a vector is multiplied by a scalar k then its
magnitude changes k times. The direction of such a vector remains same if k
> 0 however it reverses if k < 0 and becomes indeterminate if k = 0. So
by multiplying with k we obtain a vector which has same line of support as
vector . These two vectors are called
collinear vectors. Note here that null vector is collinear to all the vectors.
Collinearity and linear
two vectors and are
collinear then they are linearly dependent and vice versa. On the other hand if
two vectors are non–collinear then they are linearly
independent and vice versa. In this case
Û x = 0 and y = 0
Methods to prove
vectors and are
collinear if there exists k ∈
R such that .
points A(), B(),
C()are collinear if there exists k∈R such that that
then A, B, C are collinear.
- A(), B(),
C() are collinear if there exists
scalars l, m, n, (not all zero) such that where
l + m + n = 0.
the above methods are equivalent and any of them can be utilized to prove the
collinearity. (of 2 vectors or 3 points).
vectors (free) are always coplanar.
always determine a unique plane. Hence any vector in that plane can be uniquely
represented as a linear combination of these two vectors. .
(obviously vectors are coplanar).
Methods to prove
vectors are coplanar if there exists l, m∈R such that i.e.,
one can be uniquely expressed as a linear combination of the other two.
are coplanar (necessary and
points A, B, Cand Dlie
in the same plane if there exist l, m∈R
such that i.e. .
= 0 then points A, B, C, D are
B, C, D are coplanar if there exists scalars k, l, m, n (not all zero), such
that where k + l + m + n = 0.
If the vectors , and are
coplanar (where x, y, z > 1) then prove that
Since vectors are coplanar we can find l and m such that
() + m()
1 –x = 1 – l – m, y = ,
1 – y = , 1 – z =
= = 1.
Vectors are such that
every pair is non collinear and the vector is
collinear with and the vector is collinear with . Prove that .
Given that ....(1)
\ ( putting that value of from (1) and (2) )
\ are non collinear so they are
linearly independent hence (3) will hold only if 1 + l = 0 and 1 + m = 0
&⇒ l = m = –1
the value of l in (1) we get .