Find the absolute value of parameter `t` for which the area of the triangle whose vertices the `A(-1,1,2); B(1,2,3)a n dC(5,1,1)` is minimum.

Answer

2

Explanation

`vec(AB)=2hati+hatj +hatk,vec(AC)= (t+1)hati +0hatj -hatk`
`vec(AB)xxvec(AC)= |{:(hati,hatj,hatk),(2,1,1),(t+1,0,-1):}|`
`=- hati+ (t+3)hatj-(t-1)hatk`
` sqrt(1+(t+3)^(2)+(t+1)^(2))`
`sqrt(2t^(2)+8t +11)`
Area of `triangleABC= 1/2|vec(AB) xx vec(AC)|`
`1/2 sqrt(2t^(2)+8t +1)`
Let `f(t)=Delta^(2)=1/4(2t^(2)+8t=1)`
`f'(t)=0 Rightarrow t = -2`
At `t=-2,f'' (t)gt0`
so `Delta` is minimum at t = -2