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If two lines $L_1$ and $L_2$ in space, are defined by: $$L_1=\{x=\sqrt{\lambda}y+(\sqrt{\lambda}-1)\\z=(\sqrt{\lambda}-1)y+\sqrt{\lambda}\}\text{ and }\\L_2=\{x=\sqrt{\mu}y+(1-\sqrt{\mu})\\z=(1-\sqrt{\mu})y+\sqrt{\mu}\}$$ then $L_1$ is perpendicular to $L_2$, forall non-negative reals $\lambda$ and $\mu$, such that: It can be easily seen: $$L_1:\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{y}{1}=\frac{z-\sqrt{\lambda}}{(\sqrt{\lambda}-1)}\\ L_2:\frac{x-(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{y}{1}=\frac{z-\sqrt{\mu}}{(1-\sqrt{\mu})}\\ $$ So dot product (of those) must be zero: $$\sqrt{\lambda}\sqrt{\mu}+1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0\\ \sqrt\lambda+\sqrt\mu=0\\ \lambda=\mu=0$$ Options given are: $$\lambda=\mu,\lambda\ne\mu,\sqrt\lambda+\sqrt\mu=1,\lambda+\mu=0$$ It seems three of them are correct but only one is actually correct.

RE60K
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  • Perhaps it's only me, but I don't understand how you're defining $;L_i;$ . As far as I know, a line in space is given in the form $$(a,b,c)+t(p,q,r);,;;t\in\Bbb R$$ – Timbuc Mar 16 '15 at 14:31
  • @Timbuc your's vector form. Don't you know 3d co-ordinate geometry (mine one is that)? – RE60K Mar 16 '15 at 14:32
  • @AD I thought I did, but any other form I know is equivalente to the above one: you can also give $;x,y,z;$ in parametric form, say...I still can't understand what you did. – Timbuc Mar 16 '15 at 14:33
  • Oh, I think I missed those $;y$'s there and some curly parentheses! Let me check. It's just that, for some reason, you jumped down line at the midst of the parentheses...! – Timbuc Mar 16 '15 at 14:34
  • @Timbuc see http://www.netcomuk.co.uk/~jenolive/vect17.html – RE60K Mar 16 '15 at 14:34
  • LOL @ADG...yes, that I know, but some how the jumping in the middle of the parentheses confused me. – Timbuc Mar 16 '15 at 14:35
  • @Timbuc (x,y,z)=(a,b,c)+t(p,q,r) $\iff$ (x-a)/p=(y-b)/q=(z-c)/r (=t) Elimiates t. Firstmost is a modified form of this and the afterwards one are purely this form. Timbuc? R u there? – RE60K Mar 16 '15 at 14:38
  • I already got that, after I made sense of those broken line. Read my answer. – Timbuc Mar 16 '15 at 14:50

1 Answers1

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Let me see: the lines are

$$L_1 \;:\;\;\left\{\;\left(\;x=\sqrt\lambda\,y+\sqrt\lambda-1\,,\,\,y\,,\,\,z=(\sqrt\lambda-1)y+\sqrt\lambda\;\right)\right\}=$$

$$=\left(\sqrt\lambda-1\,,\,\,0\,,\,\,\sqrt\lambda\right)+t\left(\sqrt\lambda\,,\,\,1\,,\,\,\sqrt\lambda-1\right)\;,\;\;t\in\Bbb R$$

$$L_2\;:\;\;\left\{\;\left(\;x=\sqrt\mu\,y+1-\sqrt\lambda\,,\,\,y\,,\,\,z=(1-\sqrt\mu)y+\sqrt\mu\;\right)\right\}=$$

$$=\left(1-\sqrt\mu\,,\,0\,,\,\,\sqrt\mu\right)+t\left(\sqrt\mu\,,\,1\,,\,1-\sqrt\mu\right)$$

Thus, $\;L_1\perp L_2\iff\;$ their direction vectors are perpendicular, iff

$$0=\left(\sqrt\lambda\,,\,\,1\,,\,\,\sqrt\lambda-1\right)\cdot\left(\sqrt\mu\,,\,1\,,\,1-\sqrt\mu\right)=\sqrt{\lambda\mu}+1+\sqrt\lambda-\sqrt{\lambda\mu}-1+\sqrt\mu\iff$$

$$\iff\sqrt\lambda=-\sqrt\mu$$

Timbuc
  • 34,191
  • so which option out of four do you see as correct? – RE60K Mar 17 '15 at 03:43
  • Of the four you wrote in the bottom of the question? None, of course...but I am now wondering whether you meant to write $$x=\sqrt\lambda,y+(\sqrt\lambda-1)z+\ldots$$and not what I wrote in my answer. You really should not break definitions that way and jump lines! – Timbuc Mar 17 '15 at 07:40