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Prove that the points having position vectors \(\hat i+2\hat j+3\hat k\)\(3\hat i+4\hat j+7\hat k\) \(-3\hat i-2\hat j-5\hat k\) are collinear.

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Let us understand that, two more points are said to be collinear if they all lie on a single straight line

 Let the points be A, B and C having position vectors such that,

Position vector of A = \(\vec i+2\vec j+3\vec k\)

Position vector of B = \(3\vec i+4\vec j+7\vec k\)

 Position vector of C = \(-3\vec i-2\vec j-7\vec k\)

So, in this case if we prove that \(\vec{AB}\) and \(\vec{BC}\) are parallel to each other, then we can easily show that A, B and C are collinear.

Therefore\(\vec{AB}\) is given by

\(\vec{AB}\) = Position vector of C - Position vector of B

Let us note the relation between \(\vec{AB}\) and \(\vec{BC}\).

We know,

This relation shows that \(\vec{AB}\) and \(\vec{BC}\) are parallel to each other.

But also, \(\vec B\) is the common vector in \(\vec{AB}\) and \(\vec{BC}\).

⇒ \(\vec{AB}\) and \(\vec{BC}\) are not parallel but lies on a straight line.

Thus, A, B and C are collinear.

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