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The dr's of two lines are given by a+b+c...

The dr's of two lines are given by `a+b+c=0,2ab +2ac-bc=0`. Then the angle between the lines is

A

` pi `

B

` ( 2pi ) / ( 3 ) `

C

` ( pi )/ ( 2 ) `

D

` ( pi ) / ( 3 ) `

Text Solution

Verified by Experts

The correct Answer is:
B
We have ` a + b + c = 0 and 2 ab + 2ac - bc = 0 `
` rArr a =- ( b + c ) and 2 a ( b + c ) - bc = 0 `
` rArr 2 b ^ 2 - 5 b c + 2 c ^ 2 = 0 `
` rArr ( 2 ab + c ) ( b + 2c ) = 0 `
` 2b + c = 0 ` then a ` = - ( b + c ) rArr a = b `
` therefore a = b and c = - 2 b rArr ( a ) / ( 1 ) = ( b ) / ( 1 ) = ( c ) / ( - 2 ) `
If ` b + 2 c = 0 `, then ` a = - ( b + c ) rArr a = c `
` therefore a = c and b = - 2c rArr ( a ) / ( 1 ) = ( b ) / ( - 2 ) = ( c ) / (1 ) `
Thus, direction ratios of two lines are ` ( 1, 1, - 2 ) and ( 1 , - 2 , 1 ) ` respectively.
Angle between lines
` cos theta = ( 1 - 2 - 2 ) / ( sqrt ( 1 + 1 + 4 ) sqrt ( 1 + 1 + 4 )) = - ( 1 ) / ( 2 ) `
` rArr theta = cos^ ( - 1 ) (- ( 1 ) / ( 2 ) ) = ( 2pi ) / ( 3 ) `
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Knowledge Check

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