Show that the lines `bar (r) = ( hati + hatj - hatk) + lamda ( 3hati - hatj )` and ` bar(r) = ( 4hati - hatk ) + mu (2hati + 3hatk)` intersect. Find their point of intersection .

Explanation

The position vector of any point on the line
`bar (r) = ( hati + hatj - hatk ) + lamda ( 3hati - hatj ) ` is ` (hati + hatj - hatk ) + lamda(3hati - hatj ) = ( 1 + 3lamda)hati + (1 - lamda)hat j - hatk`
The position vector of any point on the line `bar(r) = ( 4hati - hatk) + mu ( 2hati + 3hatk ) = ( 4 + 2 mu ) hati + (-1 + 3 mu)hatk `
If the lines intersect , then they have common point . Therefore , for some values of `lamda "and mu` , we have ,
` ( 1 + 3 lamda )hati + ( 1 - lamda)hatj - hatk = ( 4 + 2 mu)hati + ( - 1 + 3mu) hatk` .
on equating the coefficients of `hati , hatj , hatk ` , we get ,
` 1 + 3 lamda = 4 + 2mu` ....(1)
` 1 - lamda = 0 ` ...(2)
and ` -1 = -1 + 3 mu` ....(3)
Solving equations (2) and (3) , we get ,
` lamda = 1 "and" mu = 0 `
These values of `lamda "and" mu` satisfy the equation (1) .
`therefore` the given lines intersect .
Putting `lamda = 1 in bar(r) = (hati + hatj - hatk) + lamda( 3 hati - hat j ) or mu = 0 in bar(r) = ( 4hati - hatk) + mu ( 2hati + 3hatk ) `, we get , the position vector of the point of intersection `bar(r) = 4(hati + o.hatj - hatk)`
`therefore` the coordinates of the point of intersection are ( 4,0, -1).