Show that the four points A, B, C and D with position vectors
and
respectively are coplanar if and only if ![]()
Given that,
![]()
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![]()
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⇒
Let A, B, C and D be coplanar.
As we know that, the vectors
,
,
and
will be coplanar if there exist scalar x, y, z, u not all zero simultaneously such that
.
Then, we can write
![]()
Where, (x + y + z + u) = 0
Provided x, y, z, u are scalars not all simultaneously zero.
Let x = 3, y = –2, z = 1 and u = –2
So, we get
![]()
Thus, A, B, C and D are coplanar if
.
⇐
If
is true.
Rearranging it, we get
![]()
Dividing this from the sum of its coefficient (that is, 4) on both sides,
![]()
Or ![]()
⇒ There is a point say P, which divides the line AC in ratio 1:3 and BD in ratio 2:2 internally.
Thus, P is the point of interaction of AC and BD.
As, vectors parallel to the same plane, or lie on the same plane are called coplanar vectors.
Hence, A, B, C and D are coplanar.
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