Find the equation of the perpendicular from point `(3,-1,11)` to line `x/2=(y-2)/3=(z-3)/4`. Also, find the coordinates of foot of perpendicular and the length of perpendicular.

Options

A. ` sqrt (66) `
B. ` sqrt ( 29 ) `
C. `sqrt(33) `
D. ` sqrt ( 53 ) `

Answer

D

Explanation

Given equation of line is ` ( x ) / ( 2 ) = ( y -2 ) / (3 ) = ( z - 3 ) / ( 4 ) = lamda `

Then coordinates of any point on the given line is
` Q ( 2lamda, 3lamda + 2, 4lamda + 3 ) ` .
` therefore ` Direction ratios of PQ is ` ( 2lamda - 3, 3lamda + 3 , 4lamda - 8 ) ` and it is perpendicular to straight line AB.
` therefore 2 ( 2 lamda - 3 ) + 3 ( 3 lamda + 3 ) + 4 ( 4 lamda - 8 ) = 0 `
` [ because a _ 1 a _ 2 + b _ 1 b _ 2 + c _ 1 c _ 2 = 0 ] `
` rArr 4 lamda - 6 + 9 lamda + 9 + 16 lamda - 32 = 0 `
` rArr 29 lamda - 29 = 0 rArr lamda = 1 `
Coordinates of Q are ( 2, 5, 7 )
Now, perpendicular distance
` |PQ| = sqrt (( 3 - 2) ^ 2 + ( - 1 - 5 ) ^ 2 + ( 7 -11) ^ 2 ) `
` = sqrt ( 1 + 36 + 16 ) = sqrt ( 53 ) `