If $\vec{a}$ , $\vec{b}$ , $\vec{c}$ are non-coplanar vectors, prove that the points having the following position vectors are collinear:
(i) $\vec{a,} \vec{b,} 3 \vec{a} - 2 \vec{b}$

(ii) $\vec{a} + \vec{b} + \vec{c}, 4 \vec{a} + 3 \vec{b}, 10 \vec{a} + 7 \vec{b} - 2 \vec{c}$

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Solution

(i) Given: $\vec{a}, \vec{b}, \vec{c}$ are non coplanar vectors.
Let the points be $A, B, C$ respectively with position vectors $\vec{a}, \vec{b,} 3 \vec{a} - 2 \vec{b} .$ Then,
$\vec{A B} =$ Position vector of B $-$ Position vector of A
$= \vec{b} - \vec{a}$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
$= 3 \vec{a} - 2 \vec{b} - \vec{b} = 3 \vec{a} - 3 \vec{b} = - 3 (\vec{b} - \vec{a})$
∴ $\vec{B C} = - 3 \vec{A B}$
$So, \vec{A B} and \vec{B C}$ are parallel vectors.
But $B$ is a point common to them.
Hence, $A, B$ and $C$ are collinear.

(ii) Given $\vec{a}, \vec{b}, \vec{c}$ are non coplanar vectors.
Let the points be $A, B, C$ respectively with the position vectors $\vec{a} + \vec{b} + \vec{c}, 4 \vec{a} + 3 \vec{b}, 10 \vec{a} + 7 \vec{b} - 2 \vec{c} .$ Then,
$\vec{A B} =$ Position vector of B $-$ Position vector of A
$= 4 \vec{a} + 3 \vec{b} - \vec{a} - \vec{b} - \vec{c} = 3 \vec{a} + 2 \vec{b} - \vec{c}$

$\vec{B C} =$ Position vector of C $-$ Position vector of B
$= 10 \vec{a} + 7 \vec{b} - 2 \vec{c} - 4 \vec{a} - 3 \vec{b} = 6 \vec{a} + 4 \vec{b} - 2 \vec{c} = 2 (3 \vec{a} + 2 \vec{b} - \vec{c})$

∴ $\vec{B C} = 2 \vec{A B}$
$So, \vec{A B} and \vec{B C}$ are parallel vectors.
But $B$ is a point common to them.
So, $\vec{A B}$ and $\vec{B C}$ are collinear.
Hence, $A, B$ and $C$ are collinear.

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