Category Archives: Vectors

November 5, 2026 · 2:23 am

3D Vector Motion Question

The particle $P$ is moving with velocity $v_{P}=(1,1,2)$ metres per second from its initial point with position vector $(-38, 31, -18)$ metres. The particle $Q$ is moving with velocity $v_{Q}=(1,\lambda,-2)$ metres per second, where $\lambda$ is a constant. At $T$ seconds, where $T>0$ , particle $Q$ passes through the point with position vector $(-35,56,\mu)$ metres, where $\mu$ is a constant.

Both particles start their motion at the same time, the paths of the particles are perpendicular, and the paths of the particles intersect at point $A$ .

(a) Determine the position vector of $A$ .

The paths of particles P and Q are perpendicular, so the dot product of their velocity vectors is zero:

$(1,1,2)\cdot(1,\lambda,-2)=0$

$1+\lambda-4=0$

$\lambda=3$

The position vectors are

$\mathbf{r}_P=\begin{pmatrix}-38+t_1\\31+t_1\\-18+2t_1\end{pmatrix}$

$\qquad\mathbf{r}_Q=\begin{pmatrix}-35+t_2\\56+3t_2\\\mu-2t_2\end{pmatrix}$

Solve simultaneously:

$-38+t_1=-35+t_2$

$31+t_1=56+3t_2$

Equation 1 subtract equation 2:

$-69=-91-2t_2$

$t_2=11$

Substitute $t_2=11$ into equation 1:

$-38+t_1=-35+11$

$t_1=14$

Hence, the position vector of A is

$\begin{pmatrix}-24\\45\\10\end{pmatrix}$

(b) Determine the value of $\mu$ and the value of $T$ so that the particles $P$ and $Q$ will collide

The $\mathbf{j}$ -component of A is 10, therefore

$\mu-2t_2=10$

$t_2=11$

$\mu=10+2(11)$

$\mu=32$

For the two particles to collide,

$t_1=t_2$

Particle Q is at the point $(-35,56,32)$ at time $T$ , and it is at point $(-24,45,10)$ 11 seconds later. Both particles start at the same time, hence

$T=3$

Both particles arrive at A at 14 seconds.

Particle $N$ is moving with constant velocity $\mathbf{v}_N)$ so that its path is perpendicular to both paths of particles $P$ and $Q$ .

(c) Determine the speed of particle $N$ , correct to 0.001 metres per second

As $\mathbf{v}_N$ is perpendicular to $P$ and $Q$ , the vector must be parallel to

$\begin{pmatrix}1\\1\\2\end{pmatrix} \times \begin{pmatrix}1\\3\\-2\end{pmatrix}=\begin{pmatrix}-8\\4\\2\end{pmatrix}$

Using the given $\mathbf{j}$ -component of $\mathbf{v}_N$ ,

$0.8=k\times2$

$k=0.4$

Therefore,

$\mathbf{v}_N=\begin{pmatrix}-3.2\\1.6\\0.8\end{pmatrix}$

and the speed of particle $N$ is

$\sqrt{(-3.2)^2+(1.6)^2+(0.8)^2}$

$\frac{2\sqrt{101}}{5}\approx 4.02\text{ metres per second}.$

Leave a Comment

Filed under Algebra, Cross Product, Simultaneous Equations, Vectors, Year 12 Specialist Mathematics

Tagged as vector motion, vectors, year 12 mathematics specialist

April 21, 2026 · 11:26 am

Find the Equation of a Plane Given Three Points

Find the Cartesian equation of the plane containing the points $A(1, 2, 3), B(-3, 0, 2)$ and $C(2, -4, -1)$

Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$
$\overrightarrow{AB}=\begin{pmatrix}-2\\0\\2\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}-3\\-2\\-1\end{pmatrix}$
$\overrightarrow{AC}=\begin{pmatrix}2\\-4\\-1\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}1\\-6\\-4\end{pmatrix}$

Find the cross product $\overrightarrow{AB}\times \overrightarrow{AC}$
$\begin{pmatrix}-3\\-2\\-1\end{pmatrix}\times\begin{pmatrix}1\\-6\\-4\end{pmatrix}=\begin{pmatrix}2\\-13\\20\end{pmatrix}$
This is the normal, $overrightarrow{n}$ , to the plane.
We know $A$ is on the plane

$\begin{equation*}\overrightarrow{n}\cdot (\overrightarrow{r}-\overrightarrow{OA})=0\end{equation}$

Hence $\overrightarrow{n}\cdot \overrightarrow{r}=\overrightarrow{n}\cdot \overrightarrow{OA}$
$\overrightarrow{n}\cdot \overrightarrow{r}=\begin{pmatrix}2\\-13\\20\end{pmatrix} \cdot \begin{pmatrix}1\\2\\3\end{pmatrix}=38$
Therefore the Cartesian equation of the plane is

$\begin{equation*}2x-13y+20z=38\end{equation}$

Leave a Comment

Filed under Algebra, Cross Product, Vectors, Year 12 Specialist Mathematics

September 24, 2025 · 9:31 am

Eigenvalues and Eigenvectors

My Year 11 Specialist students have had an investigation which involves finding eigenvalues, eigenvectors and lines that are invariant under a particular linear transformation. This is not part of the course, but I feel for teachers who have to create new investigations every year.

Let’s find the eigenvalues and eigenvectors for matrix $T=\begin{bmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}$

We want to find $\lambda$ such that

(1) $\begin{equation*}T\textbf{v}=\lambda \textbf{v}\end{equation*}$

We solve $det(T-\lambda I)=0$

$\begin{equation*}T-\lambda I=\begin{bmatrix}-\frac{1}{2}-\lambda&-\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2}-\lambda\end{bmatrix}\end{equation}$

$det\left (T-\lambda I \right )=\left (-\frac{1}{2}-\lambda \right ) \left ( \frac{1}{2}-\lambda \right )- \left (-\frac{\sqrt{3}}{2} \right ) \left ( -\frac{\sqrt{3}}{2} \right )$

Hence $0=\lambda^2-1$ and $\lambda=\pm 1$

When $\lambda=1$ , $\begin{bmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}x_1\\x_2\end{bmatrix}$

Hence, $-\frac{x_1}{2}-\frac{\sqrt{3}x_2}{2}=x_1$

$x_2=-\frac{3x_1}{\sqrt{3}}$ and the eigenvector is $\begin{bmatrix}1\\-\sqrt{3}\end{bmatrix}$

When $\lambda=-1$ , $\begin{bmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}-x_1\\-x_2\end{bmatrix}$

Hence, $-\frac{x_1}{2}-\frac{\sqrt{3}x_2}{2}=-x_1$

$x_2=\frac{x_1}{\sqrt{3}}$ and the eigenvector is $\begin{bmatrix}1\\\frac{1}{\sqrt{3}}\end{bmatrix}$

Which means the invariant lines are $y=-\sqrt{3}x$ and $y=\frac{x}{\sqrt{3}}$

A quadrilateral with vertices on our lines

The vertices after they have been transformed – A and C remain in the same place (they are on the $\lambda=1$ line)

The quadrilateral (purple) after the transformation

Leave a Comment

Filed under Co-ordinate Geometry, Eigenvalues, Matrices, Transformations, Vectors, Year 11 Specialist Mathematics

June 4, 2025 · 5:30 am

Closest Approach (Shortest Distance) e-activity (Casio Classpad)

At 1pm, object $H$ travelling with constant velocity $\begin{pmatrix}200\\10\end{pmatrix}$ km/h is sighted at the point with position vector $\begin{pmatrix}-90\\-100\end{pmatrix}$ km. At 2pm object $J$ travelling with constant velocity $\begin{pmatrix}100\\-100\end{pmatrix}$ km/h is sighted at the point with position vector $\begin{pmatrix}20\\-120\end{pmatrix}$ km. Determine the minimum distance between $H$ and $J$ and when this occurs.

OT Lee Mathematics Specialist Year 11 Unit 1 and 2 Exercise 10.1 Question 6.

(1) $\begin{equation*}\mathbf{r_H}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}\end{equation*}$

(2) $\begin{equation*}\mathbf{r_J}=\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix}\end{equation*}$

$\begin{pmatrix}-80\\-20\end{pmatrix}$ is the position vector of $J$ at 1pm.

Find the relative displacement of $H$ to $J$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\mathbf{r_H}-\mathbf{r_J}\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}-(\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix})\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

Find the relative velocity of $H$ to $J$

$\begin{equation*}\mathbf{_H}\mathbf{v_J}=\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

The relative displacement is perpendicular to the relative velocity at the closest approach.

That is

(3) $\begin{equation*}\mathbf{_H}\mathbf{r_J}\cdot\mathbf{_H}\mathbf{v_J}=0\end{equation*}$

$\begin{equation*}(\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix})\cdot(\begin{pmatrix}100\\110\end{pmatrix})=0\end{equation}$

$\begin{equation*}(-10+100t)(100)+(-80+110t)(110)=0\end{equation}$

$\begin{equation*}-1000+10 000t-8800+12100t=0\end{equation}$

$\begin{equation*}22100t=9800\end{equation}$

$\begin{equation*}t=\frac{98}{221}\end{equation}$

Substitute $t=\frac{98}{221}$ into the relative displacement and find the magnitude.

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+\frac{98}{221}\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}34\frac{76}{221}\\-31\frac{49}{221}\end{pmatrix}\end{equation}$

$\begin{equation*}\|\begin{pmatrix}34\frac{76}{221}\\-31\frac{49}{221}\end{pmatrix}\|=46.4\end{equation}$

The closest objects $H$ and $J$ get to each other is $46.4$ km at 1:27pm.

I have made an e-activity for this.

Leave a Comment

Filed under Classpad Skills, Closest Approach, Vectors, Year 11 Mathematical Methods

Tagged as casio classpad, eactivity, vectors closest approach

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. $\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}$

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

Leave a Comment

Filed under Right Trigonometry, Trigonometry, Vector Projection, Vectors, Year 11 Specialist Mathematics

Tagged as dot product, scalar product, scalar projection, vector projection

April 6, 2025 · 1:51 am

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

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

Leave a Comment

Filed under Geometry, Vectors, Year 11 Specialist Mathematics

Tagged as bisecting an angle, vectors, Year 11 Specialist Mathematics

November 14, 2024 · 4:19 am

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.

Leave a Comment

Filed under Closest Approach, Vectors, Year 11 Specialist Mathematics

Category Archives: Vectors

3D Vector Motion Question

Find the Equation of a Plane Given Three Points

Eigenvalues and Eigenvectors

Closest Approach (Shortest Distance) e-activity (Casio Classpad)

Vector and Scalar Projection

Example

Using a Vector Method to Find an Angle Bisector

Vectors – Closest Approach (2D)

Recent Posts

Categories

Archives