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Question
Find the values of p so the line `(1-x)/3 = (7y-14)/2p = (z-3)/2` and `(7-7x)/(3p) = (y -5)/1 = (6-z)/5` are at right angles.
Solution
The given equations can be written in the standard form as
`(x - 1)/-3 = (y - 2)/((2p)/7) = (z - 3)/2 "and" (x - 1)/((-3p)/7) = (y - 5)/1 = (z - 6)/-5`
The direction ratios of the lines are `<−3,(2p)/7, 2>` and `<(-3p)/7, 1,-5>` respectively.
Two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are perpendicular to each other, if a1a2 + b1 b2 + c1c2 = 0
∴ `(-3).((-3p)/7) + ((2p)/7).(1) + 2 . (-5) = 0`
⇒ `(9p)/7 + (2p)/7 - 10 = 0`
⇒ `(11p)/7 = 10`
⇒ 11p = 70
⇒ p = `70/11`
Thus, the value of p is `70/11`.
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