Question-119008

Question Number 119008 by zakirullah last updated on 21/Oct/20

Commented by zakirullah last updated on 21/Oct/20

find the initial point p or the terminal point Q whichever is missing? i. PQ^→ = [−2,3], P(1,−2) ii. PQ^→ =[4,−5], Q(−1,1) iii. PQ^→ = [−1,3,−2], P(2,−1,−3) iv. PQ^→ =[2,−3,−4], Q(3,−1,4)

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{initial}}\:\boldsymbol{\mathrm{point}}\:\boldsymbol{\mathrm{p}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{the}}\:\: \\ $$$$\boldsymbol{\mathrm{terminal}}\:\boldsymbol{\mathrm{point}}\:\boldsymbol{\mathrm{Q}}\:\boldsymbol{{whichever}}\:\boldsymbol{{is}}\:\boldsymbol{{missing}}? \\ $$$$\boldsymbol{{i}}.\:\:\boldsymbol{{P}}\overset{\rightarrow} {\boldsymbol{{Q}}}\:=\:\left[−\mathrm{2},\mathrm{3}\right],\:\boldsymbol{{P}}\left(\mathrm{1},−\mathrm{2}\right) \\ $$$$\boldsymbol{{ii}}.\:\boldsymbol{{P}}\overset{\rightarrow} {\boldsymbol{{Q}}}\:=\left[\mathrm{4},−\mathrm{5}\right],\:\boldsymbol{{Q}}\left(−\mathrm{1},\mathrm{1}\right) \\ $$$$\boldsymbol{{iii}}.\:\boldsymbol{{P}}\overset{\rightarrow} {\boldsymbol{{Q}}}\:=\:\left[−\mathrm{1},\mathrm{3},−\mathrm{2}\right],\:\boldsymbol{{P}}\left(\mathrm{2},−\mathrm{1},−\mathrm{3}\right) \\ $$$$\boldsymbol{{iv}}.\:\:\boldsymbol{{P}}\overset{\rightarrow} {\boldsymbol{{Q}}}\:=\left[\mathrm{2},−\mathrm{3},−\mathrm{4}\right],\:\boldsymbol{{Q}}\left(\mathrm{3},−\mathrm{1},\mathrm{4}\right) \\ $$

Answered by ebi last updated on 21/Oct/20

point P(p_1 ,p_2 ,...,p_n ), Q(q_1 ,q_2 ,...,q_n ) component, PQ^→ =⟨q_1 −p_1 ,q_2 −p_1 ,...,q_n −p_n ⟩=⟨u_1 ,u_2 ,...,u_n ⟩ magnitude,∣∣PQ^→ ∣∣=(√(u_1 ^2 +u_2 ^2 +...+u_n ^2 )) Question 2 i) PQ^→ =⟨−2,3⟩, P(1,−2) PQ^→ =⟨−2,3⟩=⟨q_1 −1,q_2 +2⟩ Q(−1,1) ii) PQ^→ =⟨4,−5⟩, Q(−1,1) PQ^→ =⟨4,−5⟩=⟨−1−p_1 ,1−p_2 ⟩ P(−5,6) iii) PQ^→ =⟨−1,3,−2⟩, P(2,−1,−3) PQ^→ =⟨−1,3,−2⟩=⟨q_1 −2,q_2 +2,q_3 +3⟩ Q(1,1,−5) iv) PQ^→ =⟨2,−3,−4⟩, Q(3−1,4) PQ^→ =⟨2,−3,−4⟩=⟨3−p_1 ,−1−p_2 ,4−p_3 ⟩ P(1,2,8)

$$ \\ $$$${point}\:{P}\left({p}_{\mathrm{1}} ,{p}_{\mathrm{2}} ,…,{p}_{{n}} \right),\:{Q}\left({q}_{\mathrm{1}} ,{q}_{\mathrm{2}} ,…,{q}_{{n}} \right) \\ $$$${component},\:{P}\overset{\rightarrow} {{Q}}=\langle{q}_{\mathrm{1}} −{p}_{\mathrm{1}} ,{q}_{\mathrm{2}} −{p}_{\mathrm{1}} ,…,{q}_{{n}} −{p}_{{n}} \rangle=\langle{u}_{\mathrm{1}} ,{u}_{\mathrm{2}} ,…,{u}_{{n}} \rangle \\ $$$${magnitude},\mid\mid{P}\overset{\rightarrow} {{Q}}\mid\mid=\sqrt{{u}_{\mathrm{1}} ^{\mathrm{2}} +{u}_{\mathrm{2}} ^{\mathrm{2}} +…+{u}_{{n}} ^{\mathrm{2}} } \\ $$$$ \\ $$$${Question}\:\mathrm{2} \\ $$$$\left.{i}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle−\mathrm{2},\mathrm{3}\rangle,\:\:\:{P}\left(\mathrm{1},−\mathrm{2}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle−\mathrm{2},\mathrm{3}\rangle=\langle{q}_{\mathrm{1}} −\mathrm{1},{q}_{\mathrm{2}} +\mathrm{2}\rangle \\ $$$${Q}\left(−\mathrm{1},\mathrm{1}\right) \\ $$$$ \\ $$$$\left.{ii}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle\mathrm{4},−\mathrm{5}\rangle,\:\:\:{Q}\left(−\mathrm{1},\mathrm{1}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle\mathrm{4},−\mathrm{5}\rangle=\langle−\mathrm{1}−{p}_{\mathrm{1}} ,\mathrm{1}−{p}_{\mathrm{2}} \rangle \\ $$$${P}\left(−\mathrm{5},\mathrm{6}\right) \\ $$$$ \\ $$$$\left.{iii}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle−\mathrm{1},\mathrm{3},−\mathrm{2}\rangle,\:\:\:{P}\left(\mathrm{2},−\mathrm{1},−\mathrm{3}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle−\mathrm{1},\mathrm{3},−\mathrm{2}\rangle=\langle{q}_{\mathrm{1}} −\mathrm{2},{q}_{\mathrm{2}} +\mathrm{2},{q}_{\mathrm{3}} +\mathrm{3}\rangle \\ $$$${Q}\left(\mathrm{1},\mathrm{1},−\mathrm{5}\right) \\ $$$$ \\ $$$$\left.{iv}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle\mathrm{2},−\mathrm{3},−\mathrm{4}\rangle,\:\:\:{Q}\left(\mathrm{3}−\mathrm{1},\mathrm{4}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle\mathrm{2},−\mathrm{3},−\mathrm{4}\rangle=\langle\mathrm{3}−{p}_{\mathrm{1}} ,−\mathrm{1}−{p}_{\mathrm{2}} ,\mathrm{4}−{p}_{\mathrm{3}} \rangle \\ $$$${P}\left(\mathrm{1},\mathrm{2},\mathrm{8}\right) \\ $$

Commented by zakirullah last updated on 21/Oct/20

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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