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. `nge(1)/(3)`
C. `m=n`
D. `m lt n`

Answer

A, B

Explanation

Let EB=p AB and CE =qCD
Then `0 lt p and q le1`

Since, EB+BC+CE=0
`p m(2hati-6hatj+2hatk)+(2hati-2hatj)+qn(-6hati+15hatj-3hatk)=0`
`implies(2p m+1-6qn)hati+(-6p m-2+15qn)hatj+(2p m-6qn)hatk=0`
`implies2p m-6qn+1=0`,
`-6p m-2+15qn=0`
`2 p m-6qn=0`
Solving these, we get
`p=(1)/((2m)) and q=(1)/((3n))`
`therefore 0 lt (1)/((2m)) le 1 and 0 lt (1)/((3n)) le 1`
`implies m ge (1)/(2) and n ge (1)/(3)`