Let `vec a=hat i+hat j+hat k and vec b=i and vec c=c_1 hat i + hat c_2 j +c_3 hat k` Then(a) if `c_1 = 1 and c_2=2` , find `c_3` which makes ` vec a , vec b, vec c ` coplanar(b) if `c_2 = -1 and c_3= 1` , show that no value of `c_3` can makes ` vec a , vec b, vec c ` coplanar.

Let `vec a=hat i+hat j+hat k and vec b=i and vec c=c_1 hat i + hat c_2

1.	Let `vec a=hat i+hat j+hat k and vec b=i and vec c=c_1 hat i + hat c_2 j +c_3 hat k` Then(a) if `c_1 = 1 and c_2=2` , find `c_3` which makes ` vec a , vec b, vec c ` coplanar(b) if `c_2 = -1 and c_3= 1` , show that no value of `c_3` can makes ` vec a , vec b, vec c ` coplanar.
Answer» Here, `veca = hati+hatj+hatk` `vecb = hati` `vecc = c_1hati+c_2hatj+c_3hatk` (i) As these trhree vectors are coplanar, ,their scalar product should be `0`. `:. \|[1,1,1],[1,0,0],[c_1,c_2,c_3]\| = 0` `=>-c_3+c_2 = 0` `:.c_3 = c_2 = 2` (ii)From first part, `[vecavecbvecc] = -c_3+c_2` `= -1+1 = -2` So, scalar product of these three vectors is always `-2`. So, no value of `c_1` can make these vectors coplanar.

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