Are the four points A(1, -1, 1), B(-1, 1, 1), C(1, 1, 1) and D(2, -3, 4) coplanar? Justify your answer.
Solution
The position vectors `bar"a", bar"b", bar"c", bar"d"` of the points A, B, C, D are
`bar"a" = hat"i" - hat"j" + hat"k"`, `bar"b" = -hat"i" + hat"j" + hat"k", bar"c" = hat"i" + hat"j" + hat"k", bar"d" = 2hat"i" - 3hat"j" + 4hat"k"`
∴ `bar"AB" = bar"b" - bar"a"`
`= (- hat"i" + hat"j" + hat"k") - (hat"i" - hat"j" + hat"k")`
`= - 2hat"i" + 2hat"j"`
`bar"AC" = bar"c" - bar"a"`
`= (hat"i" + hat"j" + hat"k") - (hat"i" - hat"j" + hat"k") = 2hat"j"`
and `bar"AD" = bar"d" - bar"a" = (2hat"i" - 3hat"j" + 4hat"k") - (hat"i" - hat"j" + hat"k")`
`= hat"i" - 2hat"j" + 3hat"k"`
If A, B, C, D are coplanar, then there exist scalars x, y such that
`bar"AB" = "x".bar"AC" + "y".bar"AD"`
∴ `- 2hat"i" + 2hat"j" = "x"(2hat"j") + "y"(hat"i" - 2hat"j" + 3hat"k")`
∴ `- 2hat"i" + 2hat"j" = "y"hat"i" + (2"x" - 2"y")hat"j" + "3y"hat"k"`
By equality of vectors,
y = - 2 ....(1)
2x - 2y = 2 .....(2)
3y = 0 ....(3)
From (1), y = - 2
From (3), y = 0
This is not possible.
Hence, the points A, B, C, D are not coplanar.
