A) \[\alpha \alpha '+\beta \beta '+1=0\]
B) \[(\alpha +\alpha ')+(\beta +\beta ')=0\]
C) \[\ell \ell '+mm'+nn'=1\]
D) \[\ell \ell '+mm'+nn'=0\]
Correct Answer: D
Solution :
[d] Given two lines are: \[y=\frac{mx}{\ell }+\alpha ,z=\frac{n}{\ell }x+\beta \] and \[y=\frac{m'}{\ell '}x+\alpha ',z=\frac{n'}{\ell '}x+\beta '\] These two lines can be represented as: \[\frac{y-\alpha }{m/\ell }=\frac{x-0}{1}=\frac{z-\beta }{n/\ell }\] And \[\frac{y-\alpha '}{m'/c'}=\frac{x-0}{1}=\frac{z-\beta '}{n'/\ell '}\] They are orthogonal, if \[\frac{m}{\ell }\times \frac{m'}{\ell '}+1\times 1+\frac{n}{\ell }\frac{n'}{\ell '}=-1\Rightarrow \ell \ell '+mm'+nn'=0\]
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