Scalar (Dot) Product, Angle & Projection

MATScalar (Dot) Product

The scalar (dot) product combines two vectors into a number: $\vec{a}\cdot\vec{b}=|\vec{a}|\,|\vec{b}|\cos\theta$ . It measures alignment, so it gives the angle between vectors, tests perpendicularity, and yields the projection of one vector onto another.

ISC examines it through angle-finding, perpendicularity conditions for an unknown, scalar projections, and work-done-by-a-force problems.

Dot product (definition)

\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta

\theta

is the angle between

\vec{a}

and

\vec{b}

0 \le \theta \le \pi

Dot product (components)

\vec{a}\cdot\vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3

\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}

Angle between vectors

\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}

Both vectors nonzero;

\theta = \cos^{-1}

of the right side,

\theta \in [0,\pi]

Perpendicularity condition

\vec{a} \perp \vec{b} \iff \vec{a}\cdot\vec{b} = 0

Holds for nonzero

\vec{a},\vec{b}

; gives an equation for an unknown component.

Scalar projection of a on b

\text{proj}_{\vec{b}}\,\vec{a} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}

Scalar component of

\vec{a}

along

\vec{b}

; the vector projection is this times

\dfrac{\vec{b}}{|\vec{b}|}

Work done by a force

W = \vec{F}\cdot\vec{d}

\vec{F}

is the constant force,

\vec{d}=\vec{AB}

is the displacement;

W

is in joules.

The dot product is commutative: $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}$ , and distributive over addition.
$\vec{a}\cdot\vec{a}=|\vec{a}|^2$ , so the magnitude is $|\vec{a}|=\sqrt{\vec{a}\cdot\vec{a}}$ .
For the standard basis: $\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1$ and $\hat{i}\cdot\hat{j}=\hat{j}\cdot\hat{k}=\hat{k}\cdot\hat{i}=0$ .
The sign of $\vec{a}\cdot\vec{b}$ tells the angle type: positive means acute, zero means right angle, negative means obtuse.
Scalar projection can be negative (when $\theta$ is obtuse); it is a signed length, not a magnitude.
To find work done, first compute the displacement $\vec{d}=\vec{AB}=\vec{b}-\vec{a}$ (position of $B$ minus position of $A$ ), then take the dot product with $\vec{F}$ .
The projection of $\vec{a}$ on $\vec{b}$ divides by $|\vec{b}|$ (the vector projected onto), not by $|\vec{a}|$ .

Where the marks go

Dividing by $|\vec{a}|$ instead of $|\vec{b}|$ when finding the projection of $\vec{a}$ on $\vec{b}$ .
Forgetting that $\theta$ from $\cos^{-1}$ lies in $[0,\pi]$ , so a negative cosine gives an obtuse angle, not a sign to discard.
Treating the dot product as a vector; the result is a scalar, so writing it with $\hat{i},\hat{j},\hat{k}$ is wrong.
Using the position vectors $\vec{a},\vec{b}$ directly as the displacement in work problems instead of the difference $\vec{b}-\vec{a}$ .

How the board asks it

Numerical $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$
Find the angle between the vectors $\vec{a}=2\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$ .
Numericalperpendicularity condition $\vec{a}\cdot\vec{b}=0$
Find the value of $\lambda$ for which the vectors $\vec{a}=\lambda\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}=4\hat{i}-9\hat{j}+2\hat{k}$ are perpendicular to each other.
Numericalscalar projection $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$
Find the scalar projection of the vector $\vec{a}=\hat{i}+3\hat{j}+7\hat{k}$ on the vector $\vec{b}=7\hat{i}-\hat{j}+8\hat{k}$ .
Numerical $\vec{a}\cdot\vec{a}=|\vec{a}|^2$ and distributivity
If $|\vec{a}|=3$ , $|\vec{b}|=4$ and $\vec{a}\cdot\vec{b}=6$ , evaluate $(2\vec{a}-\vec{b})\cdot(\vec{a}+3\vec{b})$ .
Derive / prove $|\vec{a}\pm\vec{b}|^2=|\vec{a}|^2\pm 2\vec{a}\cdot\vec{b}+|\vec{b}|^2$
If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$ , prove that $\vec{a}$ is perpendicular to $\vec{b}$ .
Numericalwork $\vec{F}\cdot\vec{d}$ with $\vec{d}=\vec{b}-\vec{a}$
A constant force $\vec{F}=3\hat{i}+2\hat{j}-4\hat{k}$ acts on a particle that is displaced from the point $A(1,0,2)$ to the point $B(4,3,-1)$ . Calculate the work done by the force.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.