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1.

Findthe value of `lambda`sothat the following vectors are coplanar:` vec a= hat i- hat j+ hat k , vec b=2 hat i+ hat j- hat k , vec c=lambda hat i- hat j+lambda hat k`

Answer» Three vectors are coplanar when their scalar triple product is `0.`
`=> [veca vecb vecc] = 0`
`=>|[1,-1,1],[2,1,-1],[lambda,-1,lambda]| = 0`
`=>[1(lambda-1)+1(3lambda)+1(-2lambda -1)] = 0`
`=>lambda-1+3lambda-2lambda-1 = 0`
`=> 2lambda = 2`
`=>lambda = 1`, for which these three vectors will be coplanar.
Discussion
2.

Find the volume of the parallelepiped whosecoterminous edges are represented bythe vectors:` vec a=2 hat i+3 hat j+4 hat k , vec b= hat i+2 hat j- hat k , vec c=3 hat i- hat j+2 hat k`` vec a=2 hat i+3 hat j+4 hat k , vec b= hat i+2 hat j- hat k , vec c=3 hat i- hat j-2 hat k`` vec a=11 hat i , vec b=2 hat j- hat k , vec c=13 hat k`` vec a= hat i+ hat j+ hat k , vec b= hat i- hat j+ hat k , vec c= hat i+2 hat j- hat k`

Answer» Volume of the parallelepiped can be given as triple product of three vectors.
(i)`V = [veca vecb vecc]`
`:. V = |[2,3,4],[1,2,-1],[3,-1,2]|`
`=>V = 2(3)-3(5)+4(-7)`
`=>V = -37`
But, volume can not be negative.
So, required volume is `37` cubic units.
(iii) `V = [veca vecb vecc] = [11hati` `2hatj` `13hatk] = (11hati xx 2hatj)*13hatk`
`=> V =(22hatk)*13hatk = 286` cubic units.
Similarly, we can do part(ii) and part (iv).
Discussion