Find the coordinates of the foot of perpendicular drawn from the point `2hati - hatj + hat5k` to the line `bar(r) = (11hati + 2hatj - 8hatk) + lamda ( 10 hati - 4hatj - 11hatk)`. Also , find the length of perpendicular .

Explanation

Let M be the foot of perpendicular drawn from the point
`P( 2hati - hatj + 5hatk) " on the line " bar(r) = ( 11hati - hat2j- hat8k) + lamda ( 10 hati - 4hatj - 11hatk )`.
Let the position vector of the point M be
` ( 11 hati - 2hatj - 8 hatk ) + lamda( 10 hati - 4hatj - 11hatk) = ( 11 + 10 lamda)hati + ( -2 - 4 lamda) hatj + ( -8 - 11lamda)hat k `.
Then `bar(PM)` = Position vector of M - position vector of P
` = [(11 + 10lamda)hati + (-2 - 4lamda)hatj + ( - 8 - 11 lamda )hatk ] - ( 2hati - htj + 5hatk )`
` = (9 + 10 lamda ) hat i + (-1 - 4lamda)hatj + ( -13 - 11 lamda )hatk `
since PM is perpendicular to the given line which is parrallel to
`bar(b) = 10hati - 4hatj - 11hatk , bar(PM) bot^(r) bat(b) therefore bar (PM) . bar(b) = 0`
`therefore [ (9 + 10lamda)hati + (-1 - 4lamda) hatj + ( -13 - 11 lamda )hatk].(10hati - 4hatj - 11hatk ) = 0`
`therefore 10(9 + 10 lamda) - 4(-1 - 4lamda ) - 11 (-13 - 11 lamda ) = 0 `
`therefore 90 + 100 lamda + 4 + 16lamda + 143 + 121lamda = 0`
` therefore 237 lamda + 237 = 0 therefore lamda = -1`
Putting this value of `lamda` , we get the position vector of M as ` hati + hat2j + hat3k `.
`therefore` coordinates of the foot of perpendicular M are ( 1, 2.3).
Now , `bar(PM) = (hati + 2hatj + 3 hat k ) - ( 2 hati - hatj + 5hatk ) = -hati + 3hatj - 2hatk`
`therefore |bar(PM)|= sqrt((-1)^(2) + (3)^(2) + (-2)^(2)) = sqrt( 1+ 9 + 4) = sqrt(14)`
Hence, the coordinates of the foot of perpendicular are (1,2,3) and length of perpendicular = `sqrt(14)` units.