Q.If $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}+5\hat{j}$, $3\hat{i}+2\hat{j}-3\hat{k}$ and $\hat{i}-6\hat{j}-\hat{k}$ are the position vectors of points A, B, C and D respectively, then find the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$. Deduce that $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear.

Concept: Dot Product Angle — the angle $\theta$ between two vectors is given by $\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}|\,|\vec{v}|}$.

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

$\overrightarrow{AB} = \vec{B} - \vec{A} = (2\hat{i}+5\hat{j}) - (\hat{i}+\hat{j}+\hat{k}) = \hat{i} + 4\hat{j} - \hat{k}$

$\overrightarrow{CD} = \vec{D} - \vec{C} = (\hat{i}-6\hat{j}-\hat{k}) - (3\hat{i}+2\hat{j}-3\hat{k}) = -2\hat{i} -8\hat{j} + 2\hat{k}$

Step 2: Compute dot product and magnitudes.

$\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(-2) + (4)(-8) + (-1)(2) = -2 -32 -2 = -36$

$|\overrightarrow{AB}| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{1+16+1} = \sqrt{18} = 3\sqrt{2}$

$|\overrightarrow{CD}| = \sqrt{(-2)^2 + (-8)^2 + 2^2} = \sqrt{4+64+4} = \sqrt{72} = 6\sqrt{2}$

Step 3: Find $\cos\theta$.

$\cos\theta = \frac{-36}{(3\sqrt{2})(6\sqrt{2})} = \frac{-36}{36} = -1$

Thus $\theta = \pi$ (or $180^\circ$).

Since $\theta = 180^\circ$, the vectors are opposite in direction, hence collinear.

With position vectors $A=\hat i+\hat j+\hat k$, $B=2\hat i+5\hat j$, $C=3\hat i+2\hat j-3\hat k$, $D=\hat i-6\hat j-\hat k$:

Form the vectors:

Angle:

Collinearity: $\overrightarrow{CD}=-2(\hat i+4\hat j-\hat k)=-2\,\overrightarrow{AB}$, so each vector is a scalar multiple of the other. Hence $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear (parallel, oppositely directed).

Method: Angle between two vectors, and reading off collinearity

Use the dot-product angle formula, then interpret an angle of $0^\circ$ or $180^\circ$ as the vectors being parallel — hence the segments collinear.

Steps

Step 1: Form the two vectors from the position vectors

$\overrightarrow{AB} = B - A$ and $\overrightarrow{CD} = D - C$, subtracting coordinates.

Step 2: Apply the angle formula

Divide by both magnitudes — the dot product alone is not $\cos\theta$ unless both vectors are already unit length.

Step 3: Interpret the result

$\cos\theta = 1$ means parallel and same direction ($\theta = 0^\circ$); $\cos\theta = -1$ means anti-parallel ($\theta = 180^\circ$). In either case the direction vectors are scalar multiples, so $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear. A cleaner confirmation is to spot the scalar multiple directly, e.g. $\overrightarrow{CD} = \lambda\,\overrightarrow{AB}$.

Common Mistakes

Mistake 1: Treating the dot product itself as $\cos\theta$.

Why it's wrong: $\overrightarrow{AB}\cdot\overrightarrow{CD}$ equals $\cos\theta$ only if both vectors are unit length; otherwise you must divide by both magnitudes. Correct approach: always compute $\dfrac{\overrightarrow{AB}\cdot\overrightarrow{CD}}{|\overrightarrow{AB}|\,|\overrightarrow{CD}|}$.

Mistake 2: Reading $\cos\theta = -1$ as perpendicular or as "no relation."

Why it's wrong: $\cos\theta = -1$ is $\theta = 180^\circ$ (opposite direction), while perpendicular would be $\cos\theta = 0$. Correct approach: $-1$ signals anti-parallel vectors, which are still parallel in direction.

Mistake 3: Thinking collinearity requires the same direction only.

Why it's wrong: vectors pointing exactly opposite ($180^\circ$) are also scalar multiples of each other, so the segments are still collinear. Correct approach: any $\overrightarrow{CD} = \lambda\,\overrightarrow{AB}$, with $\lambda$ positive or negative, proves collinearity.