Chapter 10 · Vector Algebra

10.3 The Scalar (Dot) Product and Projections

Multiply two vectors and get a number: the scalar (dot) product a · b = |a||b| cos θ. It measures how much two vectors point the same way — giving angles, perpendicularity tests, projections and the work done by a force.

A porter at a railway station drags a heavy suitcase along the platform by its strap, pulling at $60^\circ$ to the ground. Only part of his effort moves the suitcase forward — the part of the force along the platform. The upward part just lightens the load on the wheels. Physics measures the useful effort by the work done: (force along the motion) × (distance), that is $|\vec F|\cos\theta\times|\vec d|$. A single number, built from two vectors and the angle between them.

That combination, $|\vec a|\,|\vec b|\cos\theta$, is so useful that it gets its own name and symbol: the scalar product or dot product $\vec a\cdot\vec b$. In this section you will see that it has a remarkably simple formula in components, $a_1b_1+a_2b_2+a_3b_3$, so that you can find the angle between any two vectors in space, test whether they are perpendicular, and compute the projection of one vector on another — the length of its “shadow” along the other's direction.

The Scalar Product

📘 Definition 10.8 — Angle between two vectors
Let $\vec a$ and $\vec b$ be non-zero vectors. Draw them from the same point: $\vec a=\overrightarrow{OA}$, $\vec b=\overrightarrow{OB}$. The angle between $\vec a$ and $\vec b$ is the angle $\theta=\angle AOB$, with $0\le\theta\le\pi$.
📘 Definition 10.9 — Scalar (dot) product
The scalar product (or dot product) of non-zero vectors $\vec a$ and $\vec b$ with angle $\theta$ between them is the real number $$\vec a\cdot\vec b=|\vec a|\,|\vec b|\cos\theta.$$ If $\vec a=\vec 0$ or $\vec b=\vec 0$, we define $\vec a\cdot\vec b=0$.

Read the definition through the sign of $\cos\theta$: $\vec a\cdot\vec b\gt0$ when the angle is acute, $\vec a\cdot\vec b=0$ when it is a right angle, and $\vec a\cdot\vec b\lt0$ when it is obtuse. At the extremes, $\theta=0$ gives $\vec a\cdot\vec b=|\vec a|\,|\vec b|$ and $\theta=\pi$ gives $-|\vec a|\,|\vec b|$. Rearranged, the definition is a formula for the angle: $$\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a|\,|\vec b|}.$$

📐 Theorem 10.9 — Properties of the scalar product
For vectors $\vec a,\vec b,\vec c$ and a scalar $\lambda$:

(a) $\vec a\cdot\vec b=\vec b\cdot\vec a$ (commutative);

(b) if $\vec a\ne\vec 0$ and $\vec b\ne\vec 0$, then $\vec a\cdot\vec b=0$ if and only if $\vec a\perp\vec b$;

(c) $\vec a\cdot\vec a=|\vec a|^2$; in particular $\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1$ and $\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0$;

(d) $(\lambda\vec a)\cdot\vec b=\lambda(\vec a\cdot\vec b)=\vec a\cdot(\lambda\vec b)$;

(e) $\vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c$ (distributive).

Why. (a) The angle from $\vec a$ to $\vec b$ is the same as from $\vec b$ to $\vec a$. (b) For non-zero vectors $|\vec a|\,|\vec b|\ne0$, so the product is $0$ exactly when $\cos\theta=0$, i.e. $\theta=\tfrac\pi2$. (c) A vector makes angle $0$ with itself; $\hat i,\hat j,\hat k$ are unit vectors at right angles to each other. (d) For $\lambda\gt0$, $\lambda\vec a$ makes the same angle with $\vec b$ and is $\lambda$ times as long; for $\lambda\lt0$ the angle becomes $\pi-\theta$, whose cosine is $-\cos\theta$, and the length is $|\lambda|$ times as long — either way the product is multiplied by $\lambda$. (e) is explained with projections after Definition 10.10. $\square$

📐 Theorem 10.10 — The scalar product in components
If $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$ and $\vec b=b_1\hat i+b_2\hat j+b_3\hat k$, then $$\vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3,$$ and, for non-zero $\vec a,\vec b$, the angle $\theta$ between them is given by $$\cos\theta=\frac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{a_1^2+a_2^2+a_3^2}\,\sqrt{b_1^2+b_2^2+b_3^2}}.$$

Proof. Expand $(a_1\hat i+a_2\hat j+a_3\hat k)\cdot(b_1\hat i+b_2\hat j+b_3\hat k)$ using Theorem 10.9(d), (e). There are nine terms such as $a_1b_2(\hat i\cdot\hat j)$. By Theorem 10.9(c) the six “mixed” ones vanish and the three with matching unit vectors give $a_1b_1+a_2b_2+a_3b_3$. $\square$

✏️ Example 1 — The angle between two vectors
Find the angle between $\vec a=2\hat i-\hat j+2\hat k$ and $\vec b=\hat i+2\hat j+2\hat k$.
Dot product $\vec a\cdot\vec b=2(1)+(-1)(2)+2(2)$ $=2-2+4=4$.
Magnitudes $|\vec a|=\sqrt{4+1+4}=3$, $\ |\vec b|=\sqrt{1+4+4}=3$.
Angle $\cos\theta=\dfrac4{3\cdot3}=\dfrac49$, so $\theta=\cos^{-1}\dfrac49\approx63.61^\circ$ — acute, as the positive dot product predicted.
✏️ Example 2 — Perpendicular vectors
(a) Find $\lambda$ so that $2\hat i+\lambda\hat j+\hat k$ and $\hat i-2\hat j+3\hat k$ are perpendicular.  (b) Find the angle between $\hat i+\hat j$ and $\hat j+\hat k$.
(a) Perpendicular $\iff$ dot product $0$ (Theorem 10.9(b)): $\ 2(1)+\lambda(-2)+1(3)=5-2\lambda=0$, so $\lambda=\tfrac52$.
(b) $(\hat i+\hat j)\cdot(\hat j+\hat k)=0+1+0=1$ and both magnitudes are $\sqrt2$, so $\cos\theta=\tfrac12$ and $\theta=\tfrac\pi3$.
Note In (b) the vectors run from $O$ to $A(1,1,0)$ and $B(0,1,1)$, along diagonals of two faces of the unit cube. Since $OA=OB=AB=\sqrt2$, triangle $OAB$ is equilateral — which is why the answer is $60^\circ$.
Interactive 1 — Dot Product Lab

Set the lengths of the red vector $\vec a$ and the blue vector $\vec b$ and the angle $\theta$ between them, or press Play to sweep $\theta$ from $0^\circ$ to $180^\circ$. The thick shaded shadow is the projection of $\vec a$ on the line of $\vec b$ (light falling perpendicular to $\vec b$) — green when it falls along $\vec b$, red when it falls backwards. Watch the bar: $\vec a\cdot\vec b=|\vec a|\,|\vec b|\cos\theta$ is positive while the shadow falls along $\vec b$, zero when $\vec a\perp\vec b$, and negative when it falls backwards. The component formula $a_1b_1+a_2b_2$ always gives the same number.

$\theta$—
$|\vec a|\,|\vec b|\cos\theta$—
$a_1b_1+a_2b_2$—
projection of $\vec a$ on $\vec b$—
perpendicular?—

Projection of a Vector and Work Done

Shine a light perpendicular to the line of $\vec b$. The shadow that $\vec a$ casts on that line has signed length $|\vec a|\cos\theta$: positive if the shadow falls in the direction of $\vec b$, negative if it falls the other way.

📘 Definition 10.10 — Projection of a vector on another
Let $\vec b\ne\vec 0$ and let $\theta$ be the angle between $\vec a$ and $\vec b$. The projection of $\vec a$ on $\vec b$ (the scalar projection) is $$|\vec a|\cos\theta=\vec a\cdot\hat b=\frac{\vec a\cdot\vec b}{|\vec b|},$$ and the projection vector of $\vec a$ along $\vec b$ is $$\big(|\vec a|\cos\theta\big)\,\hat b=\frac{\vec a\cdot\vec b}{|\vec b|^2}\,\vec b.$$

So the dot product is “length of $\vec b$ times the shadow of $\vec a$ on $\vec b$”: $\ \vec a\cdot\vec b=|\vec b|\,(|\vec a|\cos\theta)$. Three useful readings follow.

✏️ Example 3 — Projection and projection vector
Find the projection of $\vec a=2\hat i+3\hat j+2\hat k$ on $\vec b=\hat i+2\hat j+\hat k$, and the projection vector of $\vec a$ along $\vec b$.
Ingredients $\vec a\cdot\vec b=2+6+2=10$ and $|\vec b|=\sqrt{1+4+1}=\sqrt6$.
Projection $\dfrac{\vec a\cdot\vec b}{|\vec b|}=\dfrac{10}{\sqrt6}=\dfrac{5\sqrt6}{3}\approx4.08$.
Projection vector $\dfrac{10}{6}\vec b=\dfrac53(\hat i+2\hat j+\hat k)=\dfrac53\hat i+\dfrac{10}3\hat j+\dfrac53\hat k$.
Check The rest of $\vec a$, namely $\vec a-\tfrac53\vec b=\tfrac13\hat i-\tfrac13\hat j+\tfrac13\hat k$, should be perpendicular to $\vec b$: $\ \tfrac13-\tfrac23+\tfrac13=0$ ✓.

Work done by a constant force. If a constant force $\vec F$ (in newtons) acts on a body while it moves through a displacement $\vec d$ (in metres), the work done is $$W=\vec F\cdot\vec d=|\vec F|\,|\vec d|\cos\theta\ \text{ joules},$$ where $\theta$ is the angle between $\vec F$ and $\vec d$. Only the projection of $\vec F$ on the direction of motion does work; a force perpendicular to the motion does none.

✏️ Example 4 — Work done
(a) A porter pulls a suitcase $20$ m along a level platform with a force of $50$ N along the strap, which makes $60^\circ$ with the ground. Find the work he does.  (b) A constant force $\vec F=4\hat i+3\hat j+2\hat k$ N moves a particle from $A(1,2,3)$ to $B(4,-1,5)$, with distances in metres. Find the work done.
(a) $W=|\vec F|\,|\vec d|\cos60^\circ=50\times20\times\tfrac12=500$ J. (The upward part $50\sin60^\circ\approx43.3$ N does no work.)
(b) Displacement $\vec d=\overrightarrow{AB}=3\hat i-3\hat j+2\hat k$ m.
(b) Work $W=\vec F\cdot\vec d=4(3)+3(-3)+2(2)$ $=12-9+4=7$ J.
Interactive 2 — Work Done by a Force

A crate is pulled along level ground by a rope. Choose the pull $|\vec F|$ and the angle $\theta$ of the rope; press Play to drag the crate $10$ m. The force splits into a green part along the ground, $|\vec F|\cos\theta$, which does all the work, and a purple part straight up, which does none. The work counter grows as $W=\vec F\cdot\vec d=|\vec F|\,|\vec d|\cos\theta$.

distance $|\vec d|$—
along ground $|\vec F|\cos\theta$—
upward $|\vec F|\sin\theta$—
work $W$—

Identities and Inequalities

Because the dot product obeys the commutative and distributive laws, you can expand products of sums exactly as in algebra — and $\vec a\cdot\vec a$ turns lengths into dot products.

📐 Theorem 10.11 — Identities and inequalities for the scalar product
For any vectors $\vec a$ and $\vec b$:

(a) $|\vec a+\vec b|^2=|\vec a|^2+2\,\vec a\cdot\vec b+|\vec b|^2$  and  $|\vec a-\vec b|^2=|\vec a|^2-2\,\vec a\cdot\vec b+|\vec b|^2$;

(b) $(\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2$;

(c) Cauchy–Schwarz inequality:

$$|\vec a\cdot\vec b|\le|\vec a|\,|\vec b|;$$

(d) triangle inequality: $|\vec a+\vec b|\le|\vec a|+|\vec b|$.

Proof. (a) $|\vec a+\vec b|^2=(\vec a+\vec b)\cdot(\vec a+\vec b)$ $=\vec a\cdot\vec a+\vec a\cdot\vec b+\vec b\cdot\vec a+\vec b\cdot\vec b$, and $\vec a\cdot\vec b=\vec b\cdot\vec a$; replace $\vec b$ by $-\vec b$ for the second identity. (b) Expand: the middle terms $-\vec a\cdot\vec b+\vec b\cdot\vec a$ cancel. (c) If either vector is $\vec 0$ both sides are $0$; otherwise $|\vec a\cdot\vec b|=|\vec a|\,|\vec b|\,|\cos\theta|\le|\vec a|\,|\vec b|$ since $|\cos\theta|\le1$. (d) By (a) and (c), $$\begin{aligned}|\vec a+\vec b|^2&\le|\vec a|^2+2|\vec a|\,|\vec b|+|\vec b|^2\\&=(|\vec a|+|\vec b|)^2;\end{aligned}$$ take square roots. $\square$

✏️ Example 5 — Lengths from dot products
$|\vec a|=3$, $|\vec b|=4$ and the angle between $\vec a$ and $\vec b$ is $60^\circ$. Find $|\vec a+\vec b|$ and $|\vec a-\vec b|$.
Dot product $\vec a\cdot\vec b=3\cdot4\cdot\cos60^\circ=6$.
Sum $|\vec a+\vec b|^2=9+2(6)+16=37$, so $|\vec a+\vec b|=\sqrt{37}\approx6.08$.
Difference $|\vec a-\vec b|^2=9-12+16=13$, so $|\vec a-\vec b|=\sqrt{13}\approx3.61$.
Check These are the two diagonals of the parallelogram on $\vec a,\vec b$. The cosine rule in the triangle with sides $3,4$ and included angle $60^\circ$ gives $9+16-2\cdot12\cdot\tfrac12=13$ ✓ — the dot-product identity is the cosine rule in vector form.
✏️ Example 6 — A vector proof: the angle in a semicircle
$AB$ is a diameter of a circle with centre $O$, and $P$ is any other point on the circle. Prove that $\angle APB=90^\circ$.
Set up Take $O$ as origin. Let $\overrightarrow{OA}=\vec a$; then $\overrightarrow{OB}=-\vec a$, and $\overrightarrow{OP}=\vec p$ with $|\vec p|=|\vec a|=r$, the radius.
The two sides $\overrightarrow{PA}=\vec a-\vec p$ and $\overrightarrow{PB}=-\vec a-\vec p$.
Dot product $\overrightarrow{PA}\cdot\overrightarrow{PB}=(\vec a-\vec p)\cdot(-\vec a-\vec p)$ $=-|\vec a|^2+|\vec p|^2$ (the cross terms $\vec a\cdot\vec p$ cancel), which is $-r^2+r^2=0$.
Conclusion Both vectors are non-zero (since $P\ne A,B$), so by Theorem 10.9(b) $\overrightarrow{PA}\perp\overrightarrow{PB}$: $\angle APB=90^\circ$. $\square$
⚠️ Common mistakes when studying alone

1. The answer is a number. $\vec a\cdot\vec b$ is a scalar, not a vector: never write $\vec a\cdot\vec b=3\hat i-\hat k$. For the same reason $\vec a\cdot\vec b\cdot\vec c$ means nothing — $(\vec a\cdot\vec b)\vec c$ is a vector, but you cannot “dot” a number with $\vec c$.

2. No cancelling. $\vec a\cdot\vec b=\vec a\cdot\vec c$ does not give $\vec b=\vec c$; it gives $\vec a\cdot(\vec b-\vec c)=0$, so $\vec b-\vec c$ could be any vector perpendicular to $\vec a$.

3. Tail to tail. The angle between two vectors is measured with both drawn from the same point. In triangle $ABC$ the angle at $B$ is between $\overrightarrow{BA}$ and $\overrightarrow{BC}$ — using $\overrightarrow{AB}$ and $\overrightarrow{BC}$ gives the exterior angle $180^\circ-B$.

4. $|\vec a+\vec b|\ne|\vec a|+|\vec b|$. Square first: $|\vec a+\vec b|^2=|\vec a|^2+2\,\vec a\cdot\vec b+|\vec b|^2$. Forgetting the middle term is the commonest error in this topic.

5. Projection direction. “Projection of $\vec a$ on $\vec b$” divides by $|\vec b|$, the vector you project onto.

Exercises

I
Scalar products and angles.
  • 1 Find $\vec a\cdot\vec b$ for $\vec a=\hat i-2\hat j+3\hat k$, $\vec b=3\hat i+\hat j+\hat k$.
  • 2 Find the angle between $\hat i-\hat j$ and $\hat j-\hat k$.
  • 3 Find the angle between $3\hat i+4\hat j$ and $4\hat i-3\hat j$.
  • 4 Find the angle between $2\hat i+2\hat j-\hat k$ and $6\hat i-3\hat j+2\hat k$.
  • 5 Find $\vec a\cdot\vec b$ if $|\vec a|=4$, $|\vec b|=3$ and the angle between them is $\tfrac\pi3$.
  • 6 Find the angle between $\vec a$ and $\vec b$ if $|\vec a|=\sqrt3$, $|\vec b|=2$ and $\vec a\cdot\vec b=\sqrt6$.
  • 7 Expand $(3\vec a-5\vec b)\cdot(2\vec a+7\vec b)$ in terms of $|\vec a|^2$, $\vec a\cdot\vec b$ and $|\vec b|^2$.
  • 8 Evaluate $\hat i\cdot(\hat j+\hat k)+\hat j\cdot(\hat k+\hat i)+\hat k\cdot(\hat i+\hat j)$.
  • 9 Use dot products with $\hat i,\hat j,\hat k$ to find the angles that $\hat i-\hat j+\sqrt2\,\hat k$ makes with the axes.
1 $3-2+3=4$.
2 Dot product $0-1+0=-1$; magnitudes $\sqrt2,\sqrt2$; $\cos\theta=-\tfrac12$, so $\theta=\tfrac{2\pi}3$.
3 $12-12=0$: the vectors are perpendicular, $\theta=\tfrac\pi2$.
4 Dot product $12-6-2=4$; magnitudes $3$ and $7$; $\ \theta=\cos^{-1}\tfrac4{21}\approx79.02^\circ$.
5 $4\cdot3\cdot\tfrac12=6$.
6 $\cos\theta=\dfrac{\sqrt6}{2\sqrt3}=\dfrac{\sqrt2}{2}$, so $\theta=\tfrac\pi4$.
7 $6|\vec a|^2+21\,\vec a\cdot\vec b-10\,\vec b\cdot\vec a-35|\vec b|^2$ $=6|\vec a|^2+11\,\vec a\cdot\vec b-35|\vec b|^2$.
8 Every term is a dot product of two different unit vectors: the total is $0$.
9 $|\vec a|=\sqrt{1+1+2}=2$. $\cos\alpha=\tfrac{\vec a\cdot\hat i}{2}=\tfrac12$, $\cos\beta=-\tfrac12$, $\cos\gamma=\tfrac{\sqrt2}2$: angles $60^\circ$, $120^\circ$, $45^\circ$.
II
Perpendicularity and projections.
  • 1 Find $\lambda$ if $3\hat i+\lambda\hat j-2\hat k$ is perpendicular to $2\hat i-\hat j+\hat k$.
  • 2 Show that $2\hat i-\hat j+\hat k$ and $\hat i+3\hat j+\hat k$ are perpendicular.
  • 3 Find the values of $x$ for which $x\hat i+2\hat j-\hat k$ and $x\hat i-\hat j+2\hat k$ are perpendicular.
  • 4 Find the projection of $\hat i+3\hat j+7\hat k$ on $7\hat i-\hat j+8\hat k$.
  • 5 Find the projection of $2\hat i-3\hat j+6\hat k$ on $\hat i+2\hat j+2\hat k$, and the projection vector.
  • 6 For $\vec a=4\hat i+3\hat j$ and $\vec b=2\hat i$, find the projection of $\vec a$ on $\vec b$ and of $\vec b$ on $\vec a$.
  • 7 Show that $A(2,-1,1)$, $B(1,-3,-5)$, $C(3,-4,-4)$ are the vertices of a right-angled triangle, and name the right angle.
  • 8 Find the angle at $A$ in the triangle with vertices $A(1,0,0)$, $B(0,1,0)$, $C(0,0,1)$.
Check: dot product, angle and projections
—
1 $6-\lambda-2=0$: $\ \lambda=4$.
2 $2-3+1=0$ ✓.
3 $x^2-2-2=0$: $\ x=\pm2$.
4 Dot product $7-3+56=60$, $|\vec b|=\sqrt{49+1+64}=\sqrt{114}$: projection $\tfrac{60}{\sqrt{114}}\approx5.62$.
5 Dot product $2-6+12=8$, $|\vec b|=3$: projection $\tfrac83$; projection vector $\tfrac89(\hat i+2\hat j+2\hat k)$.
6 $\vec a\cdot\vec b=8$. On $\vec b$: $\tfrac82=4$; on $\vec a$: $\tfrac85=1.6$ — different, because $|\vec a|\ne|\vec b|$.
7 $\overrightarrow{CA}=-\hat i+3\hat j+5\hat k$ and $\overrightarrow{CB}=-2\hat i+\hat j-\hat k$: $\ \overrightarrow{CA}\cdot\overrightarrow{CB}=2+3-5=0$, so the right angle is at $C$. Check: $|\overrightarrow{CA}|^2+|\overrightarrow{CB}|^2=35+6=41=|\overrightarrow{AB}|^2$ ✓.
8 $\overrightarrow{AB}=-\hat i+\hat j$, $\overrightarrow{AC}=-\hat i+\hat k$: dot product $1$, magnitudes $\sqrt2$: $\cos A=\tfrac12$, $A=60^\circ$ (the triangle is equilateral).
III
Identities and proofs.
  • 1 $|\vec a|=2$, $|\vec b|=3$ and $\vec a\cdot\vec b=4$. Find $|\vec a+\vec b|$ and $|\vec a-\vec b|$.
  • 2 Show that if $|\vec a+\vec b|=|\vec a-\vec b|$ then $\vec a\perp\vec b$ (for non-zero $\vec a,\vec b$). Interpret this for a parallelogram.
  • 3 $(\vec a+\vec b)\cdot(\vec a-\vec b)=27$ and $|\vec a|=2|\vec b|$. Find $|\vec a|$ and $|\vec b|$.
  • 4 $\vec a+\vec b+\vec c=\vec 0$ with $|\vec a|=3$, $|\vec b|=5$, $|\vec c|=7$. Find the angle between $\vec a$ and $\vec b$.
  • 5 For unit vectors $\hat a,\hat b$ at angle $\theta$, show that $|\hat a-\hat b|=2\sin\tfrac\theta2$.
  • 6 Prove that the diagonals of a rhombus are perpendicular.
  • 7 Verify the Cauchy–Schwarz inequality for $\vec a=\hat i+2\hat j+2\hat k$ and $\vec b=2\hat i-\hat j+2\hat k$.
1 $|\vec a+\vec b|^2=4+8+9=21$, so $\sqrt{21}$; $\ |\vec a-\vec b|^2=4-8+9=5$, so $\sqrt5$.
2 Squaring, $|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2$ $=|\vec a|^2-2\vec a\cdot\vec b+|\vec b|^2$, so $4\vec a\cdot\vec b=0$. A parallelogram whose diagonals are equal is a rectangle.
3 $|\vec a|^2-|\vec b|^2=27$ and $|\vec a|^2=4|\vec b|^2$, so $3|\vec b|^2=27$: $\ |\vec b|=3$, $|\vec a|=6$.
4 $\vec c=-(\vec a+\vec b)$, so $49=9+2\vec a\cdot\vec b+25$ and $\vec a\cdot\vec b=7.5$. Then $\cos\theta=\tfrac{7.5}{15}=\tfrac12$: $\ \theta=60^\circ$.
5 $|\hat a-\hat b|^2=1-2\cos\theta+1$ $=2(1-\cos\theta)=4\sin^2\tfrac\theta2$. Since $0\le\tfrac\theta2\le\tfrac\pi2$, $\sin\tfrac\theta2\ge0$, so $|\hat a-\hat b|=2\sin\tfrac\theta2$.
6 With sides $\vec a=\overrightarrow{AB}$, $\vec b=\overrightarrow{AD}$, the diagonals are $\vec a+\vec b$ and $\vec b-\vec a$. Their dot product is $|\vec b|^2-|\vec a|^2=0$ since the sides of a rhombus are equal.
7 $\vec a\cdot\vec b=2-2+4=4$ and $|\vec a|\,|\vec b|=3\cdot3=9$; $\ |4|\le9$ ✓.
IV
Applications.
  • 1 A constant force $\vec F=5\hat i+2\hat j+3\hat k$ N moves a particle from $(1,-1,2)$ to $(3,2,-1)$ (in metres). Find the work done.
  • 2 A child pulls a toy cart $15$ m with a force of $20$ N along a string at $60^\circ$ to the horizontal. Find the work done.
  • 3 Forces $2\hat i-\hat j+3\hat k$ N and $\hat i+2\hat j-\hat k$ N act together on a particle that moves from $A(0,1,2)$ to $B(3,2,4)$ (metres). Find the total work done.
  • 4 Crosswind (illustrative data). With $\hat i$ east and $\hat j$ north, the wind velocity is $6\hat i+8\hat j$ m/s and a runway points along $\tfrac1{\sqrt2}(\hat i+\hat j)$. Find the component of the wind along the runway and the size of the crosswind (the part perpendicular to the runway).
  • 5 A crate of weight $50$ N rests on a ramp. With $\hat i$ horizontal and $\hat k$ up, the weight is $-50\hat k$ N and the ramp rises along the unit vector $\tfrac15(4\hat i+3\hat k)$. Find the component of the weight along the ramp, and interpret its sign.
  • 6 Find the angle between two body diagonals of a cube (for example $\hat i+\hat j+\hat k$ and $-\hat i+\hat j+\hat k$).
1 $\vec d=2\hat i+3\hat j-3\hat k$ m; $\ W=10+6-9=7$ J.
2 $W=20\times15\times\cos60^\circ=150$ J.
3 Total force $3\hat i+\hat j+2\hat k$ N, $\ \vec d=3\hat i+\hat j+2\hat k$ m, so $W=9+1+4=14$ J. (Equivalently, add the work done by each force: $11+3$.)
4 Along the runway: $\tfrac{6+8}{\sqrt2}=7\sqrt2\approx9.90$ m/s. The wind speed is $10$ m/s, so the crosswind is $\sqrt{100-98}=\sqrt2\approx1.41$ m/s.
5 $(-50\hat k)\cdot\tfrac15(4\hat i+3\hat k)=-\tfrac{150}5=-30$ N. Negative: the weight pulls the crate $30$ N down the slope (the rest of the weight, $40$ N, presses into the ramp).
6 Dot product $-1+1+1=1$, magnitudes $\sqrt3$: $\cos\theta=\tfrac13$, $\theta=\cos^{-1}\tfrac13\approx70.53^\circ$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksJEE MainJEE Advanced

How this is examined: the dot product appears in almost every Board paper — angles and perpendicularity (1 mark), projections (2 marks), and identities such as $|\vec a\pm\vec b|^2$ (3 marks). JEE uses $|\vec a+\vec b+\vec c|^2$ expansions and unit-vector conditions.

E1. Board · 1 mark If $|\vec a|=2$, $|\vec b|=\sqrt3$ and $\vec a\cdot\vec b=\sqrt3$, the angle between $\vec a$ and $\vec b$ is  (A) $\tfrac\pi6$  (B) $\tfrac\pi3$  (C) $\tfrac\pi4$  (D) $\tfrac\pi2$

Idea $\cos\theta=\dfrac{\sqrt3}{2\sqrt3}=\dfrac12$.
Answer (B). Trap (A) comes from reading $\tfrac12$ as $\sin\theta$.

E2. Board · 1 mark Assertion–Reason. Assertion (A): if $\vec a\ne\vec 0$ and $\vec a\cdot\vec b=\vec a\cdot\vec c$, then $\vec b=\vec c$. Reason (R): $\vec a\cdot(\vec b-\vec c)=0$ means that $\vec b-\vec c=\vec 0$ or $\vec b-\vec c$ is perpendicular to $\vec a$.  (A) Both A and R are true and R is the correct explanation of A  (B) Both A and R are true but R is not the correct explanation of A  (C) A is true but R is false  (D) A is false but R is true

Idea R is Theorem 10.9(b) applied to $\vec a$ and $\vec b-\vec c$, so R is true. It also shows why A fails: take $\vec a=\hat i$, $\vec b=\hat j$, $\vec c=\hat k$; then $\vec a\cdot\vec b=\vec a\cdot\vec c=0$ but $\vec b\ne\vec c$.
Answer (D). You cannot “cancel” a vector from a dot product.

E3. Board · 2 marks Find the projection of $\vec a=2\hat i-\hat j+\hat k$ on $\vec b=\hat i-2\hat j+\hat k$.

Method $\vec a\cdot\vec b=2+2+1=5$ and $|\vec b|=\sqrt{1+4+1}=\sqrt6$. (1 mark)
Answer $\dfrac{\vec a\cdot\vec b}{|\vec b|}=\dfrac5{\sqrt6}=\dfrac{5\sqrt6}6\approx2.04$. (1 mark)

E4. Board · 3 marks $|\vec a|=5$, $|\vec b|=3$ and $|\vec a-\vec b|=7$. Find the angle between $\vec a$ and $\vec b$, and $|\vec a+\vec b|$.

Dot product $|\vec a-\vec b|^2=|\vec a|^2-2\vec a\cdot\vec b+|\vec b|^2$: $\ 49=25-2\vec a\cdot\vec b+9$, so $\vec a\cdot\vec b=-7.5$. (1 mark)
Angle $\cos\theta=\dfrac{-7.5}{15}=-\dfrac12$, so $\theta=\dfrac{2\pi}3$. (1 mark)
Sum $|\vec a+\vec b|^2=25-15+9=19$, so $|\vec a+\vec b|=\sqrt{19}\approx4.36$. (1 mark) Check: $|\vec a+\vec b|^2+|\vec a-\vec b|^2$ $=2(|\vec a|^2+|\vec b|^2)$: $\ 19+49=68=2(34)$ ✓.

E5. JEE Main $\hat a$ and $\hat b$ are unit vectors with $|\hat a+\hat b|=\sqrt3$. Then $(2\hat a-\hat b)\cdot(\hat a+3\hat b)$ equals  (A) $\tfrac32$  (B) $4$  (C) $\tfrac52$  (D) $\tfrac92$

Method $3=|\hat a+\hat b|^2=1+2\hat a\cdot\hat b+1$, so $\hat a\cdot\hat b=\tfrac12$. Expand: $(2\hat a-\hat b)\cdot(\hat a+3\hat b)$ $=2+6\,\hat a\cdot\hat b-\hat b\cdot\hat a-3$ $=-1+5\cdot\tfrac12=\tfrac32$.
Answer (A). Trap (B) drops the factor $2$ in $|\hat a+\hat b|^2$, giving $\hat a\cdot\hat b=1$ and $-1+5=4$; (C) writes $-\hat b\cdot\hat a$ as $+\tfrac12$; (D) forgets the term $-\hat b\cdot3\hat b=-3$.

E6. JEE Advanced (Numerical answer.) Vectors $\vec a,\vec b,\vec c$ satisfy $\vec a+\vec b+\vec c=\vec 0$, $|\vec a|=3$, $|\vec b|=4$, $|\vec c|=5$. Find $-(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)$.

Square the sum $0=|\vec a+\vec b+\vec c|^2$ $=|\vec a|^2+|\vec b|^2+|\vec c|^2$ $+\,2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)$.
Substitute $9+16+25=50$, so $\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a=-25$.
Answer $\mathbf{25}$. (Here $\vec a\perp\vec b$: from $\vec c=-(\vec a+\vec b)$, $25=9+2\vec a\cdot\vec b+16$ gives $\vec a\cdot\vec b=0$ — the 3-4-5 right triangle.)
🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. Which runway should an airport use today? Use dot products to split the wind into a headwind and a crosswind for each runway direction.

Why it matters. Aircraft take off and land into the wind, and every aircraft type has a maximum crosswind it can safely handle. Air-traffic controllers choose the runway by projecting the wind vector on each runway direction — the projection is the headwind (or tailwind), and the perpendicular part is the crosswind. Runways are named by their direction: runway 09 points to $90^\circ$ (east), runway 27 to $270^\circ$.

What to do.

  1. Find the published hourly weather reports (METAR) for an airport near you — they are freely available online — and record the wind direction (the direction the wind blows from) and speed for $8$–$12$ different hours.
  2. Look up the airport's runway directions. With $\hat i$ east and $\hat j$ north, write each runway as a unit vector $\hat r$ and each wind as a vector $\vec w$ (careful: a wind from $\phi$ blows towards $\phi+180^\circ$).
  3. For each hour compute the headwind $-\vec w\cdot\hat r$ and the crosswind $\sqrt{|\vec w|^2-(\vec w\cdot\hat r)^2}$ on each runway. Which runway would you choose each hour?
  4. Compare with the runway the airport reported in use, if the report shows it.

Tools: METAR archive (any public aviation-weather site), spreadsheet (SIN, COS, RADIANS, SQRT).

Hand in: Your table of winds, headwinds and crosswinds, the runway you would choose each hour with a reason, and a one-paragraph explanation of the projection idea for a younger student.

Going further. Light aircraft often have a crosswind limit of about $15$ knots. For which wind directions does a $25$-knot wind exceed that limit on every runway of your airport?

Instructor note: with bearing $\phi$ measured clockwise from north, a unit vector is $\sin\phi\,\hat i+\cos\phi\,\hat j$. A wind from $\phi_w$ at speed $s$ is $\vec w=-s(\sin\phi_w\hat i+\cos\phi_w\hat j)$, so the headwind on runway $\phi_r$ is $s\cos(\phi_w-\phi_r)$ and the crosswind $s|\sin(\phi_w-\phi_r)|$; students should reach this simplification themselves. 1 knot $\approx0.514$ m/s.

🧪 Self-check quiz

0/5 correct

Q1. $(\hat i+\hat j)\cdot(\hat i-\hat k)$ equals

Q2. For non-zero vectors, $\vec a\cdot\vec b=0$ means that

Q3. The projection of $3\hat i+4\hat j$ on $\hat i$ is

Q4. $\hat a$ and $\hat b$ are perpendicular unit vectors. Then $|\hat a+\hat b|$ equals

Q5. If $\vec a\cdot\vec b\lt0$ for non-zero $\vec a,\vec b$, the angle $\theta$ between them satisfies

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