Category Archives: Vectors

April 7, 2025 · 9:09 am

Vector and Scalar Projection

The vector projection (vector resolution or vector component) of $\mathbf{a}$ onto a non-zero vector $\mathbf{b}$ is splitting $\mathbf{a}$ into two vectors, one is parallel to $\mathbf{b}$ (the vector projection) and one perpendicular to $\mathbf{b}$

In the above diagram $\mathbf{a_1}$ is the vector projection of $\mathbf{a}$ onto $\mathbf{b}$ and $\mathbf{a_2}$ is perpendicular to $\mathbf{b}$ .

How do we find $\mathbf{a_1}$ and $\mathbf{a_2}$ ?

Using right trigonometry,

$cos(\theta)=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}$

Remember the scalar product (dot product) of vectors is

(1) $\begin{equation*}\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|cos(\theta)\end{equation*}$

Hence $cos(\theta)=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$

$\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}$

and, $|\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\times|\mathbf{a}|$

$|\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$

This is the scalar projection of $\mathbf{a}$ onto $\mathbf{b}$

To find the vector projection we need to multiply by $\mathbf{\hat{b}}$ , that is find a vector with the same magnitude as $\mathbf{a_1}$ in the direction of $\mathbf{b}$ .

The vector projection is

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \mathbf{\hat{b}}$

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \frac{\mathbf{b}}{|\mathbf{b}|}$

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{(|\mathbf{b}|)^2}\times \mathbf{b}$

Now for $\mathbf{a_2}$ , we know $\mathbf{a}=\mathbf{a_1}+\mathbf{a_2}$

Hence, $\mathbf{a_2}=\mathbf{a}-\mathbf{a_1}$

Example

From Cambridge Year 11 Specialist Mathematics (Chapter 3)

(a) $\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}$

$=\frac{\begin{pmatrix}4\\1\end{pmatrix}\cdot\begin{pmatrix}1\\-1\end{pmatrix}}{2}\times(\mathbf{i}-\mathbf{j})$

$=\frac{3}{2}{(\mathbf{i}-\mathbf{j})$

(b) $4\mathbf{i}+\mathbf{j}-\frac{3}{2}{(\mathbf{i}-\mathbf{j})=\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}$

(c)

The shortest distance (green vector) is the vector component of $\mathbf{a}$ perpendicular to $\mathbf{b}$ , i.e. $\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}$

$|\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}|=\sqrt{\frac{25}{4}+\frac{25}{4}}=\sqrt{\frac{50}{4}}=\frac{5\sqrt{2}}{2}$

Using a Vector Method to Find an Angle Bisector

Points $A$ and $B$ are defined by the position vectors $\mathbf{a}=3\mathbf{i}+4\mathbf{j}$ and $\mathbf{b}=12\mathbf{i}+5\mathbf{j}$ .

Find a vector that bisects $\angle{AOB}$ .

If we think about how we add vectors using the parallelogram rule

Finding the resultant vector using the parallelogram rule

we can take advantage of the geometric properties of parallelograms (or of a rhombus).

If $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, then the parallelogram is a rhombus, and the diagonal (i.e the resultant) bisects the angle.

We need to find the sum of the unit vectors.

$|\mathbf{a}|=\sqrt{3^2+4^2}=5$

$\therefore \hat{\mathbf{a}}=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}$

$|\mathbf{b}|=\sqrt{12^2+5^2}=13$

$\therefore \hat{\mathbf{b}}=\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}$

The vector that bisects $\angle{AOB}$ is

$\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}+\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}=\frac{99}{65}\mathbf{i}+\frac{64}{65}\mathbf{j}$

Vectors – Closest Approach (2D)

At 10am, object $M$ travelling with constant velocity $(20, 10)$ km/h is sighted at the point with position vector $(-100, 100)$ km. At 11am object $N$ travelling with constant velocity $v=(x,y)$ km/h is sighted at the point with position vector $(-20,-50)$ km respectively. Use a scalar product method to determine $v$ given that the two objects were closest together at a distance of $40$ km at 4pm.

OT Lee Mathematics Specialist Year 11

At 4pm $M$ is at the point with position vector $(-100+6\times 20, -100+6\times 10)=(20, -40)$

and $N$ is at the point with position vector $(-20+5\times x, -50+5\times y)=(-20+5x, -50+5y)$

We know the distance between $M$ and $N$ at 4pm is $40$ km.

Hence,

$\begin{equation*}40^2=(20-(-20+5x))^2+(-40-(-50+5y))^2\end{equation}$

$\begin{equation*}40^2=(40-5x)^2+(10-5y)^2\end{equation}$

$\begin{equation*}1600=25x^2-400x+1600+25y^2-100y+100\end{equation}$

(1) $\begin{equation*}0=x^2+y^2-16x-4y+4\end{equation*}$

In the diagram below, I have found the position vector of $N$ relative to $M$ $(_Nr_M)$ and the velocity of $N$ relative to $M$ $(N_v_M)$

We know that when $_Nr_M\cdot_Nv_M=0$ $M$ and $N$ are the closest distance apart.

$\begin{equation*}(-40+5x,-10+5y)\cdot(x-20,y-10)=0\end{equation}$

$\begin{equation*}(-40+5x)(x-20)+(-10+5y)(y-10)\end{equation}$

$\begin{equation*}5x^2+5y^2-140x-60y+900=0\end{equation}$

(2) $\begin{equation*}x^2+y^2-28x-12y+180=0\end{equation*}$

Two equations and two unknowns which we can solve simultaneously. Both equations are circles.

Equation $(1)$ becomes

(3) $\begin{equation*}(x-8)^2+(y-2)^2=64\end{equation*}$

and equation $(2)$ becomes

(4) $\begin{equation*}(x-14)^2+(y-6)^2=52\end{equation*}$

From equation $(3)$

$\begin{equation*}x=\sqrt{64-(y-2)^2}+8\end{equation}$

We will worry about the negative version later.

Substitute for $x$ into equation $(4)$

$\begin{equation*}(\sqrt{64-(y-2)^2}+8-14)^2+(y-6)^2=52\end{equation}$

$\begin{equation*}(\sqrt{64-(y-2)^2}-6)^2+(y-6)^2=52\end{equation}$

$\begin{equation*}64-(y-2)^2-12\sqrt{64-(y-2)^2}+36+y^2-12y+36=52\end{equation}$

$\begin{equation*}136-y^2+4y-4+y^2-12y-12\sqrt{64-(y-2)^2}=52\end{equation}$

$\begin{equation*}-12\sqrt{64-(y-2)^2}=-80+8y\end{equation}$

$\begin{equation*}\sqrt{64-(y-2)^2}=\frac{20}{3}-\frac{2}{3}y\end{equation}$

Square both sides of the equation

$\begin{equation*}64-(y-2)^2=\frac{400}{9}-\frac{80}{9}y+\frac{4}{9}y^2\end{equation}$

$\begin{equation*}60-y^2+4y=\frac{400}{9}-\frac{80}{9}y+\frac{4}{9}y^2\end{equation}$

$\begin{equation*}0=\frac{13}{9}y^2-\frac{116}{9}y-\frac{140}{9}\end{equation}$

$\begin{equation*}0=13y^2-116y-140\end{equation}$

$\begin{equation*}0=13y^2-130y+14y-140\end{equation}$

$\begin{equation*}0=13y(y-10)+14(y-10)\end{equation}$

$\begin{equation*}0=(y-10)(13y+14)\end{equation}$

(5) $\begin{equation*}\therefore y=10, y=-\frac{14}{13}\end{equation*}$

Substitute $y=10$ into $x=\sqrt{64-(y-2)^2}+8$

$\begin{equation*}x=\sqrt{64-(y-2)^2}+8\end{equation}$

(6) $\begin{equation*}x=8\end{equation*}$

Substitute $y=-\frac{14}{13}$ into $x$

$\begin{equation*}x=\sqrt{64-(-\frac{14}{13}-2)^2}+8\end{equation}$

(7) $\begin{equation*}x=\frac{200}{13}\end{equation*}$

Now we need to consider the negative version of $x$ . If you work through (like I did above) you end with the same equation for $y$ .

Hence our two values for $v$ are $v=(8,10)$ or $v=(\frac{200}{13},-\frac{14}{13})$ .

Would someone be expected to do this in an exam? I hope not, but I think its worth doing.

Category Archives: Vectors

Vector and Scalar Projection

Example

Using a Vector Method to Find an Angle Bisector

Vectors – Closest Approach (2D)

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