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Last updated date: 13th Jun 2024
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Answer
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Hint: We start solving by recalling the definition of collinear vectors that they line on the same line or parallel lines. We use the fact that the components of one of the collinear vectors is equal to the multiples of another vector. We use the fact that the cross product of a collinear vector is zero to prove all the conditions about the collinear vectors.

Complete step-by-step answer:
Collinear vectors: - Vectors parallel to one or lying on one line are called collinear vectors.
Condition of collinearity: - Two vectors are collinear if any of these conditions are done.
Condition-1:- Two vectors a, b are collinear if there exists a number such that the below equation will become true. $\bar{a}=n.\bar{b}$
Condition-2:- Two vectors are collinear if the relation of their coordinates are equal.
This is not valid if one of the components is zero.
Condition-3:- Two vectors are collinear if their cross product is equal to the zero vector.
This is valid only in the case where 2 vectors are three-dimensional (spatial) vectors.
Cross-product:- Cross product of vector a by vector b is the vector c, the length of which is numerically equal to the area of parallelogram constructed on vector a, b, direction is perpendicular to the plane of the vectors of a, b. If a, b vectors are written as $xi+yj+zk;\text{ pi+qj+rk}$, we get cross product a, b represented by $a\times b$ as:
$a\times b=\left|\begin{matrix}
  &i &j &k \\
  &x &y &z \\
  &p &q &r \\
\end{matrix} \right|$

Apply this definition to condition-3 we get:
Cross product a, b is 0. From condition 1, we get:
$a=nb$. If $b=xi+yj+zk,$ we get value of a as,
$a=nxi+nyj+nzk$.
Cross product of $a\times b$ is written as:
$a\times b=\left| \begin{matrix}
  &i &j &k \\
  &nx &ny &nz \\
  &x &y &z \\
\end{matrix} \right|$
By expanding this, we get it as follows:

\[\begin{align}
 & a\times b=\left( nzy-nzy \right)i-\left( nxz-nxz \right)j+\left( nxy-nxy \right)k \\
 & a\times b=oi-oj+ok\text{ = zero vector}\text{.} \\
\end{align}\]
Hence proved.

Note: Be careful with the second condition. If a term is zero in one vector that condition will go wrong. While proving the condition-3 we must take care of a, b. Alternately, assume a as (x, y, z), by this you get b as $\left( \dfrac{x}{n},\dfrac{y}{n},\dfrac{z}{n} \right)$ . Substitute these, anyways you get the same answer.