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Question Number 119155 by zakirullah last updated on 22/Oct/20

Commented by zakirullah last updated on 22/Oct/20

$$ifa=i+j+k,b=4i-2n+3k,c=i-2j+kthen$$$$findavecterofmagnitude6unit$$$$whichisparelleltothevecter2a-b+3c.$$

Commented by PRITHWISH SEN 2 last updated on 22/Oct/20

$$2\mathrm{a}-\mathrm{b}+3\mathrm{c}=(2-4+3)\mathrm{i}+(2+2-6)\mathrm{j}+(2-3+3)\mathrm{k}$$$$=\mathrm{i}-2\mathrm{j}+2\mathrm{k}...\left(\mathrm{i}\right)$$$$\mathrm{let}\mathrm{the}\mathrm{parallel}\mathrm{vecor}\mathrm{to}\left(\mathrm{i}\right)\mathrm{is}$$$$\mathrm{p}\mathbf{i}+\mathrm{q}\mathbf{j}+\mathrm{r}\mathbf{k}$$$$\therefore \frac{\mathrm{p}}{1}=\frac{\mathrm{q}}{-2}=\frac{\mathrm{r}}{2}=\lambda \mathrm{say}$$$$\mathrm{now}{\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2=6^2=36$$$${\lambda }^2(1+4+4)=36\Rightarrow \lambda =\pm 2$$$$\mathrm{the}\mathrm{required}\mathrm{vector}$$$$2\mathrm{i}-4\mathrm{j}+4\mathrm{k}$$$$\mathrm{or}$$$$-2\mathrm{i}+4\mathrm{j}-4\mathrm{k}$$$$\mathrm{sorry}\mathrm{i}\mathrm{have}\mathrm{fix}\mathrm{it}.\mathrm{i}\mathrm{am}\mathrm{making}\mathrm{typo}\mathrm{a}\mathrm{habbit}.$$

Commented by zakirullah last updated on 22/Oct/20

$$\mathit{S}\mathit{i}\mathit{r}\mathit{a}\mathit{b}\mathit{o}\mathit{u}\mathit{n}\mathit{d}\mathit{l}\mathit{e}\mathit{o}\mathit{f}\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}$$

Commented by PRITHWISH SEN 2 last updated on 22/Oct/20

$$\mathrm{welcome}$$

Commented by zakirullah last updated on 22/Oct/20

$$\mathit{s}\mathit{i}\mathit{r}\mathit{h}\mathit{o}\mathit{w}⋋=3or-3$$

Answered by Dwaipayan Shikari last updated on 22/Oct/20

$$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=4\hat{i}-2\hat{j}+3\hat{k},\overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}$$$$2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}=\hat{i}-2\hat{j}+2\hat{k}$$$$\mid 2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}\mid =\sqrt{1^2+{(-2)}^2+2^2}=3$$$$(2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}).\left(\overrightarrow{p}\right)=\mid 2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}\mid \mid \overrightarrow{p}\mid cos0°$$$$(\hat{i}-2\hat{j}+2\hat{k})\left(\overrightarrow{p}\right)=18$$$$Take\overrightarrow{p}=x\hat{i}+y\hat{j}+z\hat{k}$$$$x-2y+2z=18$$$$Onesolution$$$$x=2y=1z=9$$$$\overrightarrow{p}=2\hat{i}+\hat{j}+9\hat{k}$$$$Orx=2y=-4z=4$$$$\overrightarrow{p}=2\hat{i}-4\hat{j}+4\hat{k}$$$$manysolutions$$

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