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  • Help me understand! - Three Dimensional Geometry - JEE Main

The two lines x=ay+b,z=cy+d\; and\; x=a'y+b',z=c'y+d  will be perpendicular, if and only if

  • Option 1)

    aa'+bb'+cc'=0\; \;

  • Option 2)

    (a+a')(b+b')+(c+c')=0

  • Option 3)

    aa'+cc'+1=0

  • Option 4)

    aa'+bb'+cc'+1=0

 

Answers (1)

As we learnt in

Condition of perpendicularity -

Two lines will be perpendicular if 

\vec{b}\cdot \vec{b_{1}}= 0 or a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0 .

 

 

-

 

 Equation of lines : 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}'}

\vec{n_1}= a\hat{i}+\hat{j}+c\hat{k} ; \vec{n_2}= a'\hat{i}+\hat{j}+c'\hat{k}

\vec{n_1}.\vec{n_2}= aa'+1+ cc'=0

 


Option 1)

aa'+bb'+cc'=0\; \;

Incorrect option

Option 2)

(a+a')(b+b')+(c+c')=0

Incorrect option

Option 3)

aa'+cc'+1=0

Correct option

Option 4)

aa'+bb'+cc'+1=0

Incorrect option

Posted by

Vakul

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