Q.Find the vector joining the points $P(2, 3, 0)$ and $Q(-1, -2, -4)$ directed from P to Q.

Concept: Vector Magnitude Difference — the vector from $P$ to $Q$ is found by subtracting the coordinates of the tail from the head.

Step 1: The vector $\overrightarrow{PQ}$ is given by $Q - P$ (head minus tail).

Step 2: Subtract component-wise:

$\overrightarrow{PQ} = (-1 - 2,\; -2 - 3,\; -4 - 0)$ …

The key idea here is simple but often gets flipped around: a vector directed from one point to another is always head minus tail. If you want the vector from $P$ to $Q$, imagine standing at $P$ and walking to $Q$ — the displacement is $Q - P$.

Why does this work? The position vector of a point tells you how to get from the origin to that point. So the vector from $P$ to $Q$ is the difference between the position vector of $Q$ and the position vector of $P$. Geometrically, you're asking: "Starting at $P$, what steps do I take to reach $Q$?" Those steps are exactly the coordinate differences.

Let's work it through.

1. Write the position vectors.

   The position vector of $P(2, 3, 0)$ is $\vec{OP} = 2\hat{i} + 3\hat{j} + 0\hat{k}$.

   The position vector of $Q(-1, -2, -4)$ is $\vec{OQ} = -1\hat{i} -2\hat{j} -4\hat{k}$.

2. Apply the head-minus-tail rule.

   The vector from $P$ to $Q$ is $\vec{PQ} = \vec{OQ} - \vec{OP}$.

   This gives:

3. Simplify each component.

So $\vec{PQ} = -3\hat{i} -5\hat{j} -4\hat{k}$. …

Method: Finding the vector joining two points (head minus tail)

Use this for any "vector from $P$ to $Q$" / displacement-between-points problem.

Steps

Step 1: Fix which point is the tail and which is the head.

"Directed from $P$ to $Q$" means $P$ is the tail and $Q$ is the head. The vector is head minus tail:

Step 2: Subtract coordinates component-wise.

For $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$,

$\overrightarrow{PQ}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}.$ …

Common Mistakes

Mistake 1: Computing $P-Q$ instead of $Q-P$.

Why it's wrong: "directed from $P$ to $Q$" makes $P$ the tail and $Q$ the head, so it is $Q-P$; $P-Q$ gives $\overrightarrow{QP}$, the opposite direction. Correct approach: always use head minus tail.

Mistake 2: Sign slips when subtracting negative coordinates. …