1. (i) Show that the distinct points $\mathbf{a,b,c}$ are collinear (i.e. lie on a line) in $\mathbb R^n$ if and only if the vectors $\mathbf{b−a}$ and $\mathbf{c−a}$ are linearly dependent.
    (ii) Show that the vectors $\mathbf u=(1,2,−3)$ and $\mathbf v=(6,3,4)$ are perpendicular in $\mathbb R$. Verify directly Pythagoras’ Theorem for the right-angled triangles with vertices $0,\mathbf u,\mathbf v$ and vertices $0,\mathbf u,\mathbf u+\mathbf v$.
    (iii) Let $\mathbf{v,w}$ be vectors in $\mathbb R^n$. Show that if $\mathbf{v·x}=\mathbf{w·x}$ for all $\mathbf x$ in $\mathbb R^n$ then $\mathbf v=\mathbf w$.
    Solution.
    (i) If $\mathbf{b−a}$ and $\mathbf{c−a}$ are linearly dependent nonzero vectors, then $\mathbf{b−a}=k(\mathbf{c−a})$ for some $k\in\mathbb R$, letting $t=0,k,1$, we see that $\mathbf{a,b,c}$ lie on the line $\mathbf r=\mathbf a+t(\mathbf{c-a})$. Conversely, if $a,b,c$ lie on the line $\mathbf{r=d}+t\mathbf{v}$, we have $\mathbf{a=d}+t_1\mathbf{v}$,$\mathbf{b=d}+t_2\mathbf{v}$,$\mathbf{c=d}+t_3\mathbf{v}$ for some $t_1,t_2,t_3\in\mathbb R$, so $\mathbf{b-a}=(t_2-t_1)\mathbf{v},\mathbf{c-a}=(t_3-t_1)\mathbf{v}$, so $\mathbf{b-a}=\frac{t_2-t_1}{t_3-t_1}(\mathbf{c-a})$.
    (ii) $\mathbf u\cdot\mathbf v=1\cdot6+2\cdot3+(-3)\cdot4=0\Rightarrow \mathbf{u,v}$ are perpendicular.
    For $0,\mathbf u,\mathbf v$, we have $|\mathbf u|^2+|\mathbf v|^2=14+61=75=|\mathbf{u-v}|^2$.
    For $0,\mathbf u,\mathbf{u+v}$, the same.
    (iii) Let $\mathbf x=\mathbf e_i$, where $\mathbf e_i$ is $n$-th basis unit vector, then $\mathbf v\cdot \mathbf x$ is equal to the $n$-th coordinate of $\mathbf v$, so each pair of coordinates of $\mathbf v$ and $\mathbf w$ is equal, so $\mathbf v=\mathbf w$.
    Another way : Choose $\mathbf{x=v-w}$, we get $(\mathbf{v-w})^2=0$, so $\mathbf{v=w}$.
  2. Consider the two lines in $\mathbb R^3$ given parametrically by $\mathbf r(λ)=(1,3,0)+λ(2,3,2),\mathbf s(µ)=(2,1,0)+µ(0,2,1)$.
    Show that the shortest distance between these lines is $\sqrt{3/7}$ by solving the simultaneous equations$$(\mathbf{r}(\lambda)-\mathbf{s}(\mu)) \cdot(2,3,2)=0, \quad(\mathbf{r}(\lambda)-\mathbf{s}(\mu)) \cdot(0,2,1)=0$$What geometry do these equations encode?
    Knowledge of partial derivatives: The shortest distance could also be found by solving the equations$$\frac{\partial}{\partial \lambda}\left(|\mathbf{r}(\lambda)-\mathbf{s}(\mu)|^{2}\right)=0, \quad \frac{\partial}{\partial \mu}\left(|\mathbf{r}(\lambda)-\mathbf{s}(\mu)|^{2}\right)=0$$Determine these equations and explain why they are (essentially) the same as the previous two.
    Solution.
    Solving $\begin{cases}2(2\lambda-\mu)+3 (3\lambda-2\mu+2)+2 (2\lambda-1)=0\\2 (3\lambda-2\mu+2)+2\lambda-\mu=0\end{cases}$, we get $\lambda=\frac47,\mu=\frac{12}7$.
    So $\mathbf r(\lambda)=\left(\frac{15}{7},\frac{33}{7},\frac{8}{7}\right),\mathbf s(\mu)=\left(2,\frac{31}{7},\frac{12}{7}\right)$. So $|\mathbf {r-s}|=\sqrt{\frac37}$.
    Geometry: The common perpendicular is the shortest distance between two skew lines in $\mathbb R^3$.
    d
    Knowledge of partial derivatives: $|\mathbf{r}(\lambda)-\mathbf{s}(\mu)|^2=(-1 + 2\lambda)^2 + (2 + 3\lambda - 2\mu)^2 + (2\lambda-\mu)^2$, solving $\begin{cases}34\lambda - 16\mu+8=0\\- 16\lambda + 10\mu-8=0\end{cases}$, we also get $\lambda=\frac47,\mu=\frac{12}7$.
  3. Let $(x, y, z) = (s + t + 2, 3s − 2t + 1, 4s − 3t)$. Show that, as $s, t$ vary, the point $(x, y, z)$ ranges over a plane with equation $ax + by + cz = d$ which you should determine.
    Solution.
    We find a vector normal to the plane: $(1,3,4)\times(1,-2,-3)=(-1,7,-5)$, then $(x,y,z)\cdot(-1,7,-5)$ is constant for all points $(x,y,z)$ on the plane, and this constant is $(2,1,0)\cdot(-1,7,-5)=5$, so the equation is $-x+7y-5z=5$.
  4. Determine, in the form $\mathbf r · \mathbf n = c$, the equations of each of the following planes in $\mathbb R^3$;
    (i) the plane containing the points $(1, 0, 0), (1, 1, 0), (0, 1, 1)$;
    (ii) the plane containing the point $(2, 1, 0)$ and the line $x = y = z$;
    (iii) the two planes containing the points $(1, 0, 1), (0, 1, 1)$ and which are tangential to the unit sphere, centre $\mathbf 0$.
    Solution.
    (i) $\mathbf n=[(1,1,0)-(1,0,0)]\times[(0,1,1)-(1,0,0)]=(1,0,1)$ and $c=(1,0,0)\cdot\mathbf n=1$.
    (ii) Take a point $(1,1,1)$ on the line. $\mathbf n=[(2,1,0)-(1,1,1)]\times(1,1,1)=(1,-2,1)$ and $c=(1,-2,1)\cdot (0,0,0)=0$.
    (iii) Let $(p,q,r)$ be a point on the sphere. $\nabla(x^2+y^2+z^2)=(2x,2y,2z)$, so $(p,q,r)$ is a normal vector to the sphere at $(p,q,r)$, so the tangent plane is $px+qy+rz=1$.
    Solving $\begin{cases}p+r=1\\q+r=1\\p^2+q^2+r^2=1\end{cases}$, we get $(p,q,r)=\left(\frac23,\frac23,\frac13\right)$ or $(0,0,1)$. So the two planes are $2x+2y+z=3$ and $z=1$.
  5. Given a vector $\mathbf a∈\mathbb R^2$ and a constant $0 < λ < 1$, define $\mathbf b=\frac{\mathbf a}{1 − λ^2}$ and prove that $\frac{|\mathbf{r}-\mathbf{a}|^{2}-\lambda^{2}|\mathbf{r}|^{2}}{1-\lambda^{2}}=|\mathbf{r}-\mathbf{b}|^{2}-\lambda^{2}|\mathbf{b}|^{2}$.
    Deduce Apollonius’ Theorem which states that if $O$ and $A$ are fixed points in the plane, then the locus of all points $X$, such that $|AX| = λ |OX|$, is a circle. Find its centre and radius.
    Solution.
    Completing square wrt $r$, we get $\frac{(\mathbf{r}-\mathbf{a})^{2}-\lambda^{2}\mathbf{r}^{2}}{1-\lambda^{2}}=\left(\mathbf r-\frac{\mathbf a}{1-\lambda ^2}\right)^2-\lambda ^2 \left(\frac{\mathbf a}{1-\lambda ^2}\right)^2$.
    $|AX| = λ |OX|\Rightarrow|\mathbf{r}-\mathbf{a}|^{2}-\lambda^{2}|\mathbf{r}|^{2}=0\Rightarrow |\mathbf{r}-\mathbf{b}|^{2}-\lambda^{2}|\mathbf{b}|^{2}=0\Rightarrow|\mathbf{r}-\mathbf{b}|=\lambda|\mathbf{b}|$, therefore the locus of $X$ is a circle with center $\mathbf b$ and radius $\lambda|\mathbf b|$.
  6. A tetrahedron $ABCD$ has vertices with respective position vectors $\mathbf{a, b, c, d}$ from an origin $O$ inside the tetrahedron. The lines $AO, BO, CO, DO$ meet the opposite faces in $E, F, G$.
    (i) Show that a point lies in the plane $BCD$ if and only if it has position vector $λ\mathbf b + µ\mathbf c + ν\mathbf d$ where $λ + µ + ν = 1$.
    (ii) There are $α, β, γ, δ$, not all zero, such that $α\mathbf a + β\mathbf b + γ\mathbf c + δ\mathbf d = 0$. Show that $E$ has position vector $\frac{-\alpha \mathbf{a}}{\beta+\gamma+\delta}$.
    (iii) Deduce that $\frac{|AO|}{|AE|}+\frac{|BO|}{|BF|}+\frac{|CO|}{|CG|}+\frac{|DO|}{|DH|}=3$.
    Solution.
    (i) Let a point $\mathbf r$ lie on the plane $BCD$, then $\exists\mu,\nu:\mathbf r-\mathbf{b}=\mu(\mathbf{c}-\mathbf{b})+\nu(\mathbf{d}-\mathbf{b})\Rightarrow \mathbf r=(1-\mu-\nu) \mathbf{b}+\mu \mathbf{c}+\nu \mathbf{d}$. So $\lambda=1-\mu-\nu$.
    (ii) Because 4 vectors in $\mathbb R^3$ must be linearly dependent, $\exists α,β,γ,δ\in\mathbb R:α\mathbf a + β\mathbf b + γ\mathbf c + δ\mathbf d = 0$.
    The point $E$ is in the plane $BCD$, so having position vector $λ\mathbf b + µ\mathbf c + ν\mathbf d$ where $λ + µ + ν = 1$, and lying on the line $OA$, so having position vector $κ\mathbf a$ for some $κ$. Given that $α\mathbf a + β\mathbf b + γ\mathbf c + δ\mathbf d = 0$, we see$$\mathbf{e}=-\frac{\alpha \mathbf{a}}{\beta+\gamma+\delta}=\frac{\beta \mathbf{b}}{\beta+\gamma+\delta}+\frac{\gamma \mathbf{c}}{\beta+\gamma+\delta}+\frac{\delta \mathbf{d}}{\beta+\gamma+\delta}$$and more we see that this vector is both on $OA$ and also in the plane $BCD$ as $\frac{\beta}{\beta+\gamma+\delta}+\frac{\gamma}{\beta+\gamma+\delta}+\frac{\delta}{\beta+\gamma+\delta}=1.$Hence $\mathbf e$ is the position vector of $E$.
    (iii) Note that $\frac α{β + γ + δ}> 0$ as $E$ lies on the opposite side of $O$ from $A$. So we have$$O A=\mathbf{a} ; \quad A E=\mathbf{a}-\mathbf{e}=\frac{\alpha+\beta+\gamma+\delta}{\beta+\gamma+\delta} \mathbf{a}$$giving$$\frac{|A O|}{|A E|}=|\mathbf{a}| /\left|\frac{\alpha+\beta+\gamma+\delta}{\beta+\gamma+\delta} \mathbf{a}\right|=\frac{\beta+\gamma+\delta}{\alpha+\beta+\gamma+\delta}$$Hence,$$\frac{|A O|}{|A E|}+\frac{|B O|}{|B F|}+\frac{|C O|}{|C G|}+\frac{|D O|}{|D H|}=\frac{\beta+\gamma+\delta}{\alpha+\beta+\gamma+\delta}+\frac{\alpha+\gamma+\delta}{\alpha+\beta+\gamma+\delta}+\frac{\alpha+\beta+\delta}{\alpha+\beta+\gamma+\delta}+\frac{\alpha+\beta+\gamma}{\alpha+\beta+\gamma+\delta}=3$$