Show that the points A(2, –1, 0) B(–3, 0, 4), C(–1, –1, 4) and D(0, – 5, 2) are non coplanar
Solution
Let `bar"a", bar"b", bar"c", bar"c", bar"d"` be the position vectors of points A, B, C, D respectively.
∴ `bar"a" = 2hat"i" - hat"j", bar"b" = -3hat"i" + 4hat"k", bar"c" = -hat"i" - hat"j" + 4hat"k", bar"d" = -5hat"j" + 2hat"k"`
∴ `bar"AB" = bar"b" - bar"a"`
= `(-3hat"i" + 4hat"k") - (2hat"i" - hat"j")`
= `-5hat"i" + hat"j" + 4hat"k"`
`bar"AC" = bar"c" - bar"a"`
= `(-hat"i" - hat"j" + 4hat"k") - (2hat"i" - hat"j")`
= `-3hat"i" + 4hat"k"`
`bar"AD" = bar"d" - bar"a"`
= `bar"AD" = bar"d" - bar"a"`
= `(-5hat"j" + 2hat"k") - (2hat"i" - hat"j")`
= `-2hat"i" - 4hat"j" + 2hat"k"`
Points A, B, C, D are non-coplanar if `bar"AB", bar"AC"` and `bar"AD"` are non-coplanar.
`bar"AD"*(bar"AC" xx bar"AD") = |(-5, 1, 4),(-3, 0, 4),(-2, -4, 2)|`
= – 5(0 + 16) – 1(– 6 + 8) + 4(12 – 0)
= – 5(16) – 1(2) + 4(12)
= – 80 – 2 + 48
= – 34 ≠ 0
∴ The points A, B, C, D are non-coplanar.