find n if position vectors 3i-2j-k,2k+3j-4k,-i+j+2k,4i+5j+nk line on the same plane

Question

find n if the position vector of A,B,C,D, are 3i-2j-k,2k+3j-4k,-i+j+2k,4i+5j+nk line on the same plane

Answer 1

Position vectors of the points A,B,C,D are

Since A,B,C,D are coplanar

So,

Now

                        

                        

Now

The value of n is .

Answer 2

The position vectors are A = 3i - 2j - k, B = 2k + 3j - 4k, C = - i + j + 2k and D = 4i + 5j + nk.

The position vectors are same plane that means coplanar.

Find the all three vectors from A to B or C or D.

AB = B - A = (0i + 3j - 2k) - (3i - 2j - k) = - 3i + 5j - k.

AC = C - A = (- i + j + 2k) - (3i - 2j - k) = - 4i + 3j + 3k.

AD = D - A = (4i + 5j + nk) - (3i - 2j - k) = i + 7j + (n + 1)k.

Cross two of them and dot it with the third. If you get zero then they're coplanar, and hence the points A, B, C and D were coplanar too.

Therefore (AB×AC) · AD = 0.

Find AB×AC.

               i     j      k

AB×AC = -3    5    -1

              -4    3     3

              = i(15 + 3) - j(- 9 - 4) + (- 9 + 20)k

              = 18i + 13j + 11k.

(AB×AC) · AD = 0

(18i + 13j + 11k) · (i + 7j + (n + 1)k) = 0

18 + 91 + 11(n + 1) = 0

11(n + 1) = - 109

n + 1 = - 109/11

n = - (109/11) - 1 = - (109 + 11)/11 = - 120/11.

The value of n = - 120/11.