Solution. The distance to an axis is measured perpendicular to it, so the coordinate along that axis plays no part; the distance to a plane is the size of the one coordinate perpendicular to it.

\[\text {(a) } \sqrt {y^2+z^2},\qquad \text {(b) } \sqrt {x^2+z^2},\qquad \text {(c) } \sqrt {x^2+y^2},\qquad \text {(d) } \left |z\right | .\] In (a), for instance, the nearest point of the \(x-\)axis is \((x,0,0)\), and the distance to it is \(\sqrt {(x-x)^2+y^2+z^2}\).

Solution. Throughout, \(d_1 = \sqrt {x^2+y^2+z^2}\) and \(d_2 = \sqrt {x^2+y^2+(z-3)^2}\).

(a)

Squaring \(d_1^2 = 4d_2^2\), \[x^2+y^2+z^2 = 4\Big [x^2+y^2+(z-3)^2\Big ] ,\] \[0 = 3x^2+3y^2+3z^2 - 24z + 36 \implies x^2+y^2+z^2-8z+12 = 0 .\] Completing the square in \(z\), \[x^2+y^2+(z-4)^2 = 4 ,\] a sphere of centre \((0,0,4)\) and radius \(2\).

(b)

\(d_1+d_2 = 6\) says the sum of the distances to two fixed points is constant, which is the defining property of an ellipse; in space it is the surface obtained by revolving that ellipse about the line joining the points. The foci are \((0,0,0)\) and \((0,0,3)\), so the centre is \(\Big (0,0,\dfrac 32\Big )\), the semi-major axis is \(a = 3\) and \(c = \dfrac 32\), giving \(b^2 = 9 - \dfrac 94 = \dfrac {27}{4}\). Hence \[\frac {x^2+y^2}{\frac {27}{4}} + \frac {\big (z-\frac 32\big )^2}{9} = 1 ,\] a prolate spheroid.

(c)

\(\left |d_1-d_2\right | = 2\) is the defining property of a hyperbola, revolved about the same axis. With the same foci, \(a = 1\) and \(c = \dfrac 32\), so \(b^2 = \dfrac 94 - 1 = \dfrac 54\) and \[\frac {\big (z-\frac 32\big )^2}{1} - \frac {x^2+y^2}{\frac 54} = 1 ,\] a hyperboloid of two sheets.

Solution. Place the cube with a vertex at the origin and edges along the axes, of side \(1\). A diagonal is \(\underline {d} = \langle 1,1,1\rangle \) and an edge from the same vertex is \(\underline {e} = \langle 1,0,0\rangle \). Then \[\cos \theta = \frac {\underline {d}\cdot \underline {e}} {\left |\underline {d}\right |\left |\underline {e}\right |} = \frac {1}{\sqrt 3\cdot 1} = \frac {1}{\sqrt 3},\] so \(\theta = \cos ^{-1}\dfrac {1}{\sqrt 3}\approx 54.7^{\circ }\). The side length cancels, so the answer holds for every cube.

Solution. Orthogonality means the dot product vanishes: \[-6b + b^3 + 2b = 0 \implies b^3 - 4b = 0 \implies b\big (b-2\big )\big (b+2\big ) = 0 ,\] so \(b = 0\), \(b = 2\) or \(b = -2\). Note \(b=0\) makes the second vector the zero vector, which is orthogonal to everything; whether to admit it depends on whether the zero vector is allowed.

Solution. Take any two points \(P_1(x_1,y_1)\) and \(P_2(x_2,y_2)\) on the line, so that \[ax_1+by_1 = c \qquad \text {and}\qquad ax_2+by_2 = c .\] Subtracting, \[a\big (x_2-x_1\big ) + b\big (y_2-y_1\big ) = 0 .\] The left-hand side is exactly \(\textbf {N}\cdot \overrightarrow {P_1P_2}\), where \(\overrightarrow {P_1P_2} = \langle x_2-x_1,\ y_2-y_1\rangle \) lies along the line. So \(\textbf {N}\) is orthogonal to every vector along the line, which is what perpendicularity to the line means.

Solution. The scalar projection is \(\text {comp}_{\underline {a}}\underline {b} = \dfrac {\underline {a}\cdot \underline {b}}{\left |\underline {a}\right |}\), and \(\left |\underline {a}\right | = \sqrt {9+0+1} = \sqrt {10}\). The requirement is therefore \[\underline {a}\cdot \underline {b} = 2\sqrt {10}, \qquad \text {that is}\qquad 3b_1 - b_3 = 2\sqrt {10}.\] This is one equation in three unknowns, so there are infinitely many answers. Taking \(b_2 = b_3 = 0\) gives \[\underline {b} = \Big \langle \frac {2\sqrt {10}}{3},\ 0,\ 0\Big \rangle .\] Geometrically, every admissible \(\underline {b}\) lies on a plane perpendicular to \(\underline {a}\) at distance \(2\) from the origin along \(\underline {a}\).

Solution. \[\textbf {a}\cdot \textbf {b} = 10 + 33 - 8 = 35,\qquad \left |\textbf {a}\right |^2 = 100+121+4 = 225 .\] Hence \[\text {proj}_{\textbf {a}}\textbf {b} = \frac {\textbf {a}\cdot \textbf {b}}{\left |\textbf {a}\right |^2}\textbf {a} = \frac {35}{225}\langle 10,11,-2\rangle = \frac {7}{45}\langle 10,11,-2\rangle = \Big \langle \frac {14}{9},\ \frac {77}{45},\ -\frac {14}{45}\Big \rangle .\]

Solution. \[\textbf {a}\cdot \textbf {b} = 4 + 20 - 11 = 13,\qquad \left |\textbf {b}\right | = \sqrt {4+100+121} = \sqrt {225} = 15 ,\] so \[\text {comp}_{\textbf {b}}\textbf {a} = \frac {\textbf {a}\cdot \textbf {b}}{\left |\textbf {b}\right |} = \frac {13}{15}.\]

Solution. Let \(P_0(x_0,y_0)\) be any point on the line, so \(ax_0+by_0+c = 0\). By the question above, \(\textbf {N} = \langle a,b\rangle \) is perpendicular to the line, so the distance from \(P_1\) to the line is the length of the projection of \(\overrightarrow {P_0P_1}\) onto \(\textbf {N}\): \[d = \left |\text {comp}_{\textbf {N}}\overrightarrow {P_0P_1}\right | = \frac {\left |\textbf {N}\cdot \overrightarrow {P_0P_1}\right |}{\left |\textbf {N}\right |} = \frac {\left |a\big (x_1-x_0\big ) + b\big (y_1-y_0\big )\right |}{\sqrt {a^2+b^2}} .\] Expanding the numerator and using \(ax_0+by_0 = -c\), \[a x_1 + b y_1 - \big (ax_0+by_0\big ) = ax_1+by_1+c ,\] which gives the stated formula. Note it does not depend on which point \(P_0\) was chosen, as it must not.

Solution. The direction cosines are the components of the unit vector, and the direction angles are their inverse cosines.

(a)

\(\left |\underline {v}\right | = \sqrt {4+1+4} = 3\), so \[\cos \alpha = \frac 23,\quad \cos \beta = \frac 13,\quad \cos \gamma = \frac 23,\] giving \(\alpha \approx 48.2^{\circ }\), \(\beta \approx 70.5^{\circ }\), \(\gamma \approx 48.2^{\circ }\).

(b)

\(\left |\underline {v}\right | = \sqrt {\frac 14+1+1} = \frac 32\), so \[\cos \alpha = \frac 13,\quad \cos \beta = \frac 23,\quad \cos \gamma = \frac 23,\] giving \(\alpha \approx 70.5^{\circ }\) and \(\beta = \gamma \approx 48.2^{\circ }\).

(c)

\(\left |\underline {v}\right | = \left |c\right |\sqrt 3\), so each direction cosine is \(\dfrac {c}{\left |c\right |\sqrt 3}\). For \(c>0\) all three angles are \(\cos ^{-1}\dfrac {1}{\sqrt 3}\approx 54.7^{\circ }\); for \(c<0\) they are \(\approx 125.3^{\circ }\). The magnitude of \(c\) is irrelevant — only its sign matters, since direction cosines describe direction alone.

In every case \(\cos ^2\alpha +\cos ^2\beta +\cos ^2\gamma = 1\), which is worth using as a check.

Solution. Using \(\cos ^2\alpha + \cos ^2\beta + \cos ^2\gamma = 1\), \[\cos ^2\gamma = 1 - \Big (\frac {1}{\sqrt 2}\Big )^2 - \Big (\frac 12\Big )^2 = 1 - \frac 12 - \frac 14 = \frac 14 ,\] so \(\cos \gamma = \pm \dfrac 12\) and \[\gamma = \frac {\pi }{3} = 60^{\circ }\qquad \text {or}\qquad \gamma = \frac {2\pi }{3} = 120^{\circ }.\] Both are genuine: the two vectors are reflections of one another in the \(xy-\)plane.

Solution. Two points do not determine a plane, so the question as printed is incomplete. The two natural readings are worked below.

Reading 1: the distance from \(P\) to the line \(AB\)

\[\overrightarrow {AB} = \langle -1,3,3\rangle ,\qquad \overrightarrow {AP} = \langle 1,2,4\rangle ,\] \[\overrightarrow {AB}\times \overrightarrow {AP} = \langle 6,7,-5\rangle ,\] so \[d = \frac {\left |\overrightarrow {AB}\times \overrightarrow {AP}\right |} {\left |\overrightarrow {AB}\right |} = \frac {\sqrt {36+49+25}}{\sqrt {1+9+9}} = \frac {\sqrt {110}}{\sqrt {19}} = \frac {\sqrt {2090}}{19}\approx 2.41 .\]

Reading 2: the plane through the origin, \(A\) and \(B\)

A normal is \(\underline {n} = \overrightarrow {OA}\times \overrightarrow {OB} = \langle 3,6,-5\rangle \), and the plane passes through the origin, so it is \(3x+6y-5z = 0\). Then \[d = \frac {\left |3(0)+6(0)-5(1)\right |}{\sqrt {9+36+25}} = \frac {5}{\sqrt {70}} = \frac {\sqrt {70}}{14}\approx 0.598 .\]

Solution. Direction vectors are \[\underline {u} = \overrightarrow {AB} = \langle 1,5,2\rangle ,\qquad \underline {v} = \overrightarrow {CD} = \langle -2,3,1\rangle .\] These are not parallel, so a common perpendicular exists, with direction \[\underline {n} = \underline {u}\times \underline {v} = \langle -1,-5,13\rangle , \qquad \left |\underline {n}\right | = \sqrt {1+25+169} = \sqrt {195}.\] The distance is the length of the projection of \(\overrightarrow {AC}\) onto \(\underline {n}\). With \(\overrightarrow {AC} = \langle 1,1,4\rangle \), \[\underline {n}\cdot \overrightarrow {AC} = -1 - 5 + 52 = 46 ,\] so \[d = \frac {46}{\sqrt {195}} = \frac {46\sqrt {195}}{195}\approx 3.29 .\] Since \(d\neq 0\) the lines do not meet, so they are skew.

Solution. Take the two edges from \(P\): \[\overrightarrow {PQ} = \langle 0,1,3\rangle ,\qquad \overrightarrow {PS} = \langle 2,5,0\rangle .\] As a check that \(PQRS\) really is a parallelogram, \(\overrightarrow {PQ} + \overrightarrow {PS} = \langle 2,6,3\rangle = \overrightarrow {PR}\), so \(R\) is the opposite vertex, as required. Then \[\overrightarrow {PQ}\times \overrightarrow {PS} = \begin {vmatrix}\textbf {i}&\textbf {j}&\textbf {k}\\0&1&3\\2&5&0\end {vmatrix} = \langle -15, 6, -2\rangle ,\] and the area is \[\left |\langle -15,6,-2\rangle \right | = \sqrt {225+36+4} = \sqrt {265} \approx 16.28 .\]

Solution. \[\overrightarrow {PQ} = \langle 2,3,3\rangle ,\quad \overrightarrow {PR} = \langle -1,-1,-1\rangle ,\quad \overrightarrow {PS} = \langle 6,-2,2\rangle .\] The volume is the absolute value of the scalar triple product: \[V = \left |\begin {vmatrix}2&3&3\\-1&-1&-1\\6&-2&2\end {vmatrix}\right | .\] Expanding along the first row, \[2\big [(-1)(2)-(-1)(-2)\big ] - 3\big [(-1)(2)-(-1)(6)\big ] + 3\big [(-1)(-2)-(-1)(6)\big ]\] \[= 2(-4) - 3(4) + 3(8) = -8 - 12 + 24 = 4 ,\] so \(V = 4\).

Solution. \[\overrightarrow {PQ} = \langle 1,4,5\rangle ,\quad \overrightarrow {PR} = \langle 2,-1,1\rangle ,\quad \overrightarrow {PS} = \langle 5,2,7\rangle .\] \[\overrightarrow {PQ}\cdot \big (\overrightarrow {PR}\times \overrightarrow {PS}\big ) = \begin {vmatrix}1&4&5\\2&-1&1\\5&2&7\end {vmatrix} = 1(-7-2) - 4(14-5) + 5(4+5) = -9 - 36 + 45 = 0 .\] The triple product is zero, so the parallelepiped they span has zero volume: the four points are coplanar.

Solution.

(a)

The direction is \(\langle 2,-1,3\rangle \) and the point is the origin, so \[\overline {r} = t\langle 2,-1,3\rangle ,\qquad x = 2t,\ y = -t,\ z = 3t,\qquad \frac {x}{2} = \frac {y}{-1} = \frac {z}{3}.\]

(b)

The direction is \(\langle 4-(-1),\ -3-0,\ 3-5\rangle = \langle 5,-3,-2\rangle \), so \[\overline {r} = \langle -1,0,5\rangle + t\langle 5,-3,-2\rangle ,\qquad \frac {x+1}{5} = \frac {y}{-3} = \frac {z-5}{-2}.\]

(c)

The direction is \(\langle 2,3,-7\rangle \), so \[x = 2t,\qquad y = 2+3t,\qquad z = -1-7t .\] It meets the \(xy-\)plane where \(z=0\), that is \(t = -\dfrac 17\), at \(\Big (-\dfrac 27,\ \dfrac {11}{7},\ 0\Big )\); the \(xz-\)plane where \(y=0\), that is \(t = -\dfrac 23\), at \(\Big (-\dfrac 43,\ 0,\ \dfrac {11}{3}\Big )\); and the \(yz-\)plane where \(x=0\), that is \(t=0\), at \((0,2,-1)\).

(d)

\[x = 5+t,\qquad y = 1+4t,\qquad z = 3-2t ,\] so \(t=1\) gives \((6,5,1)\) and \(t=-1\) gives \((4,-3,5)\). The symmetric form is \[\frac {x-5}{1} = \frac {y-1}{4} = \frac {z-3}{-2}.\]

Solution.

(a)

The directions are \(\langle 2,4,-3\rangle \) and \(\langle 1,3,2\rangle \). Neither is a multiple of the other, so the lines are not parallel. Setting the parametric forms equal, \[4+2t = 2+s,\qquad -5+4t = -1+3s,\qquad 1-3t = 2s .\] The first two give \(s = 2t+2\) and \(4t-3s = 4\), hence \(4t - 6t - 6 = 4\) and \(t = -5\), \(s = -8\). Substituting into the third, \(1-3(-5) = 16\) but \(2(-8) = -16\), so it fails. The system is inconsistent: the lines neither meet nor are parallel, so they are skew.

(b)

The directions are \(\langle -6,9,-3\rangle \) and \(\langle 2,-3,1\rangle \), and \[\langle -6,9,-3\rangle = -3\langle 2,-3,1\rangle ,\] so the lines are parallel. They are not the same line: \(t=0\) gives \((0,1,0)\) on \(L_1\), and putting \(0 = 1+2s\) gives \(s = -\dfrac 12\), for which \(y = 4+\dfrac 32 = \dfrac {11}{2}\neq 1\). So they are distinct parallel lines and do not intersect.

Solution.

They are skew

The directions \(\underline {u} = \langle 1,3,-1\rangle \) and \(\underline {v} = \langle 2,1,4\rangle \) are not proportional, so the lines are not parallel. Equating the parametric forms, \[1+t = 2s,\qquad -2+3t = 3+s,\qquad 4-t = -3+4s .\] From the first, \(t = 2s-1\); substituting into the second gives \(6s-5 = 3+s\), so \(s = \dfrac 85\) and \(t = \dfrac {11}{5}\). The third then requires \(4 - \dfrac {11}{5} = \dfrac 95\) on the left and \(-3 + \dfrac {32}{5} = \dfrac {17}{5}\) on the right, which differ. So the lines do not intersect, and being neither parallel nor intersecting they are skew.

The distance

\[\underline {n} = \underline {u}\times \underline {v} = \langle 13,-6,-5\rangle , \qquad \left |\underline {n}\right | = \sqrt {169+36+25} = \sqrt {230}.\] Taking \(A(1,-2,4)\) on \(L_1\) and \(C(0,3,-3)\) on \(L_2\), \(\overrightarrow {AC} = \langle -1,5,-7\rangle \) and \[\underline {n}\cdot \overrightarrow {AC} = -13 - 30 + 35 = -8 ,\] so \[d = \frac {\left |-8\right |}{\sqrt {230}} = \frac {8}{\sqrt {230}} = \frac {4\sqrt {230}}{115}\approx 0.528 .\]

Solution.

(a)

With \(A(0,1,1)\), \(B(1,0,1)\), \(C(1,1,0)\), \[\overrightarrow {AB} = \langle 1,-1,0\rangle ,\qquad \overrightarrow {AC} = \langle 1,0,-1\rangle ,\] \[\underline {n} = \overrightarrow {AB}\times \overrightarrow {AC} = \langle 1,1,1\rangle .\] The plane through \(A\) with this normal is \(x+y+z = 2\).

(b)

A point on the line of intersection: setting \(z=0\) gives \(x=1\), \(y=3\), so \((1,3,0)\) lies on it. Its direction is \[\langle 1,0,-1\rangle \times \langle 0,1,2\rangle = \langle 1,-2,1\rangle .\] The required plane contains this direction and is perpendicular to \(x+y-2z=1\), so it also contains that plane’s normal \(\langle 1,1,-2\rangle \). Hence its own normal is \[\underline {n} = \langle 1,-2,1\rangle \times \langle 1,1,-2\rangle = \langle 3,3,3\rangle ,\] which may be divided by \(3\) to give \(\langle 1,1,1\rangle \). Through \((1,3,0)\) the plane is \[x+y+z = 4 .\]

Solution.

The formula

Take any point \(P_0(x_0,y_0,z_0)\) on the first plane, so \(ax_0+by_0+cz_0 = -d_1\). The distance from a point to the second plane is \[\frac {\left |ax_0+by_0+cz_0+d_2\right |}{\sqrt {a^2+b^2+c^2}} = \frac {\left |-d_1+d_2\right |}{\sqrt {a^2+b^2+c^2}} = \frac {\left |d_1-d_2\right |}{\sqrt {a^2+b^2+c^2}} ,\] independent of which point was chosen, which is what makes the planes parallel.

The example

The normals must be made identical before the formula applies. Multiplying the second equation by \(2\) gives \(10x+2y-2z = 2\), so in the form above \(d_1 = -5\) and \(d_2 = -2\) with \(\langle a,b,c\rangle = \langle 10,2,-2\rangle \). Hence \[d = \frac {\left |-5+2\right |}{\sqrt {100+4+4}} = \frac {3}{\sqrt {108}} = \frac {3}{6\sqrt 3} = \frac {\sqrt 3}{6}\approx 0.289 .\] Failing to rescale first is the usual error here, and would give an answer twice too large.

Solution. The angle between planes is the angle between their normals, \(\underline {n}_1 = \langle 1,1,1\rangle \) and \(\underline {n}_2 = \langle 1,-1,1\rangle \): \[\cos \theta = \frac {1-1+1}{\sqrt 3\cdot \sqrt 3} = \frac 13 ,\] so \(\theta = \cos ^{-1}\dfrac 13\approx 70.5^{\circ }\).

Solution.

The line

The line lies in both planes, so its direction is perpendicular to both normals: \[\underline {d} = \langle 3,-2,1\rangle \times \langle 2,1,-3\rangle = \langle 5,11,7\rangle .\] For a point on it, set \(z=0\) and solve \(3x-2y=1\), \(2x+y=3\), giving \(x = 1\), \(y = 1\). Hence \[x = 1+5t,\qquad y = 1+11t,\qquad z = 7t .\]

The angle

\[\cos \theta = \frac {\left |6-2-3\right |}{\sqrt {14}\sqrt {14}} = \frac {1}{14},\] so \(\theta = \cos ^{-1}\dfrac {1}{14}\approx 85.9^{\circ }\).

Solution. For a segment the parameter must be restricted. Writing \[\overline {r}(t) = (1-t)\overrightarrow {OP} + t\,\overrightarrow {OQ}, \qquad 0\leq t\leq 1 ,\] gives \(\overline {r}(0) = P\) and \(\overline {r}(1) = Q\). In components, \[\overline {r}(t) = \langle 1+t,\ 3-4t,\ -2+5t\rangle ,\qquad 0\leq t\leq 1 .\]

Solution. The cylinder is parametrised by \(x = \cos t\), \(y = \sin t\), which satisfies \(x^2+y^2=1\) automatically. The plane then fixes \(z\): \[z = 2 - y = 2 - \sin t .\] Hence \[\overline {r}(t) = \langle \cos t,\ \sin t,\ 2-\sin t\rangle , \qquad 0\leq t\leq 2\pi ,\] an ellipse, being the intersection of a cylinder with a plane oblique to its axis.

Solution.

(a)

\(\overline {r}\,' = \langle 1, -3\sin t, 3\cos t\rangle \), so \(\left |\overline {r}\,'\right | = \sqrt {1+9} = \sqrt {10}\), a constant. Hence \[L = \int _{-5}^{5}\sqrt {10}\,dt = 10\sqrt {10}\approx 31.6 .\]

(b)

\(\overline {r}\,' = \Big \langle -\sin t,\ \cos t,\ -\tan t\Big \rangle \), so \[\left |\overline {r}\,'\right | = \sqrt {\sin ^2t+\cos ^2t+\tan ^2t} = \sqrt {1+\tan ^2t} = \sec t\] on \(\Big [0,\dfrac {\pi }{4}\Big ]\), where \(\sec t>0\). Hence \[L = \int _0^{\pi /4}\sec t\,dt = \Big [\ln \left |\sec t+\tan t\right |\Big ]_0^{\pi /4} = \ln \big (\sqrt 2+1\big )\approx 0.881 .\]

Solution. Parametrise by \(x = t\). Then \(y = \dfrac {t^2}{2}\) and \(z = \dfrac {xy}{3} = \dfrac {t^3}{6}\), so \[\overline {r}(t) = \Big \langle t,\ \frac {t^2}{2},\ \frac {t^3}{6}\Big \rangle .\] The endpoint \((6,18,36)\) corresponds to \(t=6\). Now \[\overline {r}\,' = \Big \langle 1,\ t,\ \frac {t^2}{2}\Big \rangle ,\qquad \left |\overline {r}\,'\right | = \sqrt {1+t^2+\frac {t^4}{4}} = \sqrt {\Big (1+\frac {t^2}{2}\Big )^2} = 1+\frac {t^2}{2},\] the square root again being exact. Hence \[L = \int _0^6\Big (1+\frac {t^2}{2}\Big )dt = \Big [t + \frac {t^3}{6}\Big ]_0^6 = 6 + 36 = 42 .\]

Solution. Differentiating, and noting the cancellations, \[\overline {r}\,' = \langle 2t,\ t\sin t,\ t\cos t\rangle = t\langle 2,\ \sin t,\ \cos t\rangle ,\] so for \(t>0\), \(\left |\overline {r}\,'\right | = t\sqrt {4+1} = t\sqrt 5\) and \[\textbf {T}(t) = \frac {1}{\sqrt 5}\langle 2,\ \sin t,\ \cos t\rangle .\] Then \[\textbf {T}'(t) = \frac {1}{\sqrt 5}\langle 0,\ \cos t,\ -\sin t\rangle , \qquad \left |\textbf {T}'\right | = \frac {1}{\sqrt 5},\] so \[\textbf {N}(t) = \frac {\textbf {T}'}{\left |\textbf {T}'\right |} = \langle 0,\ \cos t,\ -\sin t\rangle ,\] and \[k = \frac {\left |\textbf {T}'\right |}{\left |\overline {r}\,'\right |} = \frac {1/\sqrt 5}{t\sqrt 5} = \frac {1}{5t}.\]

Solution.

(a)

The point corresponds to \(t=1\). \[\overline {r}\,' = \langle 1,2t,3t^2\rangle ,\qquad \overline {r}\,'' = \langle 0,2,6t\rangle ,\] and at \(t=1\) these are \(\langle 1,2,3\rangle \) and \(\langle 0,2,6\rangle \), with \[\overline {r}\,'\times \overline {r}\,'' = \langle 6,-6,2\rangle .\] Hence \[k = \frac {\left |\langle 6,-6,2\rangle \right |}{\left |\langle 1,2,3\rangle \right |^3} = \frac {\sqrt {76}}{14^{3/2}} = \frac {2\sqrt {19}}{14\sqrt {14}} = \frac {\sqrt {266}}{98}\approx 0.166 .\]

(b)(i)

\(y' = \sec ^2x\) and \(y'' = 2\sec ^2x\tan x\), so \[k = \frac {2\sec ^2x\left |\tan x\right |}{\big (1+\sec ^4x\big )^{3/2}} .\]

(b)(ii)

\(y' = (1+x)e^x\) and \(y'' = (2+x)e^x\), so \[k = \frac {\left |2+x\right |e^x}{\Big [1+(1+x)^2e^{2x}\Big ]^{3/2}} .\]

Solution.

The formula

Regard the plane curve as the space curve \(\overline {r}(t) = \langle x(t), y(t), 0\rangle \). Then \[\overline {r}\,' = \langle x',y',0\rangle ,\qquad \overline {r}\,'' = \langle x'',y'',0\rangle ,\] \[\overline {r}\,'\times \overline {r}\,'' = \langle 0,0,\ x'y''-y'x''\rangle .\] Substituting into \(k = \dfrac {\left |\overline {r}\,'\times \overline {r}\,''\right |} {\left |\overline {r}\,'\right |^3}\) gives the stated formula at once.

The example

\[x' = e^t\big (\cos t-\sin t\big ),\qquad y' = e^t\big (\sin t+\cos t\big ),\] \[x'' = -2e^t\sin t,\qquad y'' = 2e^t\cos t .\] Then \[x'y''-y'x'' = 2e^{2t}\Big [\cos t\big (\cos t-\sin t\big ) + \sin t\big (\sin t+\cos t\big )\Big ] = 2e^{2t},\] while \((x')^2+(y')^2 = 2e^{2t}\). Hence \[k = \frac {2e^{2t}}{\big (2e^{2t}\big )^{3/2}} = \frac {1}{\sqrt 2\,e^{t}} = \frac {\sqrt 2}{2}e^{-t} .\] The curvature decays exponentially, which is what makes this an equiangular spiral.

Solution.

(a)

The point is \(t=1\). \(\overline {r}\,' = \langle 2t, 2t^2, 1\rangle \), which at \(t=1\) is \(\langle 2,2,1\rangle \) with length \(3\), so \[\textbf {T} = \frac 13\langle 2,2,1\rangle .\] Next \(\overline {r}\,'' = \langle 2,4t,0\rangle = \langle 2,4,0\rangle \) at \(t=1\), and \[\overline {r}\,'\times \overline {r}\,'' = \langle -4,2,4\rangle ,\] of length \(6\), so \[\textbf {B} = \frac 16\langle -4,2,4\rangle = \frac 13\langle -2,1,2\rangle .\] Finally \[\textbf {N} = \textbf {B}\times \textbf {T} = \frac 19\langle -2,1,2\rangle \times \langle 2,2,1\rangle = \frac 19\langle -3,6,-6\rangle = \frac 13\langle -1,2,-2\rangle .\]

(b)

The point is \(t=0\). From the length question above, \(\overline {r}\,' = \langle -\sin t, \cos t, -\tan t\rangle \), which at \(t=0\) is \(\langle 0,1,0\rangle \), already a unit vector, so \(\textbf {T} = \langle 0,1,0\rangle \). Next \(\overline {r}\,'' = \langle -\cos t, -\sin t, -\sec ^2t\rangle \), which at \(t=0\) is \(\langle -1,0,-1\rangle \). Then \[\overline {r}\,'\times \overline {r}\,'' = \langle 0,1,0\rangle \times \langle -1,0,-1\rangle = \langle -1,0,1\rangle ,\] of length \(\sqrt 2\), so \[\textbf {B} = \frac {1}{\sqrt 2}\langle -1,0,1\rangle ,\qquad \textbf {N} = \textbf {B}\times \textbf {T} = \frac {1}{\sqrt 2}\langle -1,0,-1\rangle .\]

Solution. The point corresponds to \(t = \pi \), since then \(\sin 3\pi = 0\), \(\cos 3\pi = -1\). \[\overline {r}\,' = \langle 6\cos 3t,\ 1,\ -6\sin 3t\rangle \ \longrightarrow \ \langle -6,1,0\rangle ,\] \[\overline {r}\,'' = \langle -18\sin 3t,\ 0,\ -18\cos 3t\rangle \ \longrightarrow \ \langle 0,0,18\rangle .\]

Normal plane

This is perpendicular to \(\textbf {T}\), so it has normal \(\langle -6,1,0\rangle \): \[-6(x-0) + 1(y-\pi ) + 0(z+2) = 0 \implies -6x + y = \pi .\]

Osculating plane

This contains \(\textbf {T}\) and \(\textbf {N}\), so its normal is \(\textbf {B}\), parallel to \[\overline {r}\,'\times \overline {r}\,'' = \langle -6,1,0\rangle \times \langle 0,0,18\rangle = \langle 18,108,0\rangle ,\] which may be divided by \(18\) to give \(\langle 1,6,0\rangle \). Hence \[1(x-0) + 6(y-\pi ) + 0(z+2) = 0 \implies x + 6y = 6\pi .\]

Solution.

(a)

\[\overline {v} = \Big \langle -1,\ \frac {2}{\sqrt t}\Big \rangle ,\qquad \overline {a} = \Big \langle 0,\ -\frac {1}{t^{3/2}}\Big \rangle .\] At \(t=1\): \(\overline {v} = \langle -1,2\rangle \), \(\overline {a} = \langle 0,-1\rangle \), and the speed is \(\left |\overline {v}\right | = \sqrt 5\).

The path: \(x = 2-t\) gives \(t = 2-x\), so \(y = 4\sqrt {2-x}\), that is \(y^2 = 16(2-x)\) with \(y\geq 0\) — the upper half of a leftward parabola.

(b)

\[\overline {v} = \langle 1,\ -2\sin t,\ \cos t\rangle ,\qquad \overline {a} = \langle 0,\ -2\cos t,\ -\sin t\rangle .\] At \(t=0\): \(\overline {v} = \langle 1,0,1\rangle \), \(\overline {a} = \langle 0,-2,0\rangle \), and the speed is \(\sqrt 2\).

The path lies on the elliptical cylinder \(\dfrac {y^2}{4} + z^2 = 1\), advancing steadily in \(x\): an elliptical helix.

Solution. Integrating the acceleration, \[\overline {v}(t) = \langle 2t,\ 3t^2,\ 4t^3\rangle + \underline {c}_1 ,\] and \(\overline {v}(0) = \langle 1,0,0\rangle \) forces \(\underline {c}_1 = \langle 1,0,0\rangle \), so \[\overline {v}(t) = \langle 2t+1,\ 3t^2,\ 4t^3\rangle .\] Integrating again, \[\overline {r}(t) = \langle t^2+t,\ t^3,\ t^4\rangle + \underline {c}_2 ,\] and \(\overline {r}(0) = \langle 0,1,-1\rangle \) forces \(\underline {c}_2 = \langle 0,1,-1\rangle \), so \[\overline {r}(t) = \langle t^2+t,\ t^3+1,\ t^4-1\rangle .\] Differentiating back reproduces \(\overline {v}\), which is the check worth doing.

Solution. Take \(g = 9.8\ \text {m/s}^2\), with \(v_{0x} = 200\cos 60^{\circ } = 100\) and \(v_{0y} = 200\sin 60^{\circ } = 100\sqrt 3\).

Time of flight

The vertical motion is \(y = v_{0y}t - \dfrac 12gt^2\), which returns to zero at \[t = \frac {2v_{0y}}{g} = \frac {200\sqrt 3}{9.8}\approx 35.35\ \text {s}.\]

Range

\[R = v_{0x}t = \frac {v_0^2\sin 120^{\circ }}{g} = \frac {40000\cdot \frac {\sqrt 3}{2}}{9.8}\approx 3535\ \text {m}.\]

Maximum height

\[H = \frac {v_{0y}^2}{2g} = \frac {30000}{19.6}\approx 1531\ \text {m}.\]

Impact speed

By symmetry the vertical component returns with the same magnitude and opposite sign, and the horizontal component is unchanged, so the impact speed equals the launch speed, \(200\ \text {m/s}\).

Solution. \[\overline {v} = \langle 2t, 2t, 3t^2\rangle ,\qquad \overline {a} = \langle 2,2,6t\rangle ,\] \[\left |\overline {v}\right | = \sqrt {8t^2+9t^4} = \left |t\right |\sqrt {8+9t^2}.\] Taking \(t>0\), the tangential component is the rate of change of speed: \[a_T = \frac {d}{dt}\left |\overline {v}\right | = \frac {\overline {v}\cdot \overline {a}}{\left |\overline {v}\right |} = \frac {8t+18t^3}{t\sqrt {8+9t^2}} = \frac {8+18t^2}{\sqrt {8+9t^2}} .\] For the normal component, \[\overline {v}\times \overline {a} = \langle 12t^2-6t^2,\ 6t^2-12t^2,\ 0\rangle = \langle 6t^2,\ -6t^2,\ 0\rangle ,\] of length \(6\sqrt 2\,t^2\), so \[a_N = \frac {\left |\overline {v}\times \overline {a}\right |}{\left |\overline {v}\right |} = \frac {6\sqrt 2\,t^2}{t\sqrt {8+9t^2}} = \frac {6\sqrt 2\,t}{\sqrt {8+9t^2}} .\] As a check, \(a_T^2 + a_N^2\) should equal \(\left |\overline {a}\right |^2 = 8+36t^2\), and it does.