Problems

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Solution:

1. Suppose we have \(y = x^k\), then the tangent at \((t,t^k)\) has gradient \(\frac{\d y}{\d x} = kx^{k-1} = kt^{k-1}\) and the normal has gradient \(-\frac1k x^{1-k}\). For the push-off to be the positive \(x\)-axis, we need \(p(x)\) to be the length of the line. The line will have the equation: \begin{align*} && \frac{y - t^k}{x - t} &= -\frac1k t^{1-k} \\ y = 0: && x-t &= \frac{kt^k}{t^{1-k}} \\ && x& =t + kt^{2k-1} \end{align*} The distance from \((t,t^k)\) to \((t+kt^{2k-1},0)\) is \(\sqrt{t^{2k} + k^2t^{4k-2}} = t^k \sqrt{1+k^2t^{2k-2}} = y \sqrt{1+k^2 \frac{y^2}{x^2}} = \frac{y}{x} \sqrt{x^2 + k^2y^2}\), ie \(p(x) = \frac{y}{x}\sqrt{x^2+k^2y^2}\) If \(\left|\mathbf{r}\right|=\left|\mathbf{r-s}\right|\), then we need \(\sqrt{x^2+y^2} = \frac{y}{x}\sqrt{x^2+ky^2}\), but clearly this is satisfied when \(k = 1\).
2. The points are \((t, t^2)\) and the normal has graident \(-\frac{1}{2t}\), the normal vector is \(\frac{1}{\sqrt{1+\frac{1}{4t^2}}}\binom{1}{-\frac1{2t}} = \frac{1}{\sqrt{4t^2+1}} \binom{2t}{-1}\). If we write \(p(x) = q(x)\sqrt{4x^2+1}\) then the the new points are at \(\binom{t+q(t)2t}{t^2-\q(t)}\) an the gradient will be \(\frac{2t-q'(t)}{1+q'(t)2t+2q(t)}\). We need it to be the case that \begin{align*} && 2t &= \frac{2t-q'(t)}{1+2q(t)+2tq'(t)} \\ \Rightarrow && 4tq(t) &= -q'(t)(1+4t^2) \\ \Rightarrow && \frac{q'(t)}{q(t)} &= -\frac{4t}{1+4t^2} \\ \Rightarrow && \ln q(t) &= -\frac12 \ln(1+4t^2)+C \\ \Rightarrow && q(t) &= A(1+4t^2)^{-1/2} \\ \Rightarrow && p(x) &= A \end{align*} So the push-off's are constants.

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Solution: Set up coordinates such that \(A\) at the origin and \(\vec{AB} = \mathbf{x}\) and \(\vec{AD} = \mathbf{y}\) and so \(\vec{AC} = \mathbf{x}+\mathbf{y}\)

1. \begin{align*} AC^2 + BD^2 &= (\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y}) + (\mathbf{y}-\mathbf{x})\cdot(\mathbf{y}-\mathbf{x}) \\ &= 2\mathbf{x}\cdot\mathbf{x} + 2\mathbf{y}\cdot\mathbf{y} \\ &= AB^2 + CD^2 +AD^2 + BC^2 \end{align*}
2. \begin{align*} AC^2 -BD^2 &= (\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y}) - (\mathbf{y}-\mathbf{x})\cdot(\mathbf{y}-\mathbf{x}) \\ &= 4 \mathbf{x}\cdot \mathbf{y} \end{align*} \(\mathbf{x}\cdot\mathbf{y} = |\mathbf{x}||\mathbf{y}|\cos \theta\) which is exactly the lenth of one side mutliplied by the length of the projection to that same side.

\begin{align*} AB^2 - AD^2 &= \mathbf{x}\cdot\mathbf{x} - \mathbf{y}\cdot \mathbf{y} \\ &= (\mathbf{x}+\mathbf{y})\cdot(\mathbf{x}-\mathbf{y}) \\ &= AC \cdot BD \end{align*} So this is the area of the rectangle formed by the length of one diagonal and the projection of the other diagonal onto it.

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Solution:

1. The line \(OA\) is \(\lambda \begin{pmatrix} a^3 \\ a^2 \\ a \end{pmatrix}\). The line \(BC\) is \(\begin{pmatrix} b^3 \\ b^2 \\ b \end{pmatrix} + \mu \begin{pmatrix} c^3-b^3 \\ c^2-b^2 \\ c-b \end{pmatrix}\). If these lies are concurrent then there would be \(\mu\) and \(\lambda\) such that they are equal, and in particular, \begin{align*} && \frac{b^2 + \mu(c^2-b^2)}{b + \mu (c-b)} &= \frac{b^3 + \mu(c^3-b^3)}{b^2 + \mu (c^2-b^2)} \\ \Leftrightarrow && (b^2 + \mu(c^2-b^2))^2 &= (b+\mu(c-b))(b^3+\mu(c^3-b^3)) \\ && b^4 +2\mu b^2 (c^2-b^2) + \mu^2 (c^2-b^2) &= b^4 + \mu(c-b)b^3 + \mu b(c^3-b^3) + \mu^2 (c-b)(c^3-b^3) \\ \Leftrightarrow && 2\mu b^2 (c+b) + \mu^2(c-b)(c+b)^2 &= \mu (b^3 + b(c^2+bc+b^2)) + \mu^2 (c^3-b^3) \\ && \mu = 0 & \Rightarrow a = b \\ \Leftrightarrow && b^2c - bc^2 &= \mu (c^3-b^3-(c-b)(c+b)^2) \\ \Leftrightarrow && bc(b-c) &= \mu (c-b)(c^2+bc+b^2-c^2-2bc-b^2) \\ \Leftrightarrow && bc &= \mu (bc) \\ \Leftrightarrow && \mu &= 1 \\ && \mu = -1 & \Rightarrow a = c \end{align*} Therefore there are no solutions.
2. \begin{align*} \cos \angle AOB &= \frac{ab+a^2b^2+a^3b^3}{\sqrt{a^2+a^4+a^6}\sqrt{b^2+b^4+b^6}} \\ &= \frac{3}{\sqrt{1 + a^2 + a^4} \sqrt{1 + b^2 + b^4}} \\ &> 0 \end{align*} Therefore the angle satisfies \(\angle AOB < \tfrac12 \pi\). We cannot achieve \(0\), since that would require \(a = b = 1\), therefore it falls in the range \(0 < \angle AOB < \tfrac12 \pi\)

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Solution:

Denoting \(\overrightarrow{OY}\) by \(\mathbf{y}\) and \(\overrightarrow{OC}\) by \(\mathbf{c}\) etc, we have: \begin{align*} \mathbf{c} &= \frac23 \mathbf{a} + \frac13 \mathbf{b} \\ \mathbf{d} &= \frac13 \mathbf{a} + \frac23 \mathbf{b} \\ \mathbf{x} &= \lambda \mathbf{a} \\ \mathbf{y} &= (1-\lambda) \mathbf{b} \\ 0 &= (\mathbf{d}-\mathbf{x}) \cdot (\mathbf{c} - \mathbf{y}) \\ &=((\frac13 -\lambda)\mathbf{a} + \frac23 \mathbf{b})\cdot(\frac23 \mathbf{a} + (\frac13-1+\lambda) \mathbf{b} ) \\ &= \frac{2}{3} \cdot (\frac13-\lambda) +\frac23 \cdot(\lambda - \frac23)+(\frac{4}{9}+(\frac13-\lambda)(-\frac23+\lambda))\mathbf{a}\cdot\mathbf{b} \\ &= -\frac{2}{9} + (\frac{4}{9} - \frac{2}{9}+\lambda-\lambda^2)\mathbf{a}\cdot \mathbf{b} \\ &= - \frac{2}{9} + (\frac{2}{9} + \lambda - \lambda^2)\mathbf{a}\cdot \mathbf{b} \\ \mathbf{a}\cdot \mathbf{b} &= \frac{2/9}{2/9+\lambda - \lambda^2} \end{align*} Since \(0 \leq \lambda - \lambda^2 \leq \frac14\), \(\frac{\frac29}{\frac29+\frac14} = \frac8{17} \leq \mathbf{a}\cdot\mathbf{b} \leq 1\) If \(\angle AOB\) is as large as possible, \(\mathbf{a}\cdot\mathbf{b}\) is as small as possible, ie \(\lambda = \frac12\) and \(\mathbf{a}\cdot \mathbf{b} = \frac{8}{17}\) First notice that the length \(OM\) to the midpoint of \(AB\) is \(\sqrt{\frac14 (\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})} = \sqrt{\frac14 (2 + 2\mathbf{a}\cdot \mathbf{b})} = \sqrt{\frac12 + \frac4{17}} = \sqrt{\frac{25}{34}} = \frac{5}{\sqrt{34}}\) Notice that \(XYE\) and \(DCE\) are similar triangles, and so the heights satisfy \(\frac{h_1}{h_2} = \frac{\frac12}{\frac13} = \frac32\). Therefore the length \(OE\) is \(\frac12 \frac{5}{\sqrt{34}} + \frac{3}{5} \frac12 \frac{5}{\sqrt{34}} = \frac{8}{10} \frac{5}{\sqrt{34}} = \frac{4}{\sqrt{34}}\)

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Solution: There are two requirements (assuming they are lines not fixed points): 1. They cannot be parallel, ie \(\mathbf{u} \neq \lambda \mathbf{v}\) for any \(\lambda\) 2. They must lie in the same plane, ie \((\mathbf{b}-\mathbf{a})\cdot (\mathbf{u} \times \mathbf{v}) = 0\) The planes will not intersect in a line if they are either parallel and separate or parallel and the same. If \(b = 1\) or \(b=-1\) the planes are parallel. \begin{align*} && (x+y) + b z &= b+ 2\\ &&b(x+y) + z &= 2b + 1 \\ \Rightarrow && (1-b^2)z &= 2b+1 - b^2 -2b \\ &&&= 1-b^2 \\ \Rightarrow && z &= 1 \\ && x+ y &= 2 \\ \end{align*} Therefore our line is \(\mathbf{r} = \begin{pmatrix} 1+t \\ 1-t \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \) We must have: \(d \neq -1\) to ensure that the lines aren't parallel. We must also have: \begin{align*} 0 &= \left ( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} -\begin{pmatrix} 0 \\ 0 \\ c \end{pmatrix}\right) \cdot \left ( \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ d \\ 0 \end{pmatrix} \right) \\ &= \begin{pmatrix} 1 \\ 1 \\ 1-c \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ d+1 \end{pmatrix} \\ &= (1-c)(d+1) \end{align*} So \(c =1\)

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Solution: Set up a coordinate system so that \(x\) is E-W, \(y\) is N-S and \(z\) is the vertical direction. Also assume \(B\) is the origin, then, \(A = \begin{pmatrix} 0 \\ -100 \\ 5\end{pmatrix}, B = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}, C= \begin{pmatrix} 60 \\ 0\\ -5\end{pmatrix},\). The coal seam has points: \(\begin{pmatrix} 0 \\ -100 \\ -90\end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}, \begin{pmatrix} 60 \\ 0\\ -81\end{pmatrix},\) Therefore we can find the normal to the coal seam: \begin{align*} \mathbf{n} &= \left (\begin{pmatrix} 0 \\ -100 \\ -90\end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}\right ) \times \left ( \begin{pmatrix} 60 \\ 0\\ -81\end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}\right ) \\ &= \begin{pmatrix} 0 \\ - 100 \\ -45\end{pmatrix} \times \begin{pmatrix} 60 \\ 0 \\ -36\end{pmatrix} \\ &= \begin{pmatrix} 3600 \\ -60 \cdot 45 \\ 60 \cdot 100 \end{pmatrix} \\ &= 300\begin{pmatrix} 12 \\ -9 \\ 20\end{pmatrix} \end{align*} To measure the incline \(\theta\) to the horizontal we can take a dot with \(\hat{\mathbf{k}}\), to see: \begin{align*} \cos \theta &= \frac{20}{\sqrt{12^2+(-9)^2+20^2} \sqrt{1^2+0^2+0^2}} \\ &= \frac{20}{25} \\ &= \frac{4}{5} \end{align*} Therefore the angle is \(\cos^{-1} \tfrac 45\) The equation of the seam is \(12x - 9y + 20z = -900\). The equation of the surface is \(5x + 3y + 60z = 0\) We can compute the direction of the overlap again with a cross product: \begin{align*} \mathbf{d} &= \begin{pmatrix} 12 \\ -9 \\ 20\end{pmatrix} \times \begin{pmatrix} 5 \\ 3 \\ 60\end{pmatrix} \\ &= \begin{pmatrix} -600 \\ -620 \\ 81 \end{pmatrix} \end{align*} To get the bearing of this vector we just need to look at the \(x\) and \(y\) components, so it will be \(\tan^{-1} \frac{600}{620} = \tan^{-1} \frac{30}{31}\)

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Solution:

1. \(\overrightarrow{P_{1}P_{2}}=\overrightarrow{P_{5}P_{4}}\) is equivalent to \(z_2 - z_1 = z_4 - z_5\). \(\overrightarrow{P_{2}P_{3}}=\overrightarrow{P_{6}P_{5}}\) is equivalent to \(z_3-z_2 = z_5 - z_6\).
2. \(\overrightarrow{P_{2}P_{4}}\) is perpendicular to \(\overrightarrow{P_{3}P_{6}}\,\) is equivalent to \(\frac{z_4 - z_2}{z_6-z_3} \in i\mathbb{R}\)

If \(z_2 - z_1 =z_4 - z_5\) and \(z_3-z_2 = z_5 - z_6\) then adding we get \(z_3 - z_1 = z_4 - z_6\) or \(z_4 - z_3 = z_6-z_1\), which is equivalent to \(\overrightarrow{P_{3}P_{4}}=\overrightarrow{P_{1}P_{6}}\,\). \(\textrm{Re}(z) > 1, \textrm{Re}(w) < 0, \textrm{Re}(z) +\textrm{Re}(w)=1\). (We only need one of the first two constraints, since the other is implied by the former). Since \(\overrightarrow{P_{2}P_{4}}\) is perpendicular to \(\overrightarrow{P_{3}P_{6}}\,\) we must have that \(\textrm{Im}(z) = \textrm{Im}(w)\). Combined with the vector logic we must have that \(\textrm{Im}(z) = \frac12\). Let \(z = a + \frac12i\) and \(w = (1-a) + \frac12i\). Since \(w - 1 = k(3-2i)(z-1)\) (the angle constraint) we must have that: \begin{align*} &&-a+\frac12i &= k(3-2i)((a-1) \frac12i) \\ &&&= k( 3 a - 2+(\frac72 - 2 a)i) \\ \Rightarrow && \frac{3a-2}{-a} &= \frac{\frac72-2a}{\frac12} \\ \Rightarrow && 3a-2&= 4a^2-7a \\ \Rightarrow && 0 &= 4a^2-10a+2 \\ \Rightarrow && a &= \frac{5 \pm \sqrt{17}}{4} \end{align*} Since \(a > 1, a = \frac{5 +\sqrt{17}}{4}\) and the distance is: \begin{align*} \left | z - w \right | &= | a+\frac12i - ((1-a) +\frac12i ) | \\ &= |2a-1| \\ &= \frac{5+\sqrt{17}}{2}-1 \\ &= \frac{3+\sqrt{17}}{2} \end{align*}

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Solution: Let \(\mathbf{a}\) denote the vector position of \(A\) and similarly for \(B, C, D\). Then we know that \((\mathbf{b}-\mathbf{a})\cdot(\mathbf{b}-\mathbf{a})=(\mathbf{c}-\mathbf{d})\cdot(\mathbf{c}-\mathbf{d})\) and \((\mathbf{b}-\mathbf{c})\cdot(\mathbf{b}-\mathbf{c})=(\mathbf{a}-\mathbf{d})\cdot(\mathbf{a}-\mathbf{d})\). Subtracting these two equations we see that \(|\mathbf{a}|^2 -2\mathbf{a}\cdot\mathbf{b}+2\mathbf{c}\cdot\mathbf{b} - |\mathbf{c}|^2 = |\mathbf{c}|^2-2\mathbf{c}\cdot\mathbf{d}+2\mathbf{a}\cdot\mathbf{d}-|\mathbf{a}|^2\) or \(2|\mathbf{a}|^2 -2\mathbf{a}\cdot\mathbf{b}+2\mathbf{c}\cdot\mathbf{b} - 2|\mathbf{c}|^2 +2\mathbf{c}\cdot\mathbf{d}-2\mathbf{a}\cdot\mathbf{d}=0\) The midpoints of the diagonals \(AC\) and \(BD\) are \(\frac{\mathbf{a}+\mathbf{c}}{2}\) and \(\frac{\mathbf{b}+\mathbf{d}}{2}\), so the line is parallel to: \(\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}\). The diagonals are parallel to \(\mathbf{a}-\mathbf{c}\) and \(\mathbf{b}-\mathbf{d}\). So it suffices to prove that \((\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}) \cdot (\mathbf{a}-\mathbf{c}) = 0\) (since the other will follow by symmetry, \begin{align*} (\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}) \cdot (\mathbf{a}-\mathbf{c}) &= |\mathbf{a}|^2-\mathbf{a}\cdot\mathbf{b}-\mathbf{a}\cdot \mathbf{d}+\mathbf{b}\cdot \mathbf{c}-|\mathbf{c}|^2+\mathbf{c}\cdot \mathbf{d} \\ \end{align*} But this is exactly half the equation we determined earlier, so we are done.

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Solution: The plane is the same as the plane \((\mathbf{r} - \mathbf{a}) \cdot \mathbf{a} = 0\), ie the plane through \(\mathbf{a}\) whose normal is parallel to \(\mathbf{a}\) The distance from \(\mathbf{r}\) to the plane therefore is \(\lambda\) where \(\mathbf{r}+\lambda \frac{1}{a}\mathbf{a}\) must be on the plane, ie \((\mathbf{r}+\frac{\lambda}{a} \mathbf{a} - \mathbf{a})\cdot \mathbf{a} = 0 \Rightarrow \lambda = \frac{a^2-\mathbf{a} \cdot \mathbf{r}}{a}\) But if \(\mathbf{a} \cdot \mathbf{r} = a^2 - ar\) then \(\lambda = r\), ie the distance to the plane is the same as the distance to the origin. \(\mathbf{q} = k \mathbf{p}\) and so \(\mathbf{a} \cdot k \mathbf{p} + a |k|p = a^2\) if \(k > 0\) we will find \(k = 1\) the position vector we already know about, therefore suppose \(k < 0\) so: \begin{align*} && \mathbf{a} \cdot k \mathbf{p} - ka p &= a^2 \\ \Rightarrow && k(a^2-ap)-kap &= a^2 \\ \Rightarrow && k(a^2-2ap) &= a^2 \\ \Rightarrow && k &= \frac{a^2}{a^2-2ap} \end{align*} Therefore \(\mathbf{q} = \frac{a^2}{a^2-2ap} \mathbf{p}\) The line through \(O\) orthogonal to \(\mathbf{p}\) and \(\mathbf{a}\) will be parallel to \(\mathbf{a} \times \mathbf{p}\). Therefore we should consider points of the from \(s \mathbf{a} \times \mathbf{p}\) on the surface \(S\). \begin{align*} && s\mathbf{a} \cdot ( \mathbf{a} \times \mathbf{p}) + sa^2p |\sin \theta| &= a^2 \end{align*} The angle between \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{p}}{ap} = \frac{a^2-ap}{ap} \Rightarrow |\sin \theta| = \sqrt{1-\frac{(a-p)^2}{p^2}} = \frac{1}{p} \sqrt{2ap-a^2}\) Therefore \(sa^2 \sqrt{2ap-a^2} = a^2 \Rightarrow s = \frac{1}{\sqrt{2ap-a^2}}\) and so the points are as required. Noting that \(|\mathbf{p} \times \mathbf{t}| = |\frac{1}{p \sin \theta}\mathbf{p} \times (\mathbf{p} \times \mathbf{a}) | = |\frac{1}{p \sin \theta}p^2a \sin \theta | = pa\) The area of triangle \(PQT\) is : \begin{align*} \frac12 | (\mathbf{p} - \mathbf{t}) \times (\mathbf{q} - \mathbf{t}) | &= \frac12 |\mathbf{p} \times \mathbf{q} - \mathbf{t} \times \mathbf{q} - \mathbf{p} \times \mathbf{t} - \mathbf{t} \times \mathbf{t}| \\ &= \frac12 |\mathbf{t} \times (\mathbf{p} - \mathbf{q})| \\ &= \frac12 \cdot (1 - \frac{a^2}{a^2-2ap})| \mathbf{t} \times \mathbf{p}| \\ &= \frac12 \frac{2ap}{a^2-2ap} \cdot ap \\ &= \frac{ap^2}{a^2-ap} \end{align*}

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Solution: The path can be broken down into \(5\) sections. 1. The section from \((-R,0)\) to \((-a,0)\) which will have distance \(R-a\) and is unchangeable. 2. The distance from \((-a,0)\) to \((-b, \frac{a^2-b^2}{2a})\) whose distance we will calculate shortly. 3. The distance from \((-b, \frac{a^2-b^2}{2a})\) to \((b, \frac{a^2-b^2}{2a})\) which will have distance \(\pi b\). 4. The distance from \((b, \frac{a^2-b^2}{2a})\) to \((a,0)\) which will have the same distance as 2. 5. The distance from \((a,0)\) to \((R,0)\) which will have distance \(R-a\) and we have no control over. \begin{align*} \text{distance 2.} &= \int_b^a \sqrt{1 + \left ( \frac{x}{a}\right)^2 } \d x \end{align*} We want to minimize the total, by varying \(b\), so it makes sense to differentiate and set to zero. \begin{align*} &&0&= -2\sqrt{1+\frac{b^2}{a^2}} + \pi \\ \Rightarrow && \frac{\pi^2}{4} &= 1 + \frac{b^2}{a^2} \\ \Rightarrow && b &= a \sqrt{\frac{\pi^2}{4}-1} \end{align*} Since \(\pi \approx 3\) this point is outside our range \(0 \leq b \leq a\), and our derivative is always positive. Therefore the distance is always increasing and the king would be better off going around the hill as soon as he arrives at it.

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Solution:

Suppose \(P\) is a fraction \(0 \leq k\leq 1\) along \(AB\), then \(\overrightarrow{OP} = \overrightarrow{OA} +\overrightarrow{AP} = \overrightarrow{OA} +k\overrightarrow{AB} = \mathbf{a} + k(\mathbf{b} - \mathbf{a}) = \lambda \mathbf{b} + (1-k) \mathbf{a}\), ie an arbitrary point on \(AB\) has the position vector where \((1-k) = \rho \geq 0\) and \(k= \sigma \geq 0\) and \((1-\lambda) + \lambda = 1\). Any point on the segment \(PC\) will be of the form \(l\mathbf{c} + (1-l) (k \mathbf{b} + (1-k) \mathbf{a})\) which has the form \(\lambda \mathbf{a} + \mu \mathbf{b} + \nu \mathbf{c}\) where \(\lambda + \mu + \nu = (1-l)(1-k) + (1-l)k + l = 1\) and all coefficients are positive.

1. This is equivalent to the area inside the triangle where every point (\(\mathbf{a}, \mathbf{b}, \mathbf{c}\)) has been send to it's negative (\(-\mathbf{a}, -\mathbf{b}, -\mathbf{c}\)), ie
   

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2. Writing points as: \begin{align*} \lambda\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c} &= (1-\mu - \nu)\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c} \\ &= \mathbf{a} + (-\mu)(\mathbf{a}-\mathbf{b}) + (-\nu)(\mathbf{a} - \mathbf{c}) \\ &= \mathbf{a} + (-\mu)(\mathbf{a}-\mathbf{b}) + (-\nu)(\mathbf{a} - \mathbf{c}) + (1+\mu+\nu)\mathbf{0}\\ \end{align*} So this is a translation of \(\mathbf{a}\) of the triangle with vertices at \(\mathbf{0}, \mathbf{a-b}, \mathbf{a-c}\), or a triangle with vertices \(\mathbf{a}, 2\mathbf{a-b}, 2\mathbf{a-c}\).
   

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Solution:

1. \(\mathbf{n} \cdot \mathbf{r} = 0\) is the equation of a plane with normal \(\mathbf{n}\). \(\mathbf{n} \cdot (\mathbf{r}-\mathbf{a}) = 0\) is the equation of a plane through \(\mathbf{a}\) with normal \(\mathbf{n}\). Our expression is: \begin{align*} && \left(\mathbf{a-b}\right) \cdot \mathbf{r}&=\frac{1}{2}(\left|\mathbf{a}\right|^{2}-\left|\mathbf{b}\right|^{2}) \\ &&&=\frac{1}{2}(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}+\mathbf{b}) \\ \Leftrightarrow && \left(\mathbf{a-b}\right) \cdot \left ( \mathbf{r} - \frac12 (\mathbf{a}+\mathbf{b}) \right) &= 0 \end{align*} So this is a plane through \(\frac12 (\mathbf{a}+\mathbf{b})\) perpendicular to \(\mathbf{a}-\mathbf{b}\). ie the plane halfway between \(\mathbf{a}\) and \(\mathbf{b}\) perpendicular to the line between them.
2. \begin{align*} && 0 &= \left(\mathbf{a-r}\right)\cdot\left(\mathbf{b-r}\right) \\ &&&= \mathbf{a} \cdot \mathbf{b} - \mathbf{r} \cdot (\mathbf{a}+\mathbf{b}) + \mathbf{r}\cdot\mathbf{r} \\ &&&= \left ( \mathbf{r}- \frac12(\mathbf{a}+\mathbf{b}) \right) \cdot \left ( \mathbf{r}- \frac12(\mathbf{a}+\mathbf{b}) \right) - \frac14 \left (\mathbf{a}\cdot\mathbf{a}+2\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{b} \right) +\mathbf{a}\cdot\mathbf{b} \\ &&&= \left | \mathbf{r} - \frac12 \left (\mathbf{a}+\mathbf{b} \right) \right|^2 - \left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|^2 \end{align*} Therefore this is a sphere, centre \(\frac12 \left (\mathbf{a}+\mathbf{b} \right)\) radius \(\left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|\)
3. This is a sphere centre \(\mathbf{a}\) radius \(\frac1{\sqrt{2}} \left|\mathbf{a-b}\right|\)
4. This is a sphere centre \(\mathbf{b}\) radius \(\frac1{\sqrt{2}} \left|\mathbf{a-b}\right|\)

Suppose the first two cases are true, then by symmetry it suffices to show that we can prove either of the second cases are true. (Since everything is symmetric in \(\mathbf{a}\) and \(\mathbf{b}\)). It's useful to note that \(\mathbf{r}\cdot \mathbf{r} = \mathbf{r}\cdot \mathbf{b} + \mathbf{r}\cdot \mathbf{a} -\mathbf{a}\cdot\mathbf{b}\) from the second condition. \begin{align*} \left|\mathbf{r-a}\right|^{2} &= \mathbf{r} \cdot \mathbf{r}-2\mathbf{a}\cdot \mathbf{r} + \mathbf{a}\cdot \mathbf{a} \\ &= \mathbf{r}\cdot \mathbf{b} + \mathbf{r}\cdot \mathbf{a} -\mathbf{a}\cdot\mathbf{b} - 2\mathbf{a}\cdot \mathbf{r} + \mathbf{a}\cdot \mathbf{a} \\ &= \mathbf{r} \cdot ( \mathbf{b} - \mathbf{a}) + \mathbf{a} \cdot (\mathbf{a}-\mathbf{b}) \\ &= -\frac{1}{2}(\left|\mathbf{a}\right|^{2}-\left|\mathbf{b}\right|^{2}) + |\mathbf{a}|^2- \mathbf{a}\cdot\mathbf{b} \\ &= \frac{1}{2} |\mathbf{a}-\mathbf{b}|^2 \end{align*} as required. To show the other direction, consider Geometrically, these cases are equivalent, because together they both describe a circle of radius \(\left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|\) in the plane halfway between \(\mathbf{a}\) and \(\mathbf{b}\)