u^� × v^� = determinant (((a b c)),((2 −1 −1)))= (−b+c)i^� +(a+2c) j^� −(a+2b) k^� ⇒ −2b+2c+2a+4c−a−2b=0 a −4b+6c=0
$$\mathrm{line}\:\mathrm{passing}\:\mathrm{trought}\:\mathrm{point} \\ $$$$\left(\mathrm{2},\mathrm{2},\mathrm{1}\right)\:\Rightarrow\frac{\mathrm{x}−\mathrm{2}}{\mathrm{a}}=\frac{\mathrm{y}−\mathrm{2}}{\mathrm{b}}=\frac{\mathrm{z}−\mathrm{1}}{\mathrm{c}} \\ $$$$\mathrm{orthogonal}\:\mathrm{to}\:\mathrm{vector}\:\mathrm{2}\hat {\mathrm{i}}\:−\:\hat {\mathrm{j}}−\hat {\mathrm{k}} \\ $$$$\bar {\mathrm{u}}\:×\:\bar {\mathrm{v}}\:=\:\begin{vmatrix}{\mathrm{a}\:\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\mathrm{c}}\\{\mathrm{2}\:\:−\mathrm{1}\:\:−\mathrm{1}}\end{vmatrix}=\:\left(−\mathrm{b}+\mathrm{c}\right)\hat {\mathrm{i}}\:+\left(\mathrm{a}+\mathrm{2c}\right)\:\hat {\mathrm{j}}−\left(\mathrm{a}+\mathrm{2b}\right)\:\hat {\mathrm{k}} \\ $$$$\Rightarrow\:−\mathrm{2b}+\mathrm{2c}+\mathrm{2a}+\mathrm{4c}−\mathrm{a}−\mathrm{2b}=\mathrm{0} \\ $$$$\mathrm{a}\:−\mathrm{4b}+\mathrm{6c}=\mathrm{0}\: \\ $$
Commented by jagoll last updated on 19/Jan/20
what wrong my work ?
$$\mathrm{what}\:\mathrm{wrong}\:\mathrm{my}\:\mathrm{work}\:? \\ $$
Commented by mr W last updated on 19/Jan/20
totally wrong. you are looking for a vector u which is perpendicular to v, not looking for a vector which is perpendicular to u and v!. u×v is a vector which is perpendicular to u and v! u=(a,b,c) v=(2,−1,−1) since u⊥v, ⇒a×2+b×(−1)+c×(−1)=0
$${totally}\:{wrong}. \\ $$$${you}\:{are}\:{looking}\:{for}\:{a}\:{vector}\:\boldsymbol{{u}}\:{which}\:{is} \\ $$$${perpendicular}\:{to}\:\boldsymbol{{v}},\:{not}\:{looking}\:{for} \\ $$$${a}\:{vector}\:{which}\:{is}\:{perpendicular}\:{to}\:\boldsymbol{{u}} \\ $$$${and}\:\boldsymbol{{v}}!.\:\boldsymbol{{u}}×\boldsymbol{{v}}\:{is}\:{a}\:{vector}\:{which}\:{is}\: \\ $$$${perpendicular}\:{to}\:\boldsymbol{{u}}\:{and}\:\boldsymbol{{v}}! \\ $$$$\boldsymbol{{u}}=\left({a},{b},{c}\right) \\ $$$$\boldsymbol{{v}}=\left(\mathrm{2},−\mathrm{1},−\mathrm{1}\right) \\ $$$${since}\:\boldsymbol{{u}}\bot\boldsymbol{{v}},\:\Rightarrow{a}×\mathrm{2}+{b}×\left(−\mathrm{1}\right)+{c}×\left(−\mathrm{1}\right)=\mathrm{0} \\ $$
Commented by jagoll last updated on 19/Jan/20
oo yes sir. i understand. thanks you sir
$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{understand}.\:\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$

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