# Prove that the points having position vectors i+2j + 3k, 3i + 4j + 7k, -3i -2i - 5k are collinear.

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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.