Vectors in three dimensionsAQA A-Level Maths: Revision notes

In three dimensions a third axis, $z$, is added, perpendicular to both $x$ and $y$, with unit vector $\mathbf k$. A vector is $x\mathbf i+y\mathbf j+z\mathbf k$ or the column vector $\begin{pmatrix}x\\ y\\ z\end{pmatrix}$, and a point $(x,y,z)$ has position vector $x\mathbf i+y\mathbf j+z\mathbf k$. Addition, subtraction and multiplication by a scalar work component by component, exactly as in two dimensions. For example, with $\mathbf a=2\mathbf i-3\mathbf j+6\mathbf k$ and $\mathbf b=\mathbf i+2\mathbf j-2\mathbf k$: $2\mathbf a-3\mathbf b=(4-3)\mathbf i+(-6-6)\mathbf j+(12+6)\mathbf k=\mathbf i-12\mathbf j+18\mathbf k$.

The magnitude of $x\mathbf i+y\mathbf j+z\mathbf k$ is $\sqrt{x^2+y^2+z^2},$ from Pythagoras applied twice. So $|2\mathbf i-3\mathbf j+6\mathbf k|=\sqrt{4+9+36}=7$. The unit vector in the direction of $\mathbf a$ is $\frac{\mathbf a}{|\mathbf a|}$. The distance between two points with position vectors $\mathbf a$ and $\mathbf b$ is $|\mathbf b-\mathbf a|$, and $\overrightarrow{AB}=\mathbf b-\mathbf a$.

Two vectors are parallel if one is a scalar multiple of the other, $\mathbf v=k\mathbf u$, so all three component ratios must match. If $\mathbf u=2\mathbf i-\mathbf j+4\mathbf k$ and $\mathbf v=-6\mathbf i+3\mathbf j+\mu\mathbf k$, then $-6=2k$ gives $k=-3$, which also fits $3=-1\times(-3)$, so $\mu=4\times(-3)=-12$. Three points are collinear if $\overrightarrow{AC}=\lambda\overrightarrow{AB}$. To find a point $C$ with $\overrightarrow{AC}=2\overrightarrow{AB}$, use $\overrightarrow{OC}=\mathbf a+2\overrightarrow{AB}$.

A condition on the magnitude gives an equation for an unknown component. For $\mathbf u=2\mathbf i-\mathbf j+4\mathbf k$ and $\mathbf w=2\mathbf i+3\mathbf j+q\mathbf k$ with $|\mathbf u+\mathbf w|=6$: $\mathbf u+\mathbf w=4\mathbf i+2\mathbf j+(4+q)\mathbf k$, so $16+4+(4+q)^2=36$ and $(4+q)^2=16$. Then $4+q=\pm4$, so $q=0$ or $q=-8$. Squaring gives two roots; check each against the question.

A body starting at position $\mathbf r_0$ with constant velocity $\mathbf v$ has position $\mathbf r=\mathbf r_0+t\mathbf v$ at time $t$. The speed is the magnitude $|\mathbf v|$, a scalar; the velocity also includes direction. The displacement between two times is the change in position. For a constant acceleration $\mathbf a$ the vector forms of the equations of motion are $\mathbf v=\mathbf u+t\mathbf a$ and $\mathbf r=\mathbf r_0+t\mathbf u+\frac12t^2\mathbf a$. Example: drone $A$ at $5\mathbf i-4\mathbf j+10\mathbf k$ with velocity $2\mathbf i+3\mathbf j+6\mathbf k$ has speed $\sqrt{4+9+36}=7$ m s⁻¹. At $t=4$ it is at $13\mathbf i+8\mathbf j+34\mathbf k$. It reaches height $70$ m when the $\mathbf k$-component is $10+6t=70$, so $t=10$ s.

Two bodies are at the same position at time $t$ if $\mathbf r_A(t)=\mathbf r_B(t)$. Equate the $\mathbf i$, $\mathbf j$ and $\mathbf k$ components. All three equations must give the same $t$. Drone $B$ at $17\mathbf i+2\mathbf j+16\mathbf k$ with velocity $-2\mathbf i+\mathbf j+4\mathbf k$ gives $5+2t=17-2t$, $-4+3t=2+t$ and $10+6t=16+4t$, each solved by $t=3$. So the drones collide at $11\mathbf i+5\mathbf j+28\mathbf k$. If the equations gave different times, the paths would cross but the bodies would not be there together, and there is no collision.