The vertex A of triangle ABC is on the line `vecr=hati+hatj+lambda hatk` and the vertices B and C have respective position vectors `hati and hatj `. Let `Delta` be the area of the triangle and `Delta in [3//2,sqrt33//2]` then the range of value of `lambda` corresponding to A is

Options

A. [-8, -4]cup[4,8]`
B. `[-4,4]`
C. [-2,2]
D. `[-4,-2] cup[2,4]`

Answer

c

Explanation

` triangle=1/2|(hatj +lambdahatk) xx(hati+lambdahatk)|=1/2|=hatk+lambdahati+lambdahatj| =1/2 sqrt(2lambda^(2)+1)`
` Rightarrow 9/4 ge 1/4 (2lambda^(2) +1) ge 33/4`
` or 4 ge lambda^(2)16`
`or 2 ge |lambda|ge 4`