vectors | Racquel Sanderson

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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

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

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

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 $\angle{AOB}$ .