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The two lines x=ay+b,z=cy+d\; and x=a'y+b',z=c'y+d'\; are perpendicular to each other if

  • Option 1)

    aa'+cc'=-1\;

  • Option 2)

    \; aa'+cc'=1\;

  • Option 3)

    \; \frac{a}{a'}+\frac{c}{c'}=-1\;

  • Option 4)

    \; \frac{a}{a'}+\frac{c}{c'}=1

 

Answers (1)

As we learnt in 

Condition for perpendicularity -

\vec{n}.\vec{n_{1}}= 0 or a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0

-

 

 L_{1}:\frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}

L_{2}:\frac{{x-b}'}{{a}'}=\frac{y}{-1}=\frac{{z-d}'}{{c}'}

{aa}'+1+{cc}'= 0


Option 1)

aa'+cc'=-1\;

Incorrect Option

 

Option 2)

\; aa'+cc'=1\;

Correct Option

 

Option 3)

\; \frac{a}{a'}+\frac{c}{c'}=-1\;

Incorrect Option

 

Option 4)

\; \frac{a}{a'}+\frac{c}{c'}=1

Incorrect Option

 

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Vakul

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