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CBSE Class 12 Math 2020 Outside Delhi Set 1 Solved Paper

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Question : 6 of 36
Marks: +1, -0
The two lines x=ay+b,  z=cy+dx=ay+b,\;z=cy+d; and x=a′y+b′,  z=c′y+d′x=a' y+b',\;z=c' y+d' are perpendicular to each other, if
Solution:     👈: Video Solution
Fact: Two lines   x−x1a1=  x−y1b1=  z−z1c1\;\frac{x-x_1}{a_1}=\;\frac{x-y_1}{b_1}=\;\frac{z-z_1}{c_1}
and   x−x2a2=  y−y2b2=  z−z2c2\;\frac{x-x_2}{a_2}=\;\frac{y-y_2}{b_2}=\;\frac{z-z_2}{c_2}
are ⊥\perp if a1a2+b1b2+c1c2=0a_1 a_2+b_1 b_2+c_1 c_2=0
Given lines can be written as   x−ba=  y1=  z−dc\;\frac{x-b}{a}=\;\frac{y}{1}=\;\frac{z-d}{c} (i)
and   x−b′a′=  y1=  z−d′c′\;\frac{x-b'}{a'}=\;\frac{y}{1}=\;\frac{z-d'}{c'}
As lines are perpendicular
∴  aa′+1+cc′=0\therefore\; a a'+1+c c'=0
⇒  aa′+cc′=−1\Rightarrow\; a a'+c c'=-1
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