Vector joining two points

From a picture to a formula. Draw the triangle formed by $O$, $A$, and $B$. Walking from $O$ to $A$ and then from $A$ to $B$ lands you at the same place as walking directly from $O$ to $B$ — that is the triangle law of vector addition, and it is the entire proof, stated in the derivation block below.

In coordinates. If $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, the rule becomes purely arithmetic: subtract corresponding coordinates.

where: $(x_1,y_1,z_1)$ is the tail point $A$, $(x_2,y_2,z_2)$ is the head point $B$, and $\hat i, \hat j, \hat k$ are the unit vectors along the axes.

Length and direction. The magnitude of this vector is the straight-line distance $AB$, and its coefficients are direction ratios of the line through $A$ and $B$ — divide each by the magnitude to get direction cosines. This single computation therefore answers three different-sounding exam questions at once: "find the vector," "find the distance," and "find the direction ratios of the line."

where: the right side is the 3D distance formula applied to $A$ and $B$.

Reversing the trip. Swapping which point is the tail flips the sign: $\overrightarrow{BA} = \vec a - \vec b = -\overrightarrow{AB}$. Same line, same length, opposite arrow.

Why the origin never shows up in the answer. It is worth pausing on this, because it looks strange at first: $\vec a$ and $\vec b$ both depend on where $O$ is placed, yet $\vec b - \vec a$ does not. Move the origin anywhere you like and both position vectors shift by exactly the same amount, so that shift cancels in the subtraction. This is precisely why $\overrightarrow{AB}$ is a genuine geometric property of the two points $A$ and $B$ alone — a physical arrow you could draw on paper without ever mentioning an origin — even though computing it algebraically happens to route through one.

Find the vector $\overrightarrow{PQ}$, the distance $PQ$, and the unit vector along $\overrightarrow{PQ}$, where $P(1,-2,3)$ and $Q(4,0,-1)$.

Step 1. Apply the rule — head $Q$ minus tail $P$, coordinate by coordinate.

Step 2. Find the magnitude, which is the distance $PQ$.

Step 3. Divide the vector by its own magnitude to get the unit vector along $\overrightarrow{PQ}$.

So $\overrightarrow{PQ} = 3\hat i+2\hat j-4\hat k$, the distance is $\sqrt{29}$ units, and the direction ratios of line $PQ$ are $3, 2, -4$. Final answer: $\overrightarrow{PQ}=3\hat i+2\hat j-4\hat k$, $PQ=\sqrt{29}$.

Competitive stretch

If $A$, $B$, $C$ have position vectors $\vec a = \hat i + 2\hat j + 3\hat k$, $\vec b = 3\hat i + 4\hat j + 7\hat k$, $\vec c = -3\hat i - 2\hat j - 5\hat k$, show they are collinear.

Step 1. Compute $\overrightarrow{AB} = \vec b-\vec a = 2\hat i+2\hat j+4\hat k$.

Step 2. Compute $\overrightarrow{AC} = \vec c-\vec a = -4\hat i-4\hat j-8\hat k$.

Step 3. Check proportionality: $\overrightarrow{AC} = -2\left(2\hat i+2\hat j+4\hat k\right) = -2\,\overrightarrow{AB}$. Since $\overrightarrow{AC}$ is a scalar multiple of $\overrightarrow{AB}$ and they share the common point $A$, the points $A, B, C$ lie on one straight line — collinear.

• Wrong subtraction order. $\overrightarrow{AB}=\vec b-\vec a$ is always head minus tail; swapping the order silently gives $\overrightarrow{BA}$, the negative of what was asked.
• Mixing origins. The formula only works when $\vec a$ and $\vec b$ are both measured from the same fixed origin $O$; using position vectors quoted from different reference points gives a meaningless result.
• Picking unrelated vector pairs for a collinearity proof. You must use two vectors that share a common point, like $\overrightarrow{AB}$ and $\overrightarrow{AC}$, and show they are parallel — not any two of $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{CA}$ chosen at random without checking the shared vertex.
• Stopping at "parallel" instead of "collinear," and reading direction ratios off the wrong vector. Proportional vectors alone prove the lines are parallel; only a shared point (such as $A$ in both $\overrightarrow{AB}$ and $\overrightarrow{AC}$) upgrades that to "the points lie on the same line." Similarly, the direction ratios of line $AB$ come from the coordinate differences in $\overrightarrow{AB}$ itself, not from the coordinates of one endpoint quoted alone.

Ask the tutor: Want to see why $\overrightarrow{AB}=\vec b-\vec a$ feeds directly into the section formula, or try the collinearity proof with a fresh set of points?