Let `veca = hati + hatj + hatk = hati` and `vecc = c_1veci + c_2hatj + c_3hatk` then

1) Let `c_1 = 1` and `c_2 = 2`, find `c_3` which makes `veca, vecb "and" vecc`coplanar

2) if `c_2 = -1` and `c_3 = 1`, show that no value of `c_1`can make `veca, vecb and vecc` coplanar

#### Solution

`veca = hati + hatj + hatk = hati` and `vecc = c_1veci + c_2hatj + c_3hatk`

Let c_{1} = 1 and c_{2} = 2

`vecc = hati + 2hatj + c_3hatk`

For vectors to be coplanar scalar triple product should be equal to zero.

`:. veca*(vecb xx vecc) = 0`

`=> (hati + hatj + hatk)*[hati xx (hati + 2hatj+c_3hatk)] = 0`

`=> (hati + hatj + hatk)*(-c_3hatj + 2hatk) = 0`

⇒ 0 - c_{3}+ 2 =0

⇒ c_{3} = 2

2) If c_{2} = –1 and c_{3} = 1

Let `veca, vecb and vecc` be coplanar

For vectors to be coplanar scalar triple product should be equal to zero.

`:. veca*(vecb xx vecc) = 0`

`(veci + hatj + hatk)*[hati xx c_1hati - hatj + hatk)]`

`[hati xx (c_1hati - hatj + hatk)] = [(hati, hatj, hatk),(1,0,0),(c_1, -1, 1)]`

`= hati(0 - 0) - hatj(1-0) + hatk(-1-0)`

`= -hatj - hatk`

So, here we can see that the value of the vector product of `vecb` and `vec c` does involve `c_1`

Therefore we can say that there is no value of `c_1` can make `veca, vecb " and " vecc` coplanar