Show that the following points are collinear:
P = (4, 5, 2), Q = (3, 2, 4), R = (5, 8, 0).
Solution
Let `bar"a" , bar"b" , bar"c"` be position vectors of the points.
P = (4, 5, 2), Q = (3, 2, 4), R = (5, 8, 0) respectively.
Then `bar"a" = 4hat"i" + 5hat"j" + 2hat"k" , bar"b" = 3hat"i" + 2hat"j" + 4hat"k" , bar"c" = 5hat"i" + 8hat"j" + 0hat"k"`
`bar"AB" = bar"b" - bar"a"`
`= (3hat"i" + 2hat"j" + 4hat"k") - (4hat"i" + 5hat"j" + 2hat"k")`
`= - hat"i" - 3hat"j" + 2hat"k"` i.e.
`= - (hat"i" + 3hat"j" - 2hat"k")` .....(1)
and `bar"BC" = bar"c" - bar"b"`
`= (5hat"i" + 8hat"j" + 0hat"k") - (3hat"i" + 2hat"j" + 4hat"k")`
`= 2hat"i" + 6hat"j" - 4hat"k"`
`= 2(hat"i" + 3hat"j" - 2hat"k")`
`= 2 . bar"AB"` ....[By(1)]
∴ `bar"BC"` is a non-zero scalar multiple of `bar"AB"`
∴ they are parallel to each other.
But they have point B in common.
∴ `bar"BC"` and `bar"AB"` are collinear vectors.
Hence, the points A, B, and C are collinear.
