Analytic Geometry in Space: Difficult Problems with Solutions

Solution:
Introduce an orthonormal coordinate system with origin [tex]A[/tex] and axes along [tex]AB[/tex], [tex]AD[/tex] and [tex]AA_1[/tex]. Then
[tex]M(2\sqrt{2}, \sqrt{2}, 0),\quad C_1(2\sqrt{2}, 2\sqrt{2}, 2),\quad A_1(0, 0, 2),\quad D(0, 2\sqrt{2}, 0)[/tex]
The direction vectors of the two lines are
[tex]\overrightarrow{MC_1}=(0, \sqrt{2}, 2),\qquad \overrightarrow{A_1D}=(0, 2\sqrt{2}, -2)[/tex]
and their dot product is
[tex]\overrightarrow{MC_1}\centerdot\overrightarrow{A_1D}=0\times 0+\sqrt{2}\times 2\sqrt{2}+2\times (-2)=4-4=0[/tex]
So the direction vectors are perpendicular. The angle between two lines (intersecting or skew) is determined by the angle between their direction vectors, hence it equals [tex]90^\circ[/tex].

Solution:
The line [tex]g[/tex] passes through [tex]P_1(1, -2, 5)[/tex] and has direction vector [tex]\vec{v}_1=(\lambda, -3, 4)[/tex], and the line [tex]p[/tex] passes through [tex]P_2(3\lambda+1, 2, 1)[/tex] and has direction vector [tex]\vec{v}_2=(3, 2, -2)[/tex]. Two lines lie in one plane exactly when the vectors [tex]\overrightarrow{P_1P_2}=(3\lambda, 4, -4)[/tex], [tex]\vec{v}_1[/tex] and [tex]\vec{v}_2[/tex] are coplanar, i.e. when their mixed product is zero:
[tex]\begin{vmatrix}3\lambda&4&-4\\ \lambda&-3&4\\ 3&2&-2\end{vmatrix}=3\lambda(6-8)-4(-2\lambda-12)-4(2\lambda+9)=12-6\lambda=0\ \Rightarrow\ \lambda=2[/tex]
For [tex]\lambda=2[/tex] the direction vectors [tex](2, -3, 4)[/tex] and [tex](3, 2, -2)[/tex] are not collinear, so the lines intersect (at the point [tex](1, -2, 5)[/tex], obtained for [tex]t=-2[/tex]) and indeed determine a plane.

Solution:
Let [tex]D(0, y, 0)[/tex]. The volume of a tetrahedron is one sixth of the absolute value of the mixed product of the three edge vectors issuing from one vertex:
[tex]V=\frac{1}{6}\left|\left(\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}\right)\right|[/tex]
Here [tex]\overrightarrow{AB}=(-1, 1, -2)[/tex], [tex]\overrightarrow{AC}=(0, 2, -4)[/tex], [tex]\overrightarrow{AD}=(2, y+1, -1)[/tex], and expanding along the first row we get
[tex]\begin{vmatrix}-1&1&-2\\ 0&2&-4\\ 2&y+1&-1\end{vmatrix}=-1\times (-2+4y+4)-1\times (0+8)-2\times (0-4)=-4y-2[/tex]
The condition [tex]V=5[/tex] gives
[tex]\frac{1}{6}\left|4y+2\right|=5\iff 4y+2=\pm 30\iff y=7\text{ or }y=-8[/tex]
Hence there are two possible vertices: [tex]D(0, 7, 0)[/tex] and [tex]D(0, -8, 0)[/tex].

Solution:
Since [tex]\overrightarrow{OA}=-\overrightarrow{AO}=-(\vec{a}+\vec{b})[/tex], the volume is
[tex]V=\frac{1}{6}\left|\left(\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC}\right)\right|[/tex]
The cross product is bilinear and anticommutative ([tex]\vec{a}×\vec{a}=\vec{b}×\vec{b}=\vec{0}[/tex], [tex]\vec{b}×\vec{a}=-\vec{a}×\vec{b}[/tex]), so
[tex](\vec{a}+\vec{b})×(\vec{a}-\vec{b})=\vec{a}×\vec{a}-\vec{a}×\vec{b}+\vec{b}×\vec{a}-\vec{b}×\vec{b}=-2\,\vec{a}×\vec{b}\ \Rightarrow\ \overrightarrow{OA}×\overrightarrow{OB}=2\,\vec{a}×\vec{b}[/tex]
Hence
[tex]\left(\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC}\right)=\left(2\,\vec{a}×\vec{b}\right)\centerdot\left(\vec{a}×\vec{b}\right)=2\left|\vec{a}×\vec{b}\right|^2[/tex]
where
[tex]\left|\vec{a}×\vec{b}\right|=|\vec{a}|\,|\vec{b}|\sin \frac{\pi}{3}=3\times 2\times \frac{\sqrt{3}}{2}=3\sqrt{3}[/tex]
Thus the mixed product equals [tex]2\times 27=54[/tex] and
[tex]V=\frac{54}{6}=9[/tex]

Solution:
When a point [tex](x_0, 0, z_0)[/tex] of the hyperbola rotates about [tex]Ox[/tex], it describes a circle in the plane [tex]x=x_0[/tex] with centre [tex](x_0, 0, 0)[/tex] and radius [tex]|z_0|[/tex], i.e. the set of points with [tex]x=x_0[/tex] and [tex]y^2+z^2=z_0^2[/tex]. Hence a point [tex](x, y, z)[/tex] lies on the surface exactly when the point [tex]\left(x, 0, \sqrt{y^2+z^2}\right)[/tex] lies on the hyperbola: we keep the coordinate along the axis of rotation and replace [tex]z^2[/tex] by [tex]y^2+z^2[/tex]:
[tex]\frac{x^2}{9}-\frac{y^2+z^2}{4}=1[/tex]
The axis of rotation [tex]Ox[/tex] is the real (transverse) axis of the hyperbola, so the surface is a two-sheeted hyperboloid of revolution.

Solution:
Complete the squares:
[tex]6(x^2-2x)=6(x-1)^2-6,\quad -2(y^2-2y)=-2(y-1)^2+2,\quad 4(z^2+2z)=4(z+1)^2-4[/tex]
The equation becomes
[tex]6(x-1)^2-2(y-1)^2+4(z+1)^2-6+2-4+8=0\iff 6(x-1)^2-2(y-1)^2+4(z+1)^2=0[/tex]
The translation [tex]X=x-1,\ Y=y-1,\ Z=z+1[/tex] moves the origin to [tex]O'(1, 1, -1)[/tex]; after dividing by [tex]12[/tex] we obtain
[tex]\frac{X^2}{2}+\frac{Z^2}{3}-\frac{Y^2}{6}=0[/tex]
This is an elliptic cone with vertex [tex](1, 1, -1)[/tex] and axis parallel to [tex]Oy[/tex]: its sections by the planes [tex]Y=h\ne 0[/tex] are the ellipses [tex]\frac{X^2}{2}+\frac{Z^2}{3}=\frac{h^2}{6}[/tex], and the section by [tex]Y=0[/tex] is only the vertex.

Solution:
Since [tex](\vec{u},\vec{v},\vec{x})=(\vec{u}×\vec{v})\centerdot\vec{x}[/tex], we first compute the cross products:
[tex]\vec{a}×\vec{b}=(2, 1, -1),\quad \vec{b}×\vec{c}=(3, 3, -1),\quad \vec{c}×\vec{a}=(-2, -2, 1)[/tex]
With [tex]\vec{x}=(x, y, z)[/tex] the conditions become the linear system
[tex]\begin{cases}2x+y-z=1\\ 3x+3y-z=2\\ -2x-2y+z=0\end{cases}[/tex]
Its determinant equals [tex]1\ne 0[/tex] (in general it equals [tex](\vec{a},\vec{b},\vec{c})^2[/tex], and here [tex](\vec{a},\vec{b},\vec{c})=\vec{a}\centerdot(\vec{b}×\vec{c})=3-2=1[/tex]), so the solution is unique. Adding the last two equations gives [tex]x+y=2[/tex]; then the third equation gives [tex]z=2(x+y)=4[/tex], and the first gives [tex]2x+y=5[/tex]. Hence
[tex]x=3,\quad y=-1,\quad z=4,\qquad \vec{x}=(3, -1, 4)[/tex]
Check: [tex]\vec{x}=2\vec{a}+\vec{c}[/tex], so [tex](\vec{a},\vec{b},\vec{x})=(\vec{a},\vec{b},\vec{c})=1[/tex], [tex](\vec{b},\vec{c},\vec{x})=2(\vec{b},\vec{c},\vec{a})=2[/tex] and [tex](\vec{c},\vec{a},\vec{x})=0[/tex].

Solution:
A plane contains the axis [tex]Oz[/tex] if and only if its equation has the form [tex]Ax+By=0[/tex], [tex](A, B)\ne(0, 0)[/tex]: it passes through the origin and its normal vector [tex]\vec{n}=(A, B, 0)[/tex] is perpendicular to the axis. The cosine of the angle between two planes is [tex]\frac{|\vec{n}_1⋅\vec{n}_2|}{|\vec{n}_1|\,|\vec{n}_2|}[/tex], and the normal vector of [tex]\beta[/tex] is [tex]\vec{n}_\beta=(2, 1, -\sqrt{5})[/tex], [tex]|\vec{n}_\beta|=\sqrt{10}[/tex]. Hence
[tex]\cos\varphi=\frac{|2A+B|}{\sqrt{A^2+B^2}\,\sqrt{10}}=\frac{1}{2}\ \Rightarrow\ 4(2A+B)^2=10\left(A^2+B^2\right)[/tex]
Expanding and dividing by [tex]2[/tex]:
[tex]3A^2+8AB-3B^2=0\iff(3A-B)(A+3B)=0[/tex]
Hence [tex]B=3A[/tex], e.g. [tex](A, B)=(1, 3)[/tex], or [tex]A=-3B[/tex], e.g. [tex](A, B)=(3, -1)[/tex]. Squaring added no solutions, because both sides were non-negative; directly:
[tex]\frac{|2\times 1+3|}{\sqrt{10}\,\sqrt{10}}=\frac{|2\times 3-1|}{\sqrt{10}\,\sqrt{10}}=\frac{1}{2}[/tex]
So the required planes are [tex]x+3y=0[/tex] and [tex]3x-y=0[/tex].

Solution:
First we find the common point of [tex]b[/tex] and [tex]\alpha[/tex]. Substituting the parametric equations of [tex]b[/tex] into the equation of [tex]\alpha[/tex]:
[tex](2-3s)-2s+5(6-5s)-2=30-30s=0\ \Rightarrow\ s=1,\quad P(-1, 1, 1)[/tex]
The points of [tex]\alpha[/tex] are fixed by the symmetry, so [tex]P\in b'[/tex]. Take one more point of [tex]b[/tex], [tex]Q(2, 0, 6)[/tex] (for [tex]s=0[/tex]), and find its symmetric point [tex]Q'[/tex]. The normal vector of [tex]\alpha[/tex] is [tex]\vec{n}=(1, -2, 5)[/tex], and the perpendicular from [tex]Q[/tex] to [tex]\alpha[/tex] consists of the points [tex]Q+\lambda\vec{n}=(2+\lambda, -2\lambda, 6+5\lambda)[/tex]. Its foot [tex]H[/tex] is the midpoint of [tex]QQ'[/tex]:
[tex](2+\lambda)+4\lambda+5(6+5\lambda)-2=30+30\lambda=0\ \Rightarrow\ \lambda=-1,\quad H(1, 2, 1),\quad Q'=2H-Q=(0, 4, -4)[/tex]
The line [tex]b'[/tex] passes through [tex]P[/tex] and [tex]Q'[/tex]:
[tex]\overrightarrow{PQ'}=(1, 3, -5)\ \Rightarrow\ b':\ \frac{x+1}{1}=\frac{y-1}{3}=\frac{z-1}{-5}[/tex]
Check: reflecting the direction vector [tex]\vec{d}=(-3, 1, -5)[/tex] of [tex]b[/tex] (here [tex]\vec{d}⋅\vec{n}=-30[/tex], [tex]|\vec{n}|^2=30[/tex]) gives a vector parallel to [tex]\overrightarrow{PQ'}[/tex]:
[tex]\vec{d}-2\,\frac{\vec{d}⋅\vec{n}}{|\vec{n}|^2}\,\vec{n}=\vec{d}+2\vec{n}=(-1, -3, 5)\parallel(1, 3, -5)[/tex]

Solution:
The normal vector of [tex]\beta[/tex] is [tex]\vec{n}_\beta=(3, 4, -1)[/tex]. The vector [tex]\overrightarrow{AB}[/tex] is not parallel to [tex]\vec{n}_\beta[/tex], so the plane [tex]\gamma[/tex] is unique; its normal vector is perpendicular to both [tex]\overrightarrow{AB}[/tex] and [tex]\vec{n}_\beta[/tex]:
[tex]\overrightarrow{AB}=(-2, -8, -2)\parallel(1, 4, 1),\qquad (1, 4, 1)×(3, 4, -1)=(-8, 4, -8)\parallel(2, -1, 2)[/tex]
[tex]\gamma:\ 2x-y+2(z+1)=0\iff 2x-y+2z+2=0[/tex]
This plane passes through [tex]A[/tex], and [tex]B\in\gamma[/tex] as well, since [tex]-4+8-6+2=0[/tex]. The perpendicular from [tex]B[/tex] to [tex]\beta[/tex] has direction [tex]\vec{n}_\beta[/tex], which is parallel to [tex]\gamma[/tex] because [tex]\gamma\perp\beta[/tex]; since [tex]B\in\gamma[/tex], this perpendicular lies in [tex]\gamma[/tex]. So its foot [tex]F[/tex] lies in both [tex]\beta[/tex] and [tex]\gamma[/tex], i.e. [tex]F\in g[/tex]. Every point of [tex]g[/tex] lies in [tex]\beta[/tex] and therefore is at distance at least [tex]BF[/tex] from [tex]B[/tex]. Hence the distance from [tex]B[/tex] to [tex]g[/tex] equals the distance from [tex]B[/tex] to [tex]\beta[/tex]:
[tex]BF=\frac{|3\times (-2)+4\times (-8)-(-3)+1|}{\sqrt{9+16+1}}=\frac{34}{\sqrt{26}}=\frac{17\sqrt{26}}{13}[/tex]

Solution:
We have [tex]\overrightarrow{AB}=(4, 0, -2)[/tex] and [tex]\overrightarrow{CD}=(3, 3, -3)[/tex]. Arbitrary points of the two lines and the vector between them:
[tex]M=A+p\,\overrightarrow{AB}=(4p, 0, -2-2p),\qquad N=C+q\,\overrightarrow{CD}=(2+3q, 3q, -3q)[/tex]
[tex]\overrightarrow{MN}=(2+3q-4p, \ 3q, \ 2+2p-3q)[/tex]
The conditions [tex]MN\perp AB[/tex] and [tex]MN\perp CD[/tex], written with the parallel vectors [tex](2, 0, -1)[/tex] and [tex](1, 1, -1)[/tex]:
[tex]\begin{cases}2(2+3q-4p)-(2+2p-3q)=0\\ (2+3q-4p)+3q-(2+2p-3q)=0\end{cases}\iff\begin{cases}9q-10p+2=0\\ 9q-6p=0\end{cases}[/tex]
Subtracting the second equation from the first gives [tex]-4p+2=0[/tex]; the system has a unique solution:
[tex]p=\frac{1}{2},\quad q=\frac{1}{3}\ \Rightarrow\ M(2, 0, -3),\quad N(3, 1, -1)[/tex]
Check: [tex]\overrightarrow{MN}[/tex] is parallel to [tex]\overrightarrow{AB}×\overrightarrow{CD}[/tex], and since [tex]\overrightarrow{MN}\ne\vec{0}[/tex], the lines are skew and [tex]MN[/tex] is their common perpendicular:
[tex]\overrightarrow{MN}=(1, 1, 2),\qquad \overrightarrow{AB}×\overrightarrow{CD}=(6, 6, 12)=6\,\overrightarrow{MN}[/tex]

Solution:
Choose an orthonormal coordinate system with [tex]A(0, 0, 0)[/tex] and [tex]B(d, 0, 0)[/tex]. For [tex]M(x, y, z)[/tex] both sides of [tex]AM=k\,BM[/tex] are non-negative, so the condition is equivalent to [tex]AM^2=k^2BM^2[/tex]:
[tex]x^2+y^2+z^2=k^2\left((x-d)^2+y^2+z^2\right)\iff\left(1-k^2\right)\left(x^2+y^2+z^2\right)+2k^2dx-k^2d^2=0[/tex]
Since [tex]k\ne1[/tex], we can divide by [tex]1-k^2[/tex] and complete the square in [tex]x[/tex]:
[tex]\left(x+\frac{k^2d}{1-k^2}\right)^2+y^2+z^2=\frac{k^4d^2}{\left(1-k^2\right)^2}+\frac{k^2d^2}{1-k^2}=\frac{k^2d^2}{\left(1-k^2\right)^2}[/tex]
This is a sphere (the sphere of Apollonius) with centre [tex]\left(\frac{k^2d}{k^2-1}, 0, 0\right)[/tex] on the line [tex]AB[/tex] and radius
[tex]R=\frac{kd}{\left|1-k^2\right|}[/tex]

Solution:
Put [tex]\vec{a}=\overrightarrow{OA}[/tex], [tex]\vec{b}=\overrightarrow{OB}[/tex], [tex]\vec{c}=\overrightarrow{OC}[/tex]; these vectors are non-zero and pairwise orthogonal. The area of a triangle is half the length of the cross product of two of its side vectors, so
[tex]S_{OAB}=\frac{1}{2}|\vec{a}×\vec{b}|,\quad S_{OBC}=\frac{1}{2}|\vec{b}×\vec{c}|,\quad S_{OCA}=\frac{1}{2}|\vec{c}×\vec{a}|,\quad S_{ABC}=\frac{1}{2}\left|\overrightarrow{CA}×\overrightarrow{CB}\right|[/tex]
Since [tex]\overrightarrow{CA}=\vec{a}-\vec{c}[/tex] and [tex]\overrightarrow{CB}=\vec{b}-\vec{c}[/tex], bilinearity, anticommutativity and [tex]\vec{c}×\vec{c}=\vec{0}[/tex] give
[tex]\overrightarrow{CA}×\overrightarrow{CB}=(\vec{a}-\vec{c})×(\vec{b}-\vec{c})=\vec{a}×\vec{b}-\vec{a}×\vec{c}-\vec{c}×\vec{b}=\vec{a}×\vec{b}+\vec{b}×\vec{c}+\vec{c}×\vec{a}[/tex]
The vector [tex]\vec{a}×\vec{b}[/tex] is perpendicular to [tex]\vec{a}[/tex] and [tex]\vec{b}[/tex], and so is [tex]\vec{c}[/tex]; in space the vectors perpendicular to two non-collinear vectors form one line, hence [tex]\vec{a}×\vec{b}\parallel\vec{c}[/tex]. Similarly [tex]\vec{b}×\vec{c}\parallel\vec{a}[/tex] and [tex]\vec{c}×\vec{a}\parallel\vec{b}[/tex]. Since [tex]\vec{a}[/tex], [tex]\vec{b}[/tex], [tex]\vec{c}[/tex] are pairwise orthogonal, so are the vectors [tex]\vec{a}×\vec{b}[/tex], [tex]\vec{b}×\vec{c}[/tex], [tex]\vec{c}×\vec{a}[/tex]; therefore all mixed terms vanish when we square the sum (the Pythagorean theorem):
[tex]\left|\vec{a}×\vec{b}+\vec{b}×\vec{c}+\vec{c}×\vec{a}\right|^2=|\vec{a}×\vec{b}|^2+|\vec{b}×\vec{c}|^2+|\vec{c}×\vec{a}|^2[/tex]
Dividing by [tex]4[/tex], we obtain
[tex]S_{ABC}^2=S_{OAB}^2+S_{OBC}^2+S_{OCA}^2[/tex]