Chapter 10 · Vector Algebra

10.2 Addition, Scalar Multiples and the Section Formula

Walk along one arrow, then along another: where you end up is their sum. Add the triangle and parallelogram laws, stretch vectors by scalars, and you can find the vector joining two points and the point that divides a segment in any ratio.

A ferry on the Hooghly heads straight across the river at $4$ m/s, but the current carries it downstream at $3$ m/s. Where does it actually go, and how fast? Neither $4$ nor $3$ nor $7$ is the answer: the two velocities have to be combined as arrows. Put the second arrow where the first one ends, and the arrow from the start to the finish — $5$ m/s, at about $37^\circ$ to the straight-across line — is the ferry's true velocity. That is vector addition, and it is the same rule whether the arrows are velocities, forces or the legs of a drone's flight.

In this section you will add and subtract vectors by the triangle law and the parallelogram law, multiply them by scalars, and then do all of it with components — which turns every geometric construction into a line of arithmetic. Two tools come out at once: the vector $\overrightarrow{AB}=\vec b-\vec a$ joining two points, and the section formula for the point that divides a segment in a given ratio. Both are used again and again in Section 10.3, Section 10.4 and in Chapter 11 — Three Dimensional Geometry.

Adding Vectors: the Triangle and Parallelogram Laws

📘 Definition 10.6 — Sum of two vectors (triangle law)

Let $\vec a$ and $\vec b$ be vectors. Draw $\vec a=\overrightarrow{AB}$ and then draw $\vec b$ starting at the terminal point of $\vec a$: $\vec b=\overrightarrow{BC}$. The sum $\vec a+\vec b$ is the vector from the initial point of $\vec a$ to the terminal point of $\vec b$: $$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}.$$ The difference is $\vec a-\vec b=\vec a+(-\vec b)$.

Going round a closed triangle brings you back to the start, so $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{AA}=\vec 0$. The same works for any closed polygon: $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}=\vec 0$.

📐 Theorem 10.3 — Parallelogram law

If two vectors $\vec a$ and $\vec b$ are drawn from the same point, $\vec a=\overrightarrow{OA}$ and $\vec b=\overrightarrow{OB}$, and $OACB$ is the parallelogram with sides $OA$ and $OB$, then their sum is the diagonal through $O$: $$\vec a+\vec b=\overrightarrow{OC}.$$

Proof. In the parallelogram $OACB$ the sides $AC$ and $OB$ are parallel, equal in length and point the same way, so $\overrightarrow{AC}=\overrightarrow{OB}=\vec b$. By the triangle law in triangle $OAC$: $\ \overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AC}=\vec a+\vec b$. $\square$ The other diagonal is the difference: $\overrightarrow{BA}=\overrightarrow{BO}+\overrightarrow{OA}=-\vec b+\vec a=\vec a-\vec b$.

📐 Theorem 10.4 — Properties of vector addition

For any vectors $\vec a,\vec b,\vec c$:

(a) commutative: $\vec a+\vec b=\vec b+\vec a$; (b) associative: $(\vec a+\vec b)+\vec c=\vec a+(\vec b+\vec c)$;

Why. (a) In the parallelogram $OACB$, going $O\to A\to C$ gives $\vec a+\vec b$ and going $O\to B\to C$ gives $\vec b+\vec a$; both end at $C$. (b) Draw $\vec a=\overrightarrow{PQ}$, $\vec b=\overrightarrow{QR}$, $\vec c=\overrightarrow{RS}$ head to tail. Then $(\vec a+\vec b)+\vec c=\overrightarrow{PR}+\overrightarrow{RS}=\overrightarrow{PS}$ and $\vec a+(\vec b+\vec c)=\overrightarrow{PQ}+\overrightarrow{QS}=\overrightarrow{PS}$. (c) and (d) are $\overrightarrow{AB}+\overrightarrow{BB}=\overrightarrow{AB}$ and $\overrightarrow{AB}+\overrightarrow{BA}=\overrightarrow{AA}$. $\square$ Because of (b) we can write $\vec a+\vec b+\vec c$ without brackets.

✏️ Example 1 — Head-to-tail simplification

(a) Simplify $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}$. (b) Simplify $\overrightarrow{AB}-\overrightarrow{CB}$. (c) $ABCD$ is a parallelogram. Show that $\overrightarrow{AC}+\overrightarrow{BD}=2\overrightarrow{BC}$.

💡 Show solution▶

(a) Each arrow starts where the last one ended: $(\overrightarrow{AB}+\overrightarrow{BC})+\overrightarrow{CD}=\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}$.

(b) $-\overrightarrow{CB}=\overrightarrow{BC}$, so $\overrightarrow{AB}-\overrightarrow{CB}=\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$.

(c) $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$ and $\overrightarrow{BD}=\overrightarrow{BC}+\overrightarrow{CD}$. In a parallelogram $\overrightarrow{CD}=-\overrightarrow{AB}$, so the sum is $\overrightarrow{AB}+2\overrightarrow{BC}-\overrightarrow{AB}=2\overrightarrow{BC}$.

Check Take the square $A(0,0)$, $B(1,0)$, $C(1,1)$, $D(0,1)$: $\overrightarrow{AC}+\overrightarrow{BD}=(1,1)+(-1,1)=(0,2)=2\overrightarrow{BC}$ ✓.

Drag the tips of the red vector $\vec a$ and the blue vector $\vec b$. In triangle law mode, Play slides $\vec b$ until its tail sits on the tip of $\vec a$; the green arrow from the start to the finish is the sum. The parallelogram mode shows the same sum as a diagonal, the difference mode shows $\vec a-\vec b$ as the other diagonal, and the last mode adds the scalar multiples $\lambda\vec a+\mu\vec b$.

mode:$\lambda$:$\mu$:

$\vec a$—

$\vec b$—

result—

lengths—

Scalar Multiples and Components

📘 Definition 10.7 — Multiplying a vector by a scalar

Let $\vec a$ be a vector and $\lambda$ a real number. The scalar multiple $\lambda\vec a$ is the vector with magnitude $|\lambda\vec a|=|\lambda|\,|\vec a|$, pointing in the same direction as $\vec a$ if $\lambda\gt0$ and in the opposite direction if $\lambda\lt0$. If $\lambda=0$ or $\vec a=\vec 0$, then $\lambda\vec a=\vec 0$.

So $2\vec a$ is twice as long as $\vec a$, $-\vec a=(-1)\vec a$ is its negative, and $\tfrac12\vec a$ is half of it. In particular, for $\vec a\ne\vec 0$, the vector $\dfrac1{|\vec a|}\vec a$ has magnitude $1$ and the direction of $\vec a$: it is the unit vector $\hat a=\dfrac{\vec a}{|\vec a|}$ of Section 10.1.

📐 Theorem 10.5 — Scalar multiples and collinearity

For vectors $\vec a,\vec b$ and scalars $\lambda,\mu$:

(a) $\lambda(\vec a+\vec b)=\lambda\vec a+\lambda\vec b$, $(\lambda+\mu)\vec a=\lambda\vec a+\mu\vec a$, $\lambda(\mu\vec a)=(\lambda\mu)\vec a$.

(b) If $\vec a\ne\vec 0$, then $\vec b$ is collinear with $\vec a$ if and only if $\vec b=\lambda\vec a$ for some scalar $\lambda$.

Why (b). If $\vec b=\lambda\vec a$, it lies along $\vec a$ by Definition 10.7. Conversely, if $\vec b$ is parallel to $\vec a$, take $\lambda=\pm\dfrac{|\vec b|}{|\vec a|}$ with the sign $+$ when they point the same way and $-$ when they point opposite ways; then $\lambda\vec a$ has the length and direction of $\vec b$. (Part (a) is the statement that scaling a whole triangle of vectors by $\lambda$ gives a similar triangle.) $\square$

Now combine Theorem 10.5(a) with the component form $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$: every operation acts on the components one at a time.

📐 Theorem 10.6 — Operations in components

Let $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$, $\ \vec b=b_1\hat i+b_2\hat j+b_3\hat k$ and let $\lambda$ be a scalar. Then $$\vec a\pm\vec b=(a_1\pm b_1)\hat i+(a_2\pm b_2)\hat j+(a_3\pm b_3)\hat k,$$ $$\lambda\vec a=(\lambda a_1)\hat i+(\lambda a_2)\hat j+(\lambda a_3)\hat k.$$ In particular, for $\vec a\ne\vec 0$: $\ \vec b$ is collinear with $\vec a$ if and only if $b_1=\lambda a_1$, $b_2=\lambda a_2$, $b_3=\lambda a_3$ for one and the same $\lambda$ — that is, the components are proportional, $\dfrac{b_1}{a_1}=\dfrac{b_2}{a_2}=\dfrac{b_3}{a_3}=\lambda$ (where a component of $\vec a$ is $0$, the matching component of $\vec b$ must be $0$ too).

Proof. Using the commutative and associative laws to regroup, and Theorem 10.5(a) to collect, $\vec a+\vec b$ $=(a_1\hat i+b_1\hat i)+(a_2\hat j+b_2\hat j)+(a_3\hat k+b_3\hat k)$ $=(a_1+b_1)\hat i+(a_2+b_2)\hat j+(a_3+b_3)\hat k$. The difference and $\lambda\vec a$ are the same. The collinearity test is Theorem 10.5(b) read with Theorem 10.2(b) (equal vectors have equal components). $\square$

📐 Theorem 10.7 — The vector joining two points

If $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ have position vectors $\vec a$ and $\vec b$, then $$\overrightarrow{P_1P_2}=\vec b-\vec a,$$ $$\overrightarrow{P_1P_2}=(x_2-x_1)\hat i+(y_2-y_1)\hat j+(z_2-z_1)\hat k,$$ $$\begin{aligned}|\overrightarrow{P_1P_2}|^2&=(x_2-x_1)^2+(y_2-y_1)^2\\&\quad+(z_2-z_1)^2.\end{aligned}$$

Proof. By the triangle law, $\overrightarrow{OP_1}+\overrightarrow{P_1P_2}=\overrightarrow{OP_2}$, so $\overrightarrow{P_1P_2}=\overrightarrow{OP_2}-\overrightarrow{OP_1}=\vec b-\vec a$; now use Theorem 10.6 and Theorem 10.2(a). $\square$ The magnitude is the 3-D distance formula from Class 11 — now it comes for free. Remember the order: terminal minus initial.

✏️ Example 2 — Arithmetic with components

Let $\vec a=2\hat i-\hat j+2\hat k$ and $\vec b=-\hat i+\hat j-\hat k$. Find (a) $\vec a+\vec b$ and the unit vector in its direction; (b) $2\vec a-3\vec b$ and its magnitude.

💡 Show solution▶

(a) $\vec a+\vec b=(2-1)\hat i+(-1+1)\hat j+(2-1)\hat k$ $=\hat i+\hat k$. Its magnitude is $\sqrt{1+0+1}=\sqrt2$, so the unit vector is $\tfrac1{\sqrt2}\hat i+\tfrac1{\sqrt2}\hat k$.

(b) $2\vec a=4\hat i-2\hat j+4\hat k$ and $-3\vec b=3\hat i-3\hat j+3\hat k$, so $2\vec a-3\vec b=7\hat i-5\hat j+7\hat k$.

Magnitude $|2\vec a-3\vec b|=\sqrt{49+25+49}$ $=\sqrt{123}\approx11.09$.

✏️ Example 3 — The vector joining two points

Find the vector $\overrightarrow{PQ}$ joining $P(2,3,0)$ to $Q(-1,-2,-4)$, its magnitude, and its direction cosines.

💡 Show solution▶

Vector Terminal minus initial: $\overrightarrow{PQ}=(-1-2)\hat i+(-2-3)\hat j+(-4-0)\hat k$ $=-3\hat i-5\hat j-4\hat k$.

Magnitude $|\overrightarrow{PQ}|=\sqrt{9+25+16}=\sqrt{50}=5\sqrt2$.

Direction cosines $-\tfrac3{5\sqrt2},\ -\tfrac5{5\sqrt2}=-\tfrac1{\sqrt2},\ -\tfrac4{5\sqrt2}$. Check: $\tfrac{9+25+16}{50}=1$ ✓.

Note $\overrightarrow{QP}=3\hat i+5\hat j+4\hat k$ has the same length and all three direction cosines change sign.

✏️ Example 4 — Collinearity: points and parameters

(a) Show that $A(1,-2,-8)$, $B(5,0,-2)$ and $C(11,3,7)$ are collinear, and find the ratio in which $B$ divides $AC$. (b) Find $p$ and $q$ so that $2\hat i+p\hat j+3\hat k$ and $4\hat i-6\hat j+q\hat k$ are collinear.

💡 Show solution▶

(a) Vectors $\overrightarrow{AB}=4\hat i+2\hat j+6\hat k$ and $\overrightarrow{BC}=6\hat i+3\hat j+9\hat k=\tfrac32\overrightarrow{AB}$.

(a) Conclusion $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are collinear (Theorem 10.5(b)) and share the point $B$, so $A,B,C$ lie on one line. Since $\overrightarrow{BC}=\tfrac32\overrightarrow{AB}$ with a positive multiple, $B$ lies between $A$ and $C$ and $AB:BC=2:3$. (Lengths: $|\overrightarrow{AB}|=\sqrt{56}=2\sqrt{14}$, $|\overrightarrow{BC}|=\sqrt{126}=3\sqrt{14}$.)

(b) The $\hat i$-components give $\lambda=\tfrac42=2$. Then $-6=2p$ and $q=2\cdot3$: $\ p=-3$, $q=6$.

⚠️ Collinear vectors, collinear points

$\overrightarrow{AB}=\lambda\overrightarrow{CD}$ only says the two segments are parallel: $ABDC$ could be a trapezium. To prove that three points $A,B,C$ are collinear, use two vectors that share a point — $\overrightarrow{AB}$ and $\overrightarrow{AC}$, or $\overrightarrow{AB}$ and $\overrightarrow{BC}$ — and show one is a multiple of the other.

The Section Formula

Where is the point two-thirds of the way from $A$ to $B$? Or the point beyond $B$ that is twice as far from $A$ as from $B$? In Class 10 you answered such questions in the plane with coordinates; with vectors the answer is one line, valid in space as well.

A point $R$ on the line $AB$ divides $AB$ internally in the ratio $m:n$ ($m,n\gt0$) if $R$ lies between $A$ and $B$ and $AR:RB=m:n$. It divides $AB$ externally in the ratio $m:n$ ($m\ne n$) if $R$ lies on the line $AB$ outside the segment and $AR:RB=m:n$.

📐 Theorem 10.8 — Section formula

Let $A$ and $B$ have position vectors $\vec a$ and $\vec b$, and let $m,n\gt0$.

(a) The point $R$ dividing $AB$ internally in the ratio $m:n$ has position vector

$$\vec r=\frac{m\vec b+n\vec a}{m+n}.$$

(b) If $m\ne n$, the point $R$ dividing $AB$ externally in the ratio $m:n$ has position vector

$$\vec r=\frac{m\vec b-n\vec a}{m-n}.$$

Proof. (a) $R$ is between $A$ and $B$, so $\overrightarrow{AR}$ and $\overrightarrow{RB}$ point the same way, and their lengths are in the ratio $m:n$. Hence $n\,\overrightarrow{AR}=m\,\overrightarrow{RB}$ (both sides have the same direction and the length $n\cdot AR=m\cdot RB$). By Theorem 10.7, $n(\vec r-\vec a)=m(\vec b-\vec r)$, so $(m+n)\vec r=m\vec b+n\vec a$. Divide by $m+n\gt0$.

(b) Now $R$ is outside the segment, so $\overrightarrow{AR}$ and $\overrightarrow{BR}$ point the same way (both away from the segment, or both towards it), with lengths in the ratio $m:n$: $\ n\,\overrightarrow{AR}=m\,\overrightarrow{BR}$. So $n(\vec r-\vec a)=m(\vec r-\vec b)$, i.e. $(m-n)\vec r=m\vec b-n\vec a$; divide by $m-n\ne0$. (c) is (a) with $m=n=1$. $\square$

A memory aid for (a): each position vector is weighted by the part of the ratio next to the other end — $\vec b$ gets $m$ (the part touching $A$) and $\vec a$ gets $n$. For (b), replace $n$ by $-n$.

✏️ Example 5 — Internal and external division

$A(1,2,-3)$ and $B(4,-1,3)$ are given. Find the points $R$ and $S$ that divide $AB$ in the ratio $2:1$ internally and externally, and check the ratios.

💡 Show solution▶

Internal $\vec r=\dfrac{2\vec b+\vec a}{3}$ $=\dfrac{(8,-2,6)+(1,2,-3)}{3}$ $=\dfrac{(9,0,3)}{3}$, so $R=(3,0,1)$.

External $\vec s=\dfrac{2\vec b-\vec a}{2-1}=(8-1,\,-2-2,\,6+3)$, so $S=(7,-4,9)$.

Check $R$ $\overrightarrow{AR}=2\hat i-2\hat j+4\hat k$ and $\overrightarrow{RB}=\hat i-\hat j+2\hat k$: $\ \overrightarrow{AR}=2\overrightarrow{RB}$ ✓ ($AR:RB=2:1$).

Check $S$ $\overrightarrow{AS}=6\hat i-6\hat j+12\hat k$ and $\overrightarrow{BS}=3\hat i-3\hat j+6\hat k$: $\ \overrightarrow{AS}=2\overrightarrow{BS}$ ✓, and $S$ lies beyond $B$.

✏️ Example 6 — The centroid of a triangle

(a) Show that the centroid $G$ of the triangle with vertices of position vectors $\vec a,\vec b,\vec c$ has position vector $\dfrac{\vec a+\vec b+\vec c}{3}$. (b) Find the centroid of the triangle with vertices $A(1,0,2)$, $B(3,-2,4)$, $C(-1,5,0)$.

💡 Show solution▶

Median The midpoint $D$ of $BC$ has position vector $\vec d=\dfrac{\vec b+\vec c}2$ (Theorem 10.8(c)).

Divide 2 : 1 The centroid divides the median $AD$ internally in the ratio $2:1$, so $\vec g=\dfrac{2\vec d+\vec a}{3}=\dfrac{\vec b+\vec c+\vec a}{3}$.

Symmetry The answer is symmetric in $\vec a,\vec b,\vec c$, so the same point lies on all three medians — which proves that the medians are concurrent.

(b) $\vec g=\tfrac13\big((1,0,2)+(3,-2,4)+(-1,5,0)\big)$ $=\tfrac13(3,3,6)$, so $G=(1,1,2)$.

Drag the points $A$ and $B$. Choose the ratio $m:n$ and internal or external division; the lab marks the point $R$ given by the section formula and measures $AR$ and $RB$ to show that their ratio really is $m:n$. Play steps $m$ through $1,2,\dots,6$ with $n$ fixed, so you can watch $R$ slide towards $B$ (internal) — or race away to infinity as $m$ approaches $n$ (external). The plane is used to keep the picture clear; in space the formula works component by component in exactly the same way.

division:$m$:$n$:

$\vec a,\ \vec b$—

ratio—

$\vec r$—

$AR : RB$—

⚠️ Common mistakes when studying alone

1. Initial minus terminal. $\overrightarrow{AB}=\vec b-\vec a$ (position vector of $B$ minus that of $A$), not $\vec a-\vec b$, which is $\overrightarrow{BA}$.

2. Swapping the weights. For $AR:RB=m:n$ the formula is $\dfrac{m\vec b+n\vec a}{m+n}$: the $m$ goes with $\vec b$. Check with a quick sanity test — for a large $m$ the point should be close to $B$.

3. External division with the wrong sign. $\dfrac{m\vec b-n\vec a}{m-n}$. If $m\gt n$, $R$ lies beyond $B$; if $m\lt n$, it lies beyond $A$. There is no external point for $m=n$.

5. Parallel is not collinear for points. See the warning after Example 4.

Exercises

The triangle and parallelogram laws.

1 Simplify $\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RS}$.
2 Simplify $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}$.
3 Simplify $\overrightarrow{AB}-\overrightarrow{AC}$.
4 $ABCD$ is a parallelogram with $\overrightarrow{AB}=\vec a$ and $\overrightarrow{AD}=\vec b$. Express the diagonals $\overrightarrow{AC}$ and $\overrightarrow{BD}$ in terms of $\vec a$ and $\vec b$.
5 $ABCDEF$ is a regular hexagon. Show that $\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=3\overrightarrow{AD}$.
6 Use the triangle law to explain why $|\vec a+\vec b|\le|\vec a|+|\vec b|$ for all vectors $\vec a,\vec b$. When does equality hold?

💡 Show solutions▶

1 $\overrightarrow{PS}$.

2 $\vec 0$ — the path returns to $A$.

3 $\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}$.

4 $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\vec a+\vec b$ (since $\overrightarrow{BC}=\overrightarrow{AD}$); $\ \overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}=\vec b-\vec a$.

5 Key fact (Theorem 10.8(c) with $A$ as origin): $\overrightarrow{AU}+\overrightarrow{AV}=2\overrightarrow{AM}$, where $M$ is the midpoint of $UV$. Let the side be $s$, so $AD=2s$. By symmetry in the line $AD$, the midpoint $N$ of $BF$ lies on $AD$ at distance $\tfrac s2$ from $A$, and the midpoint $M$ of $CE$ lies on $AD$ at distance $\tfrac{3s}2$. So $\overrightarrow{AB}+\overrightarrow{AF}=2\overrightarrow{AN}=\tfrac12\overrightarrow{AD}$ and $\overrightarrow{AC}+\overrightarrow{AE}=2\overrightarrow{AM}=\tfrac32\overrightarrow{AD}$. Adding $\overrightarrow{AD}$: total $\left(\tfrac12+\tfrac32+1\right)\overrightarrow{AD}=3\overrightarrow{AD}$ ✓.

6 Draw $\vec a=\overrightarrow{PQ}$, $\vec b=\overrightarrow{QR}$. Then $|\vec a+\vec b|=PR$, and in triangle $PQR$ one side is at most the sum of the other two: $PR\le PQ+QR$. Equality holds exactly when $Q$ lies on the segment $PR$, i.e. when $\vec a$ and $\vec b$ point in the same direction (or one of them is $\vec 0$).

Components and scalar multiples. In parts 1–5, $\vec a=2\hat i-\hat j+3\hat k$, $\ \vec b=\hat i+2\hat j-\hat k$, $\ \vec c=-3\hat i+\hat j+2\hat k$.

1 Find $\vec a+\vec b$.
2 Find $\vec a-\vec b$.
3 Find $2\vec a+3\vec b-\vec c$.
4 Find $|\vec a+\vec b+\vec c|$.
5 Find the unit vector in the direction of $\vec a+\vec b$.
6 Find the vector of magnitude $5$ in the direction opposite to $\hat i+2\hat j-\hat k$.
7 For which $\lambda$ is $\lambda(2\hat i-\hat j+2\hat k)$ a unit vector?
8 Show that $2\hat i-3\hat j+4\hat k$ and $-4\hat i+6\hat j-8\hat k$ are collinear, and find the ratio of their magnitudes.
9 Find $x$ and $y$ if $x(\hat i+\hat j)+y(\hat i-\hat j)=5\hat i+\hat j$.

Check: linear combinations $\lambda\vec a+\mu\vec b$

$\vec a$:

$\vec b$:

$\lambda,\ \mu$:

—

💡 Show solutions▶

1 $3\hat i+\hat j+2\hat k$.

2 $\hat i-3\hat j+4\hat k$.

3 $(4+3+3)\hat i+(-2+6-1)\hat j$ $+(6-3-2)\hat k$ $=10\hat i+3\hat j+\hat k$.

4 $\vec a+\vec b+\vec c=0\hat i+2\hat j+4\hat k$, so the magnitude is $\sqrt{20}=2\sqrt5$.

5 $|3\hat i+\hat j+2\hat k|=\sqrt{14}$: $\ \tfrac1{\sqrt{14}}(3\hat i+\hat j+2\hat k)$.

6 The unit vector is $\tfrac1{\sqrt6}(\hat i+2\hat j-\hat k)$; reverse and scale: $-\tfrac5{\sqrt6}(\hat i+2\hat j-\hat k)=-\tfrac5{\sqrt6}\hat i-\tfrac{10}{\sqrt6}\hat j+\tfrac5{\sqrt6}\hat k$.

7 $|\lambda|\cdot3=1$: $\ \lambda=\pm\tfrac13$.

8 $-4\hat i+6\hat j-8\hat k=-2(2\hat i-3\hat j+4\hat k)$, so they are collinear (opposite directions) and the second is twice as long: ratio $2:1$ (magnitudes $2\sqrt{29}$ and $\sqrt{29}$).

9 $(x+y)\hat i+(x-y)\hat j=5\hat i+\hat j$: $\ x+y=5$, $x-y=1$, so $x=3$, $y=2$.

III

Joining points, collinearity and the section formula.

1 Find the vector joining $P(1,2,3)$ to $Q(4,6,3)$ and its magnitude.
2 Find $\overrightarrow{AB}$ for $A(-1,0,2)$, $B(3,-4,4)$, and the unit vector along $\overrightarrow{AB}$.
3 Find the direction cosines of the vector joining $(1,-1,-2)$ to $(-1,1,0)$.
4 Show that $A(2,6,3)$, $B(1,2,7)$, $C(3,10,-1)$ are collinear. Which point lies between the other two?
5 Are $A(1,1,1)$, $B(2,3,4)$, $C(3,5,6)$ collinear?
6 Find the position vector of the point dividing the join of $\vec a=\hat i+2\hat j-\hat k$ and $\vec b=-\hat i+\hat j+\hat k$ internally in the ratio $2:1$.
7 Find the position vector of the point dividing the same join externally in the ratio $2:1$.
8 Find the midpoint of $P(3,-2,5)$ and $Q(-1,4,1)$.
9 In what ratio does $R(5,0,2)$ divide the segment joining $A(1,-4,-2)$ and $B(7,2,4)$?
10 Find the centroid of the triangle with vertices $(2,-1,3)$, $(0,4,-2)$, $(1,0,5)$.
11 $ABCD$ is a parallelogram with $A(1,2,3)$, $B(4,5,6)$, $C(7,9,10)$. Find $D$.

💡 Show solutions▶

1 $\overrightarrow{PQ}=3\hat i+4\hat j+0\hat k$, magnitude $5$.

2 $\overrightarrow{AB}=4\hat i-4\hat j+2\hat k$, $|\overrightarrow{AB}|=6$; unit vector $\tfrac23\hat i-\tfrac23\hat j+\tfrac13\hat k$.

3 The vector is $-2\hat i+2\hat j+2\hat k$, magnitude $2\sqrt3$: direction cosines $-\tfrac1{\sqrt3},\tfrac1{\sqrt3},\tfrac1{\sqrt3}$.

4 $\overrightarrow{AB}=-\hat i-4\hat j+4\hat k$ and $\overrightarrow{AC}=\hat i+4\hat j-4\hat k=-\overrightarrow{AB}$: collinear, with $A$ between $B$ and $C$ — in fact $A$ is the midpoint of $BC$, since $\tfrac12(B+C)=(2,6,3)$.

5 $\overrightarrow{AB}=\hat i+2\hat j+3\hat k$, $\overrightarrow{AC}=2\hat i+4\hat j+5\hat k$; ratios $\tfrac21=\tfrac42\ne\tfrac53$, so not collinear.

6 $\tfrac{2\vec b+\vec a}{3}=\tfrac13(-2+1,\,2+2,\,2-1)$ $=-\tfrac13\hat i+\tfrac43\hat j+\tfrac13\hat k$.

7 $\tfrac{2\vec b-\vec a}{1}=(-2-1)\hat i+(2-2)\hat j+(2+1)\hat k$ $=-3\hat i+3\hat k$.

8 $\tfrac12(3-1,\,-2+4,\,5+1)=(1,1,3)$.

9 $\overrightarrow{AR}=4\hat i+4\hat j+4\hat k$ and $\overrightarrow{RB}=2\hat i+2\hat j+2\hat k$, so $\overrightarrow{AR}=2\overrightarrow{RB}$: internally in the ratio $2:1$.

10 $\tfrac13(2+0+1,\,-1+4+0,\,3-2+5)$ $=(1,1,2)$.

11 The diagonals bisect each other, so $\vec a+\vec c=\vec b+\vec d$ and $\vec d=\vec a+\vec c-\vec b$ $=(1+7-4,\,2+9-5,\,3+10-6)$: $\ D=(4,6,7)$.

Applications and proofs.

1 Ferry. Relative to the water a ferry moves at $4$ m/s due east; the river flows due north at $3$ m/s. Find the ferry's velocity relative to the bank (as a vector, with $\hat i$ east and $\hat j$ north), its speed, and its direction.
2 Drone survey (illustrative data). A drone flies three legs (in km) $\vec d_1=2\hat i+3\hat j+\hat k$, $\vec d_2=-4\hat i+\hat j$, $\vec d_3=2\hat i-4\hat j-\hat k$. How far from the start is it after two legs, and where is it after all three?
3 Prove with vectors that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
4 Prove with position vectors that the diagonals of a parallelogram bisect each other.
5 Equilibrium (illustrative forces). Forces $\vec F_1=3\hat i-2\hat j+\hat k$ N and $\vec F_2=-\hat i+4\hat j+2\hat k$ N act on a particle. Find the force $\vec F_3$ that keeps the particle in equilibrium ($\vec F_1+\vec F_2+\vec F_3=\vec 0$) and its magnitude.
6 For any vectors $\vec a,\vec b,\vec c$, show that the points with position vectors $\vec a-2\vec b+3\vec c$, $\ -2\vec a+3\vec b-\vec c$ and $4\vec a-7\vec b+7\vec c$ are collinear.

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1 $\vec v=4\hat i+3\hat j$ m/s; speed $\sqrt{16+9}=5$ m/s; direction $\tan^{-1}\tfrac34\approx36.87^\circ$ north of east.

2 After two legs: $\vec d_1+\vec d_2=-2\hat i+4\hat j+\hat k$, distance $\sqrt{4+16+1}=\sqrt{21}\approx4.58$ km. After three: $\vec d_1+\vec d_2+\vec d_3=\vec 0$ — it is back at the start.

3 Let the vertices have position vectors $\vec a,\vec b,\vec c$ and let $D,E$ be the midpoints of $AB,AC$. Then $\overrightarrow{DE}=\tfrac{\vec a+\vec c}2-\tfrac{\vec a+\vec b}2=\tfrac12(\vec c-\vec b)=\tfrac12\overrightarrow{BC}$: a positive multiple of $\overrightarrow{BC}$, so $DE\parallel BC$, and $DE=\tfrac12BC$.

4 In parallelogram $ABCD$, $\overrightarrow{AB}=\overrightarrow{DC}$, i.e. $\vec b-\vec a=\vec c-\vec d$, so $\vec a+\vec c=\vec b+\vec d$. Hence $\tfrac{\vec a+\vec c}2=\tfrac{\vec b+\vec d}2$: the midpoints of $AC$ and $BD$ coincide.

5 $\vec F_3=-(\vec F_1+\vec F_2)=-(2\hat i+2\hat j+3\hat k)$ $=-2\hat i-2\hat j-3\hat k$ N; $|\vec F_3|=\sqrt{4+4+9}=\sqrt{17}\approx4.12$ N.

6 Call the points $P,Q,R$. $\overrightarrow{PQ}=-3\vec a+5\vec b-4\vec c$ and $\overrightarrow{PR}=3\vec a-5\vec b+4\vec c=-\overrightarrow{PQ}$. The vectors share the point $P$, so $P,Q,R$ are collinear ($P$ is the midpoint of $QR$).

🎯 Exam corner Board · 1 markBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: 1-mark questions test $\overrightarrow{AB}=\vec b-\vec a$, collinearity and unit vectors; 2–3 mark questions ask for internal and external division or a short vector proof; case studies wrap the same ideas in a real setting. JEE favours collinearity conditions and ratio problems where two lines meet.

E1. Board · 1 mark If $\overrightarrow{AB}=2\hat i-\hat j+\hat k$ and the position vector of $A$ is $\hat i+3\hat j$, then the position vector of $B$ is (A) $3\hat i+2\hat j+\hat k$ (B) $\hat i-4\hat j+\hat k$ (C) $-\hat i+4\hat j-\hat k$ (D) $3\hat i+4\hat j+\hat k$

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Idea $\overrightarrow{AB}=\vec b-\vec a$, so $\vec b=\vec a+\overrightarrow{AB}$ $=(1+2)\hat i+(3-1)\hat j+(0+1)\hat k$.

Answer (A) $3\hat i+2\hat j+\hat k$. Trap (B) is $\overrightarrow{AB}-\vec a$ (subtracting instead of adding); (C) is $\vec a-\overrightarrow{AB}$; (D) adds $+1$ instead of $-1$ in the $\hat j$-component.

E2. Board · 1 mark The value of $\lambda$ for which $2\hat i-3\hat j+4\hat k$ and $\lambda\hat i+6\hat j-8\hat k$ are collinear is (A) $4$ (B) $-4$ (C) $2$ (D) $-2$

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Idea Proportional components: $\tfrac{6}{-3}=\tfrac{-8}{4}=-2$, so the second vector is $-2$ times the first and $\lambda=-2\cdot2=-4$.

Answer (B). Trap (D) gives the multiplier $-2$ instead of the component $\lambda$; (A) forgets the sign.

E3. Board · 3 marks $P$ and $Q$ have position vectors $3\hat i+\hat j-2\hat k$ and $-\hat i+5\hat j+2\hat k$. Find the position vectors of the points $R$ and $S$ dividing $PQ$ internally and externally in the ratio $3:1$, and verify that $P$, $R$, $S$ are collinear.

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Internal $\vec r=\dfrac{3\vec q+\vec p}{4}$ $=\dfrac{(-3+3)\hat i+(15+1)\hat j+(6-2)\hat k}{4}$ $=4\hat j+\hat k$. (1 mark)

External $\vec s=\dfrac{3\vec q-\vec p}{2}=\dfrac{-6\hat i+14\hat j+8\hat k}{2}$ $=-3\hat i+7\hat j+4\hat k$. (1 mark)

Collinear $\overrightarrow{PR}=-3\hat i+3\hat j+3\hat k$ and $\overrightarrow{PS}=-6\hat i+6\hat j+6\hat k=2\overrightarrow{PR}$, with common point $P$ ✓. (1 mark) Also $\overrightarrow{RQ}=-\hat i+\hat j+\hat k$, so $PR:RQ=3:1$ ✓.

E4. Board · Case study Drone over a garden (illustrative coordinates, in metres; $z$ is height). Three lamp posts stand at $A(2,0,0)$, $B(6,3,0)$ and $C(1,9,0)$, and a drone hovers at $D(3,4,12)$. (i) Find $\overrightarrow{AB}$ and the distance $AB$. (1 mark) (ii) Find the midpoint $M$ of $BC$. (1 mark) (iii) Find the centroid $G$ of triangle $ABC$, show that the drone is directly above $G$, and find the distance from the drone to lamp post $A$. (2 marks)

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(i) $\overrightarrow{AB}=4\hat i+3\hat j$, $\ AB=\sqrt{16+9}=5$ m.

(ii) $M=\tfrac12(6+1,\,3+9,\,0)=(3.5,\,6,\,0)$.

(iii) $G=\tfrac13(2+6+1,\,0+3+9,\,0)=(3,4,0)$. $\overrightarrow{GD}=12\hat k$ is vertical, so $D$ is directly above $G$, $12$ m up. $\overrightarrow{DA}=-\hat i-4\hat j-12\hat k$, so $DA=\sqrt{1+16+144}=\sqrt{161}\approx12.69$ m.

E5. JEE Main The points with position vectors $10\hat i+3\hat j$, $\ 12\hat i-5\hat j$ and $a\hat i+11\hat j$ are collinear. Then $a$ equals (A) $-8$ (B) $4$ (C) $8$ (D) $12$

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Method $\overrightarrow{AB}=2\hat i-8\hat j$ and $\overrightarrow{AC}=(a-10)\hat i+8\hat j$ must be proportional: $\tfrac{a-10}{2}=\tfrac{8}{-8}=-1$, so $a=8$.

Answer (C). Trap (D) $=12$ comes from $\tfrac{a-10}{2}=+1$ (dropping the sign); (A) negates the answer.

⏱ Fast route Going from $A$ to $B$ the $\hat j$-component drops by $8$ while $\hat i$ rises by $2$; going from $A$ to $C$ the $\hat j$-component rises by $8$, so $\hat i$ must fall by $2$: $a=10-2=8$.

E6. JEE Advanced (Numerical answer.) In triangle $OAB$, $\overrightarrow{OA}=\vec a$ and $\overrightarrow{OB}=\vec b$. $M$ is the midpoint of $OA$ and $N$ divides $OB$ internally in the ratio $2:1$. The lines $AN$ and $BM$ meet at $P$, and $\overrightarrow{OP}=x\vec a+y\vec b$. Find $4(x+y)$.

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Two descriptions A point of $AN$ is $\vec a+t(\vec n-\vec a)=(1-t)\vec a+\tfrac{2t}3\vec b$ (since $\vec n=\tfrac23\vec b$). A point of $BM$ is $\vec b+s(\vec m-\vec b)=\tfrac s2\vec a+(1-s)\vec b$.

Compare $\vec a$ and $\vec b$ are not collinear (they are two sides of a triangle), so the coefficients must match: $1-t=\tfrac s2$ and $\tfrac{2t}3=1-s$. From the first, $s=2-2t$; then $\tfrac{2t}3=2t-1$, so $t=\tfrac34$, $s=\tfrac12$.

Answer $x=1-t=\tfrac14$, $y=\tfrac{2t}3=\tfrac12$, so $4(x+y)=\mathbf 3$. Check with $s$: $\tfrac s2=\tfrac14$ ✓, $1-s=\tfrac12$ ✓.

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. If you walk a closed loop round your school grounds and add up every leg as a vector, do you really get $\vec 0$? How big is the “closing error”, and where does it come from?

Why it matters. Surveyors mapping a plot of land, and ships and aircraft navigating without satellite signals (“dead reckoning”), add displacement vectors leg by leg. Because the path is closed, the sum should be $\vec 0$; the size of the actual sum — the closing error of the traverse — tells the surveyor how good the measurements were, and it is shared back over the legs to correct them.

What to do.

Choose a closed route with 4–6 straight legs. For each leg, measure its length (pace it out after calibrating your step length over $20$ m, or use a measuring tape) and its compass bearing (a phone compass app is fine).
Convert each leg to components: with $\hat i$ east and $\hat j$ north, a leg of length $d$ on bearing $\theta$ is $d\sin\theta\,\hat i+d\cos\theta\,\hat j$.
Add all the legs in a spreadsheet. Find the closing error vector, its magnitude, and the ratio (closing error) ÷ (total distance walked).
Repeat the walk once more and compare. Which legs do you think caused most of the error?

Tools: Phone compass, tape or calibrated pacing, spreadsheet (SIN, COS, RADIANS, SQRT).

Hand in: A sketch of the route, the table of legs and components, the closing error with its magnitude and ratio, and a short paragraph explaining how a surveyor would spread the correction over the legs.

Going further. Find out how a phone's step counter and compass combine for indoor navigation in a large building where GPS does not work, and why its position “drifts”.

Instructor note: with paced distances and phone bearings, closing ratios of about $1:30$ to $1:100$ are typical; tape-measured legs reach about $1:200$. A simple correction (Bowditch's rule) shares the closing error between legs in proportion to their lengths. Remind students that bearings are measured clockwise from north, which is why the east component uses $\sin\theta$.

🧪 Self-check quiz

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Q1. For any triangle $ABC$, $\ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}$ equals

Q2. The vector joining $A(1,2,3)$ to $B(3,2,1)$ is

Q3. The midpoint of the segment joining the points with position vectors $\vec a$ and $\vec b$ has position vector

Q4. If $|\vec a|=4$, then $|-3\vec a|$ equals

Q5. The point dividing the join of $\vec a$ and $\vec b$ externally in the ratio $3:1$ has position vector

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