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

Let’s unpack this carefully. The question tests two separate ideas: what it means for vectors to be collinear, and what the dot product of a vector with its scalar multiple actually gives. Many students mix these up, so let’s build the intuition first.

Collinear vectors are vectors that lie along the same line or parallel lines. The key property: two non-zero vectors are collinear if and only if one is a scalar multiple of the other. That is, $\vec{b} = \lambda \vec{a}$ for some real number $\lambda$. No angle condition or dot product zero is required — that’s for perpendicular vectors.

Now look at the assertion: $\vec{a}$ and $(-2\vec{a})$ are clearly related by $\lambda = -2$. Since $-2$ is a real number, they are indeed collinear. The negative sign just means they point in opposite directions, but they’re still on the same line.

The reason claims $\vec{a} \cdot (-2\vec{a}) = 0$. Let’s compute that:

$\vec{a} \cdot (-2\vec{a}) = -2 (\vec{a} \cdot \vec{a}) = -2 |\vec{a}|^2$.

Since $\vec{a} \neq \vec{0}$, $|\vec{a}|^2 > 0$, so this product is negative, not zero. The reason is completely wrong.

So the assertion is true, the reason is false. In exam language, this means option (C) — Assertion true, Reason false.

Let’s walk through the reasoning step by step.

1. Check the assertion: Are $\vec{a}$ and $-2\vec{a}$ collinear? …

Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions. Equivalently, $\vec{a}$ and $\vec{b}$ are collinear if $\vec{a}=\lambda\vec{b}$ for …