Category Archives: Simultaneous Equations

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

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

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

Simultaneous Equation (or is it?)

Solve simultaneously

$X+Y+Z=10$
$XYZ=30$
$XY+YZ+XZ=31$

We could attempt to solve this simultaneously, but I think the algebra would be tricky.

The three equations are related to the roots of a cubic polynomial.

If the general equation of the polynomial is $ax^3+bx^2+cx+d$ , then we know

The sum of the roots

The product of the roots

and

\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}

So from our three equations we have

(1) $\begin{equation*}X+Y+Z=10=\frac{-b}{a}\end{equation*}$

(2) $\begin{equation*}XYZ=30=\frac{-d}{a}\end{equation*}$

(3) $\begin{equation*}XY+YZ+XZ=31=\frac{c}{a}\end{equation*}$

Let $a=1$ , then $b=-10, c=31$ , and $d=-30$

Our cubic is $t^3-10t^2+31t-30$ and we can try to solve it.

The roots will be factors of $-30$ , so $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$

Try $t=2$

$\begin{equation*}(2)^3-10(2)^2+31(2)-30=0\end{equation}$

Hence $t=2$ is a root.

Use synthetic division to find the quadratic factor

$2$	$1$	$-10$	$31$	$-30$
		$2$	$-16$	$30$
	$1$	$-8$	$15$	$0$

The quadratic factor is $x^2-8x+15$ , which factorises to $(x-3)(x-5)$

Hence the solutions are $X=2, Y=3$ , and $Z=5$

We could assume the solutions are natural numbers, then we can look at factors of 30.

Hence the solutions are $X=2, Y=3,$ and $Z=5$

But with this approach we might not be able to find the solutions.

Age Question (Year 8 equation solving)

Eight years ago my father was three times as old as I shall be in five years time. When I was born he was 41 years old. How old am I now?

I always find these age questions a bit weird – a bit of a riddle, and contrived (just so we can solve some equations)

Let $x$ be my age now, and $y$ be my fathers age now.

(1) $\begin{equation*}y=x+41\end{equation*}$

Because my father was 41 when I was born.

(2) $\begin{equation*}y-8=3(x+5)\end{equation*}$

$y-8$ for 8 years ago, and $3(x+5)$ for three times my age in 5 years.

Solve the equations simultaneously. Substitute $y=x+41$ into equation $2$

$\begin{equation*}x+41-8=3(x+5)\end{equation}$

$\begin{equation*}x-33=3x+15\end{equation}$

$\begin{equation*}18=2x\end{equation}$

$\begin{equation*}x=9\end{equation}$

Hence my current age is $9$ .

Factors of Thirty	$X+Y+Z$	$XY+YZ+XZ$
$1, 1, 30$	$1+1+30=32$	$1+30+30=61$
$1, 2, 15$	$1+2+15=18$	$2+15+30=47$
$1, 5, 6$	$1+5+6=12$	$5+6+30=41$
$2, 3, 5$	$2+3+5=10$	$6+10+15=31$

Category Archives: Simultaneous Equations

3D Vector Motion Question

Simultaneous Equation (or is it?)

Age Question (Year 8 equation solving)

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