Find the vector and the cartesian equations of the line that passes through the points`(3, 2, 5), (3, 2, 6)`.

Options

A. ` r = 3 hati - 2 hatj - 5 hatk , x - 3 = y + 2 = ( z + 5 ) / ( 11 ) `
B. ` r = 3 hati - 2 hatj - 5 hatk , ( x - 3 ) / ( 0 ) = ( y + 2 ) /( 0 ) = ( z + 5 )/ ( 11 ) `
C. ` r = 3hati - 2 hatj - 5 hatk + lamda ( 11 hatk ) , x - 3 = y + 2 = ( z + 5 ) / (11 ) `
D. ` r = 3 hati - 2 hatj - 5 hatk + lamda ( 11 hatk ) , ( x -3 ) /( 0 ) = ( y + 2 ) /( 0 ) = ( z + 5 ) /( 0 ) `

Answer

D

Explanation

We have, ` a= 3hati - 2 hatj - 5hatk `
and ` b = 3hati - 2 hatj + 6 hatk `
we know that, the vector equation of a line passing through the point having position vectors a and b is
` r = 3 hati - 2hatj - 5 hatk + lamda [ ( 3 hati - 2 hatj + 6 hatk ) - ( 3 hati - 2hatj - 5 hatk ) ] `
` rArr r = 3hati - 2 hatj - 5hatk + lamda ( 11 hatk ) " " ...(i)`
[which is the vector equation]
Putting ` r = x hati + y hatj + z` hatk , in Eq. (i) , we have
` x hati + y hatj + z hatk = 3 hati - 2 hatj + (11 lamda - 5 ) hatk `
On comparing coefficients of ` hati , hatj and hatk ` on both sides, we get
` x= 3, y = - 2 `
and ` z = 11 lamda - 5 `
` therefore (x - 3 ) / ( 0 ) = ( y + 2 ) / 0 = ( z + 5 ) / ( 11 ) `
which is the cartesian form of required line.