A,B C and dD are four points such that `vec (AB) = m(2 hati - 6 hatj + 2hatk) vec(BC) = (hati - 2hatj) and vec(CD) = n (-6 hati + 15 hatj - 3 hatk)`. If CD intersects AB at some points E, then

Options

A. ` m ge 1//2`
B. `n ge 1//3`
C. m= n
D. `m lt n`

Answer

a,b

Explanation

Let `vec(EB) =p, vec(AB) and vec(CE) =q vec(CD)`
Then ` 0 lt p and q le 1`
since ` vec(EB) + vec(BC) + vec(CE) =vec0`
`pm (2hati - 6hatj + 2hatk ) + (hati -2hatj)`
`+ qn (- 6hati + 15 hatj - 3hatk ) =0`
`or (2pm + 1 - 6qn) hati + ( - 6pm -2 + 15 qn) hatj`
` + 2pm - 6qn= vec0`
` Rightarrow 2pm - 6qn + 1 = vec0 , - 6pm - 2+ 15 qn = vec0`
` 2pm -6qn = vec0`
solving these, we get
p = 1/ (2m) and q =1 /(3n)
` 0 lt 1 // (2m) le 1 and 0 lt 1//(3n) le 1`
` Rightarrow m ge 1//2 and n ge 1//3`