Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 119013 by zakirullah last updated on 21/Oct/20

Commented by zakirullah last updated on 21/Oct/20

      find the component and         the magnitude of PQ^→   i.  P(−1,2), Q(2,−1)  ii.  P(−2,1), Q(2,3)  iii.  P(−1,1,2), Q(2,−1,3)  iv.  P(2,4,6),  Q(1,−2,3)

$$\:\:\:\:\:\:\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{component}}\:\boldsymbol{{and}}\: \\ $$$$\:\:\:\:\:\:\boldsymbol{{the}}\:\boldsymbol{{magnitude}}\:\boldsymbol{{of}}\:\boldsymbol{{P}}\overset{\rightarrow} {\boldsymbol{{Q}}} \\ $$$${i}.\:\:\boldsymbol{{P}}\left(−\mathrm{1},\mathrm{2}\right),\:\boldsymbol{{Q}}\left(\mathrm{2},−\mathrm{1}\right) \\ $$$${ii}.\:\:\boldsymbol{{P}}\left(−\mathrm{2},\mathrm{1}\right),\:\boldsymbol{{Q}}\left(\mathrm{2},\mathrm{3}\right) \\ $$$${iii}.\:\:\boldsymbol{{P}}\left(−\mathrm{1},\mathrm{1},\mathrm{2}\right),\:\boldsymbol{{Q}}\left(\mathrm{2},−\mathrm{1},\mathrm{3}\right) \\ $$$${iv}.\:\:\boldsymbol{{P}}\left(\mathrm{2},\mathrm{4},\mathrm{6}\right),\:\:\boldsymbol{{Q}}\left(\mathrm{1},−\mathrm{2},\mathrm{3}\right) \\ $$

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 ))    i)  PQ^→ =⟨3,−3⟩,   ∣∣PQ^→ ∣∣=3(√2)    ii)  PQ^→ =⟨4,2⟩,   ∣∣PQ^→ ∣∣=2(√5)    iii)  PQ^→ =⟨3,−2,1⟩,   ∣∣PQ^→ ∣∣=(√(14))    iv)  PQ^→ =⟨−1,−6,−3⟩,   ∣∣PQ^→ ∣∣=(√(46))

$${point}\:{P}\left({p}_{\mathrm{1}} ,{p}_{\mathrm{2}} ,...,{p}_{{n}} \right),\:{Q}\left({q}_{\mathrm{1}} ,{q}_{\mathrm{2}} ,...,{q}_{{n}} \right) \\ $$$${component},\:{P}\overset{\rightarrow} {{Q}}=\langle{q}_{\mathrm{1}} −{p}_{\mathrm{1}} ,{q}_{\mathrm{2}} −{p}_{\mathrm{1}} ,...,{q}_{{n}} −{p}_{{n}} \rangle=\langle{u}_{\mathrm{1}} ,{u}_{\mathrm{2}} ,...,{u}_{{n}} \rangle \\ $$$${magnitude},\mid\mid{P}\overset{\rightarrow} {{Q}}\mid\mid=\sqrt{{u}_{\mathrm{1}} ^{\mathrm{2}} +{u}_{\mathrm{2}} ^{\mathrm{2}} +...+{u}_{{n}} ^{\mathrm{2}} } \\ $$$$ \\ $$$$\left.{i}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle\mathrm{3},−\mathrm{3}\rangle,\:\:\:\mid\mid{P}\overset{\rightarrow} {{Q}}\mid\mid=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\left.{ii}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle\mathrm{4},\mathrm{2}\rangle,\:\:\:\mid\mid{P}\overset{\rightarrow} {{Q}}\mid\mid=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$ \\ $$$$\left.{iii}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle\mathrm{3},−\mathrm{2},\mathrm{1}\rangle,\:\:\:\mid\mid{P}\overset{\rightarrow} {{Q}}\mid\mid=\sqrt{\mathrm{14}} \\ $$$$ \\ $$$$\left.{iv}\right) \\ $$$${P}\overset{\rightarrow} {{Q}}=\langle−\mathrm{1},−\mathrm{6},−\mathrm{3}\rangle,\:\:\:\mid\mid{P}\overset{\rightarrow} {{Q}}\mid\mid=\sqrt{\mathrm{46}} \\ $$$$ \\ $$

Commented by zakirullah last updated on 21/Oct/20

thanks sir

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com