Find the equation of the perpendicular drawn from (2,4,-1) to the line `(x+5)/1=(y+3)/4=(z-6)/3`.

Options

A. ` ( x - 2 ) / ( 6 ) = ( y - 4 ) /( 3 ) = ( z+ 1 ) / ( 2 ) `
B. ` ( x + 2 ) / ( 6 ) = ( y - 4 ) /( 3 ) = ( z + 1 ) /( 2 ) `
C. ` ( x + 2 ) /( - 6 ) = ( y - 4 ) / ( 3) = ( z + 1 ) / ( 2 ) `
D. ` ( x + 2 ) /( 6 ) = ( y + 4 ) / ( 3) = ( z+ 1 ) / ( 2 ) `

Answer

A

Explanation

Given equation of line in
` ( x + 5 ) / ( 1 ) = ( y+ 3 ) / ( 4 ) = ( z- 6 ) /( - 9 ) = lamda ` [say]
Any point on the line is ` ( lamda - 5, 4lamda - 3, - 9 lamda + 6 ) `
Now, directions ratio of PQ are
` ( lamda - 5 - 2, 4lamda - 3 - 4, - 9 lamda + 6 + 1 ) `
i.e, ` ( lamda - 7, 4lamda - 7, 9lamda + 7 ) `
Since, PQ is perpendicular to the given line,
` therefore 1 ( lamda - 5) + 4 ( 4lamda - 7 ) - 9 ( - 9 lamda + 7 ) = 0 `
` " "[ because a_ 1 a _ 2 + b _ 1 b _ 2 + c _ 1 c_ 2 = 0 ] `
` rArr 9 lamda - 96 = 0 rArr lamda = 1 `
` therefore ` Foot of perpendicular ` Q ( 1 - 5, 4 - 3 , - 9 + 6 ) `
i.e, `Q ( - 4, 1, - 3 ) `
The equation of the line passing through ` P ( 2, 4 , -1 ) and N ( -4, 1, - 3 ) ` is
` ( x - 2 ) / ( 6 ) = ( y - 4 ) /( 3 ) = ( z + 1)/( 2 ) `
or ` ( x - 2 ) / ( 6 ) = ( y - 4 ) / ( 3 ) = ( z + 1 ) / ( 2 ) `