Given that `vecA,vecB,vecC` form triangle such that `vecA=vecB+vecC`. Find a,b,c,d such that area of the triangle is `5sqrt(6)` where `vecA=aveci+bveci+cveck. vecB=dveci+3vecj+3veck and vecC=3veci+vecj-2veck`.

Answer

a= -8, b=4, c=2, d =-11

Explanation

Here, `vecA,vecB and vecC` are the vectors representing the sides of triangle ABC, where `vecA= ahati + bhatj +chatk`
`vecB = hati + 3hatj + 4hatk and vecC = 3hati + hatj - 2hatk`
Given that `vecA =vecB +vecC`, Therefore,
`ahati + bhatj + chatk = (d+3) hati + 4hatj + 2hatk`
` Rightarrow a=d +3, b=4,c=2`
`vecB xxvecC= |{:(hati ,hatj ,hatk),(d,3,4),(3,1,-2):}|`
`10 hati + (2d + 12) hatj + (d -9) hatk`
Area of `triangle = 1/2 |vecB xx vecC|`
`1/2 sqrt([100 + (2d + 12)^(2) + (d-9)^(2)])`
`5 sqrt6`
` or sqrt((5d^(2)+30d + 325)) = 10 sqrt6`
`or 5d^(2) + 30d - 275 = 0 or d^(2) + 6 d - 55 =0`
`or (d +11) (d-5) =0
`Rightarrow d=5 or -11 `
where d= 5 a= 8, b= 4 and c=2 and where d = -11 , a = -8 ,b =4 and c= 2