JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 12)
If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :
ab' + bc' + 1 = 0
cc' + a + a' = 0
bb' + cc' + 1 = 0
aa' + c + c' = 0
Explanation
Equation of 1st line is
$${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$$
Dr's of 1st line = ($$a$$, 1 , c)
Equation of 2nd line is
$${{x - b'} \over {a'}} = {{y - b'} \over {c'}} = {z \over 1}$$
Dr's of 2nd line = ($$a'$$, c' , 1)
Lines are perpendicular, so the dot product of the Dr's of two lines are zero.
$$ \therefore $$ $$aa$$' + c + c' = 0
$${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$$
Dr's of 1st line = ($$a$$, 1 , c)
Equation of 2nd line is
$${{x - b'} \over {a'}} = {{y - b'} \over {c'}} = {z \over 1}$$
Dr's of 2nd line = ($$a'$$, c' , 1)
Lines are perpendicular, so the dot product of the Dr's of two lines are zero.
$$ \therefore $$ $$aa$$' + c + c' = 0
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