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Question Number 62517 by Tawa1 last updated on 22/Jun/19

Commented by $@ty@m last updated on 22/Jun/19

$$whatismeantbyBC=\left(\begin{array}{c}3\\ 1\end{array}\right)$$$$pl.clarifythisnotation.$$

Commented by Prithwish sen last updated on 22/Jun/19

$$\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{is}$$$$\mathrm{BC}=3\hat{\mathbf{i}}+1\hat{\mathbf{j}}$$

Answered by tanmay last updated on 22/Jun/19

$$Twovectors\overrightarrow{P}and\overrightarrow{Q}are\parallel toeachotherif$$$$onlyif\overrightarrow{P}=\lambda \overrightarrow{Q}$$$$nowhere..B\overrightarrow{C}=3i+jD\overrightarrow{A}=-3i-j$$$$soB\overrightarrow{C}=(-1)D\overrightarrow{A}henceB\overrightarrow{C}\parallel D\overrightarrow{A}$$$$again..$$$$B\overrightarrow{C}=O\overrightarrow{C}-O\overrightarrow{B}$$$$C\overrightarrow{D}=O\overrightarrow{D}-O\overrightarrow{C}$$$$D\overrightarrow{A}=O\overrightarrow{A}-O\overrightarrow{D}$$$$Addthem$$$$B\overrightarrow{C}+C\overrightarrow{D}+D\overrightarrow{A}=O\overrightarrow{A}-O\overrightarrow{B}$$$$butlookhereO\overrightarrow{A}-O\overrightarrow{B}=B\overrightarrow{A}$$$$soB\overrightarrow{A}=(3i+j)+(-i-2j)+(-3i-j)$$$$B\overrightarrow{A}=-i-2jA\overrightarrow{B}=i+2j$$$$nowlookA\overrightarrow{B}=i+2jandC\overrightarrow{D}=-i-2j$$$$soA\overrightarrow{B}=(-1)\times C\overrightarrow{D}$$$$henceA\overrightarrow{B}\parallel C\overrightarrow{D}$$$$soABCDis\parallel gm$$$$$$

Commented by Tawa1 last updated on 22/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by $@ty@m last updated on 22/Jun/19

$$\because B\overrightarrow{C}=-D\overrightarrow{A}$$$$\Rightarrow BC\parallel AD...\left(i\right)$$$$B\overrightarrow{A}=B\overrightarrow{C}+C\overrightarrow{D}+D\overrightarrow{A}=\left(\begin{array}{c}-1\\ -2\end{array}\right)$$$$\therefore B\overrightarrow{A}=C\overrightarrow{D}$$$$\Rightarrow BA\parallel CD...\left(ii\right)$$$$From\left(i\right)\mathrm{\&}\left(ii\right),weconcludethat$$$$ABCDisaparallelogram.$$

Commented by Tawa1 last updated on 22/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by tanmay last updated on 22/Jun/19

$$Q\overrightarrow{R}=O\overrightarrow{R}-O\overrightarrow{Q}=(2i+2j)-(i+6j)=i-4j$$$$T\overrightarrow{S}=O\overrightarrow{S}-O\overrightarrow{T}=(5i+4j)-(xi+yj)=(5-x)i+(4-y)j$$$$midpointofQS=(\frac{1+5}{2},\frac{6+4}{2})=(3,5)$$$$midpointR\overrightarrow{T}=(\frac{2+x}{2},\frac{2+y}{2})$$$$so\frac{2+x}{2}=\frac{1+5}{2}\to x=4\frac{2+y}{2}=\frac{6+4}{2}\to y=8$$$$(x,y)=(4,8)$$$$T\overrightarrow{S}=(5-4)i+(4-8)j=i-4j$$$$$$

Commented by Tawa1 last updated on 22/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by tanmay last updated on 22/Jun/19

$$R\overrightarrow{S}=(5-2)i+(4-2)j=3i+2j$$$$\mid R\overrightarrow{S}\mid =\sqrt{3^2+2^2}=\sqrt{13}$$$$cos\theta =\frac{3}{\sqrt{13}}sin\theta =\frac{2}{\sqrt{13}}$$$$R\overrightarrow{S}=\sqrt{13}(icos\theta +jsin\theta )$$

Commented by Tawa1 last updated on 22/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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