Q.Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.

Concept: Vector Magnitude Properties — the vector from $P$ to $Q$ is the difference of their position vectors, and its magnitude is the distance between the points.

The vector joining $P$ to $Q$ is

Its scalar components are simply the differences in each coordinate:

The magnitude (length) of $\vec{PQ}$ is the distance formula in 3D:

The core idea here is simple: a vector in space is just a directed displacement. If you want the vector that takes you from point $P$ to point $Q$, you need to find how far you move along each axis. That "how far" along each axis is exactly the difference in the corresponding coordinates.

Think of it this way: if you're at $P$ and you want to reach $Q$, you first move along the $x$-axis by $(x_2 - x_1)$, then along the $y$-axis by $(y_2 - y_1)$, and finally along the $z$-axis by $(z_2 - z_1)$. These three numbers are the scalar components — they tell you the signed length of the projection of $\overrightarrow{PQ}$ onto each coordinate axis.

The magnitude of the vector is then just the straight-line distance between the two points, which comes from applying the Pythagorean theorem in three dimensions.

Let's break it down step by step.

1. Write the position vectors of the points.

   The position vector of a point is simply the vector from the origin to that point.

   For $P(x_1, y_1, z_1)$, the position vector is $\vec{OP} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.

   For $Q(x_2, y_2, z_2)$, it is $\vec{OQ} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$.

2. Find the vector from $P$ to $Q$.

   By the triangle law of vector addition, the vector from $P$ to $Q$ is the difference between the position vectors of $Q$ and $P$:

Substituting the expressions:

Grouping like components:

3. Identify the scalar components.

   The scalar components are the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$ respectively. So:

   • Component along $x$-axis: $x_2 - x_1$
   • Component along $y$-axis: $y_2 - y_1$
   • Component along $z$-axis: $z_2 - z_1$
   Watch out

   A common mistake is to write the components as $x_1 - x_2$ instead of $x_2 - x_1$. Remember: the vector goes from $P$ to $Q$, so it's "head minus tail" — $Q$ minus $P$.

4. Find the magnitude.

   The magnitude of a vector $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$ is given by $|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}$. Applying this to $\overrightarrow{PQ}$:

This is exactly the distance formula between two points in 3D space.

Method: Scalar Components and Length of the Vector Joining Two Points

Use this to turn two points into the vector between them and find that vector's length.

Steps

Step 1: Subtract head minus tail

The vector from $P$ to $Q$ is

Always take terminal point $Q$ minus initial point $P$ — the direction $P\to Q$ dictates the order.

Step 2: Read off the scalar components

The scalar components are the coefficients of $\hat i,\hat j,\hat k$: the coordinate differences $x_2-x_1,\;y_2-y_1,\;z_2-z_1$.

Step 3: Apply the magnitude formula

the 3D distance between the points. Squaring removes any sign, so the length is the same either way — but the components depend on doing $Q-P$.

Common Mistakes

Mistake 1: Reversing the subtraction to $P-Q$

Why it's wrong: $\overrightarrow{PQ}$ goes from $P$ to $Q$, so it is $Q-P$; using $P-Q$ flips the sign of every component. Correct approach: head minus tail, i.e. $x_2-x_1$, not $x_1-x_2$.

Mistake 2: Confusing scalar components with direction cosines

Why it's wrong: scalar components are the raw coordinate differences; direction cosines are those differences divided by the magnitude. Correct approach: report $x_2-x_1,\,y_2-y_1,\,z_2-z_1$ unnormalised.

Mistake 3: Forgetting to square each difference

Why it's wrong: the magnitude is $\sqrt{\sum(\text{difference})^2}$, not $\sqrt{\sum|\text{difference}|}$. Correct approach: square, add, then take the root.