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Question Number 78522 by jagoll last updated on 18/Jan/20

$$whatisthe$$$$linepassingthrough(2,2,1)$$$$andparallelto2\hat{i}-\hat{j}-\hat{k}?$$

Commented by mr W last updated on 18/Jan/20

$$\frac{x-2}{2}=\frac{y-2}{-1}=\frac{z-1}{-1}$$

Commented by jagoll last updated on 18/Jan/20

$$howaboutiflineorthogonalsir?$$

Commented by mr W last updated on 18/Jan/20

$$\frac{x-2}{10}=\frac{y-2}{13}=\frac{z-1}{7}$$

Commented by jagoll last updated on 19/Jan/20

$$\mathrm{how}\mathrm{get}(10,13,7)\mathrm{sir}?$$

Commented by mr W last updated on 19/Jan/20

$$line1\left(given\right):$$$$vector2i-j-k$$$$i.e.\frac{x}{2}=\frac{y}{-1}=\frac{z}{-1}$$$$$$$$line2throughpoint(2,2,1)and\mathrm{\perp }line1:$$$$say\frac{x-2}{a}=\frac{y-2}{b}=\frac{z-1}{c}$$$$line2\mathrm{\perp }line1:$$$$2\times a-1\times b-1\times c=0$$$$\Rightarrow a=\frac{b+c}{2}$$$$apointshouldbothonline1andon$$$$line2:$$$$pointonline1:\frac{x}{2}=\frac{y}{-1}=\frac{z}{-1}=t$$$$\Rightarrow x=2t$$$$\Rightarrow y=-t$$$$\Rightarrow z=-t$$$$pointonline2:\frac{x-2}{a}=\frac{y-2}{b}=\frac{z-1}{c}=s$$$$\Rightarrow x=2+sa=2t$$$$\Rightarrow y=2+sb=-t$$$$\Rightarrow z=1+sc=-t$$$$$$$$\Rightarrow sc=-t-1$$$$\Rightarrow sb=-t-2$$$$\Rightarrow sa=2t-2$$$$a=\frac{b+c}{2}\Rightarrow sa=\frac{sb+sc}{2}\Rightarrow 2t-2=\frac{-t-2-t-1}{2}$$$$\Rightarrow 4t-4=-2t-3$$$$\Rightarrow t=\frac{1}{6}$$$$\Rightarrow sa=2\times \frac{1}{6}-2=-\frac{10}{6}$$$$\Rightarrow sb=-\frac{1}{6}-2=-\frac{13}{6}$$$$\Rightarrow sc=-\frac{1}{6}-1=-\frac{7}{6}$$$$\Rightarrow a:b:c=10:13:7$$$$\Rightarrow line2:\frac{x-2}{10}=\frac{y-2}{13}=\frac{z-1}{7}$$

Commented by mr W last updated on 19/Jan/20

$$clear?$$$$maybethereareothereasierways.$$

Commented by jagoll last updated on 19/Jan/20

$$\mathrm{line}\mathrm{passing}\mathrm{trought}\mathrm{point}$$$$(2,2,1)\Rightarrow \frac{\mathrm{x}-2}{\mathrm{a}}=\frac{\mathrm{y}-2}{\mathrm{b}}=\frac{\mathrm{z}-1}{\mathrm{c}}$$$$\mathrm{orthogonal}\mathrm{to}\mathrm{vector}2\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$$$$\overline{\mathrm{u}}\times \overline{\mathrm{v}}=\left|\begin{array}{c}\mathrm{a}\mathrm{b}\mathrm{c}\\ 2-1-1\end{array}\right|=(-\mathrm{b}+\mathrm{c})\hat{\mathrm{i}}+(\mathrm{a}+2\mathrm{c})\hat{\mathrm{j}}-(\mathrm{a}+2\mathrm{b})\hat{\mathrm{k}}$$$$\Rightarrow -2\mathrm{b}+2\mathrm{c}+2\mathrm{a}+4\mathrm{c}-\mathrm{a}-2\mathrm{b}=0$$$$\mathrm{a}-4\mathrm{b}+6\mathrm{c}=0$$

Commented by jagoll last updated on 19/Jan/20

$$\mathrm{what}\mathrm{wrong}\mathrm{my}\mathrm{work}?$$

Commented by mr W last updated on 19/Jan/20

$$totallywrong.$$$$youarelookingforavector\mathit{u}whichis$$$$perpendicularto\mathit{v},notlookingfor$$$$avectorwhichisperpendicularto\mathit{u}$$$$and\mathit{v}!.\mathit{u}\times \mathit{v}isavectorwhichis$$$$perpendicularto\mathit{u}and\mathit{v}!$$$$\mathit{u}=(a,b,c)$$$$\mathit{v}=(2,-1,-1)$$$$since\mathit{u}\mathrm{\perp }\mathit{v},\Rightarrow a\times 2+b\times (-1)+c\times (-1)=0$$

Commented by jagoll last updated on 19/Jan/20

$$\mathrm{oo}\mathrm{yes}\mathrm{sir}.\mathrm{i}\mathrm{understand}.\mathrm{thanks}\mathrm{you}\mathrm{sir}$$

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