Category Archives: Vectors

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}

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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

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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 46.4km at 1:27pm.

I have made an e-activity for this.

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

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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}

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

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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}

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Filed under Geometry, Vectors, Year 11 Specialist Mathematics

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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.

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Filed under Closest Approach, Vectors, Year 11 Specialist Mathematics