Q.Show that the points $A(-2\hat{i}+3\hat{j}+5\hat{k})$, $B(\hat{i}+2\hat{j}+3\hat{k})$ and $C(7\hat{i}-\hat{k})$ are collinear.

Concept: Collinearity Condition — three points are collinear if the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ (or $\overrightarrow{AC}$) are parallel, i.e., one is a scalar multiple of the other.

Step 1: Find $\overrightarrow{AB}$.

Step 2: Find $\overrightarrow{BC}$.

Step 3: Check if $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{AB}$. …

The idea behind collinearity is simple: if three points lie on the same straight line, then the vector from the first to the second must point in exactly the same (or exactly opposite) direction as the vector from the second to the third. In other words, $\overrightarrow{AB}$ and $\overrightarrow{BC}$ must be parallel. And two vectors are parallel precisely when one is a scalar multiple of the other.

Let’s check this condition step by step.

1. Write the position vectors clearly.

   We have:

   $A = -2\hat{i} + 3\hat{j} + 5\hat{k}$

   $B = \hat{i} + 2\hat{j} + 3\hat{k}$

   $C = 7\hat{i} + 0\hat{j} - \hat{k}$

   (Notice that $C$ has no $\hat{j}$ component — it’s $0\hat{j}$, which is fine.)

2. Find $\overrightarrow{AB}$.

   $\overrightarrow{AB} = \text{position of } B - \text{position of } A$

   $= (\hat{i} + 2\hat{j} + 3\hat{k}) - (-2\hat{i} + 3\hat{j} + 5\hat{k})$

   $= \hat{i} + 2\hat{j} + 3\hat{k} + 2\hat{i} - 3\hat{j} - 5\hat{k}$

   $= (1+2)\hat{i} + (2-3)\hat{j} + (3-5)\hat{k}$

   $= 3\hat{i} - \hat{j} - 2\hat{k}$

3. Find $\overrightarrow{BC}$.

   $\overrightarrow{BC} = C - B$

   $= (7\hat{i} + 0\hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})$

   $= 7\hat{i} - \hat{i} + 0\hat{j} - 2\hat{j} - \hat{k} - 3\hat{k}$

   $= 6\hat{i} - 2\hat{j} - 4\hat{k}$

4. Check if $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{AB}$.

   Compare components:

   $\overrightarrow{AB} = 3\hat{i} - \hat{j} - 2\hat{k}$

   $\overrightarrow{BC} = 6\hat{i} - 2\hat{j} - 4\hat{k}$

   Notice that $6 = 2 \times 3$, $-2 = 2 \times (-1)$, and $-4 = 2 \times (-2)$.

   So $\overrightarrow{BC} = 2 \cdot (3\hat{i} - \hat{j} - 2\hat{k}) = 2 \overrightarrow{AB}$. …

Method: Testing three points for collinearity

To decide whether three points lie on one straight line, turn them into two vectors sharing a point and check whether one is a scalar multiple of the other.

Steps

Step 1: Form two vectors from a shared point

From points $A$, $B$, $C$, build two direction vectors that have a point in common, e.g. $\overrightarrow{AB} = \vec{b}-\vec{a}$ and $\overrightarrow{BC} = \vec{c}-\vec{b}$ (both meet at $B$). Subtract position vectors component by component.

Step 2: Test for a scalar multiple

The points are collinear precisely when the two vectors are parallel, i.e. one is a scalar multiple of the other:

$\overrightarrow{BC} = \lambda\,\overrightarrow{AB}\quad\text{for some scalar }\lambda.$ …

Common Mistakes

Mistake 1: Checking only that the two vectors have equal (or proportional) length.

Why it's wrong: collinearity is about direction, not magnitude — two vectors can have a neat length ratio yet point different ways. Correct approach: require one vector to be a scalar multiple of the other, so the ratio holds component-by-component.

Mistake 2: Matching the component ratio for only two coordinates and assuming the third.

Why it's wrong: two matching ratios do not force the third; a single mismatched component means the points are not collinear. Correct approach: verify the same scalar $\lambda$ works for all three components (or that the cross product is $\vec{0}$). …