Analytic Geometry in Space: Problems with Solutions

Solution:
A sphere is tangent to a plane exactly when the distance from its center to the plane equals its radius. By the formula for the distance from a point to a plane,
[tex]r=\frac{|6\times 1+6\times 4-7\times (-7)+42|}{\sqrt{6^2+6^2+(-7)^2}}=\frac{|6+24+49+42|}{\sqrt{121}}=\frac{121}{11}=11[/tex]
So the sphere is [tex](x-1)^2+(y-4)^2+(z+7)^2=121[/tex] and its radius is [tex]11[/tex].

Solution:
The direction cosines are the coordinates of the unit vector [tex]\vec{e}=(\cos\alpha, \cos\beta, \cos\gamma)[/tex] along the line ([tex]\alpha[/tex], [tex]\beta[/tex], [tex]\gamma[/tex] are the angles between the line and the coordinate axes), so
[tex]|\vec{e}|^2=\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1[/tex]
Using the double-angle formula [tex]\cos 2\varphi=2\cos^2\varphi-1[/tex], we get
[tex]\cos 2\alpha+\cos 2\beta+\cos 2\gamma=2\left(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\right)-3=2-3=-1[/tex]

Solution:
We have [tex]\overrightarrow{AB}=(-1, -2, 3)[/tex] and [tex]\overrightarrow{BA}=-\overrightarrow{AB}[/tex]. Put [tex]\vec{u}=\overrightarrow{AB}[/tex]. The cross product is bilinear, [tex]\vec{a}×\vec{a}=\vec{u}×\vec{u}=\vec{0}[/tex] and [tex]\vec{u}×\vec{a}=-\vec{a}×\vec{u}[/tex], hence
[tex]\left(\vec{a}-2\vec{u}\right)×\left(3\vec{a}-\vec{u}\right)=3\,\vec{a}×\vec{a}-\vec{a}×\vec{u}-6\,\vec{u}×\vec{a}+2\,\vec{u}×\vec{u}=-\vec{a}×\vec{u}+6\,\vec{a}×\vec{u}=5\,\vec{a}×\vec{u}[/tex]
We compute
[tex]\vec{a}×\vec{u}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-1&-1&2\\-1&-2&3\end{vmatrix}=\left((-1)\times 3-2\times (-2), \ 2\times (-1)-(-1)\times 3, \ (-1)\times (-2)-(-1)\times (-1)\right)=(1, 1, 1)[/tex]
Hence the required product is [tex]5\,(1, 1, 1)=(5, 5, 5)[/tex]. Direct check: [tex]\vec{a}-2\overrightarrow{AB}=(1, 3, -4)[/tex], [tex]3\vec{a}+\overrightarrow{BA}=(-2, -1, 3)[/tex] and
[tex](1, 3, -4)×(-2, -1, 3)=(9-4, \ 8-3, \ -1+6)=(5, 5, 5)[/tex]

Solution:
The vectors [tex]\overrightarrow{AB}=(2, 2, 2)[/tex] and [tex]\overrightarrow{AC}=(2, 0, -2)[/tex] are not collinear, so the three points determine a unique plane. A normal vector of this plane is
[tex]\overrightarrow{AB}×\overrightarrow{AC}=\left(2\times (-2)-2\times 0, \ 2\times 2-2\times (-2), \ 2\times 0-2\times 2\right)=(-4, 8, -4)=-4\,(1, -2, 1)[/tex]
so we can take [tex]\vec{n}=(1, -2, 1)[/tex]. The plane through [tex]A[/tex] with normal vector [tex]\vec{n}[/tex] is
[tex](x-1)-2(y-2)+(z-3)=0\iff x-2y+z=0[/tex]
Check: [tex]B:\ 3-8+5=0[/tex] and [tex]C:\ 3-4+1=0[/tex].

Solution:
The plane meets each coordinate axis at the point where the other two coordinates are zero:
[tex]y=z=0\Rightarrow x=8,\qquad x=z=0\Rightarrow y=-4,\qquad x=y=0\Rightarrow z=6[/tex]
So the pyramid has vertices [tex]O(0, 0, 0)[/tex], [tex](8, 0, 0)[/tex], [tex](0, -4, 0)[/tex] and [tex](0, 0, 6)[/tex]. Its three edges from [tex]O[/tex] lie on the coordinate axes, so they are pairwise perpendicular and have lengths [tex]8[/tex], [tex]4[/tex] and [tex]6[/tex]. Taking the right triangle with legs [tex]8[/tex] and [tex]4[/tex] in the plane [tex]Oxy[/tex] as the base, the height is [tex]6[/tex]:
[tex]V=\frac{1}{3}\times \frac{8\times 4}{2}\times 6=32[/tex]

Solution:
The equation does not contain [tex]y[/tex]. Hence, if a point [tex](x_0, y_0, z_0)[/tex] satisfies it, so does every point [tex](x_0, y_0+s, z_0)[/tex], [tex]s\in\mathbb{R}[/tex]. Thus, together with each of its points, the surface contains the whole line through that point parallel to [tex]\vec{p}=(0, 1, 0)[/tex], i.e. to the axis [tex]Oy[/tex].
In the plane [tex]y=0[/tex] the equation describes the circle with center [tex](-2, 0, 0)[/tex] and radius [tex]3[/tex]:
[tex]k:\ (x+2)^2+z^2=9,\quad y=0[/tex]
Therefore the surface is a circular cylinder with rulings parallel to [tex]\vec{p}=(0, 1, 0)[/tex] and directrix [tex]k[/tex]. It is not a circle: one equation in space defines a surface, while a circle needs two equations.

Solution:
We have [tex]\overrightarrow{AB}=(3, 2, 1)[/tex], [tex]\overrightarrow{AC}=(0, 2, 1)[/tex] and
[tex]\overrightarrow{AB}×\overrightarrow{AC}=\left(2\times 1-1\times 2, \ 1\times 0-3\times 1, \ 3\times 2-2\times 0\right)=(0, -3, 6)[/tex]
Hence the area of the triangle and the length of the side [tex]AC[/tex] are
[tex]S=\frac{1}{2}\left|\overrightarrow{AB}×\overrightarrow{AC}\right|=\frac{1}{2}\sqrt{0+9+36}=\frac{3\sqrt{5}}{2},\qquad |AC|=\sqrt{0+4+1}=\sqrt{5}[/tex]
Since [tex]S=\frac{1}{2}\,|AC|\times BH[/tex], we get
[tex]BH=\frac{2S}{|AC|}=\frac{3\sqrt{5}}{\sqrt{5}}=3[/tex]

Solution:
A normal vector of [tex]\alpha[/tex] is [tex]\vec{n}=(1, -4, 3)[/tex], so the line through [tex]M[/tex] perpendicular to [tex]\alpha[/tex] is
[tex]x=5+t,\quad y=-1-4t,\quad z=2+3t,\qquad t\in\mathbb{R}[/tex]
Substituting into the equation of the plane:
[tex](5+t)-4(-1-4t)+3(2+3t)+11=26+26t=0\Rightarrow t=-1[/tex]
So the foot of the perpendicular, i.e. the required projection, is [tex]M_0(4, 3, -1)[/tex]. Check: [tex]4-12-3+11=0[/tex], and [tex]\overrightarrow{MM_0}=(-1, 4, -3)=-\vec{n}[/tex] is collinear with [tex]\vec{n}[/tex].

Solution:
The line [tex]g_2[/tex] passes through [tex]P(2, 2, 0)[/tex] and has direction vector [tex]\vec{v}_2=(1, -1, 1)[/tex]; the line [tex]g_1[/tex] has direction vector [tex]\vec{v}_1=(1, 0, 1)[/tex]. These vectors are not collinear and both are parallel to [tex]\alpha[/tex], so the plane is unique and a normal vector of it is
[tex]\vec{n}=\vec{v}_2×\vec{v}_1=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&-1&1\\1&0&1\end{vmatrix}=(-1, 0, 1)[/tex]
The plane through [tex]P[/tex] with this normal vector is
[tex]-(x-2)+(z-0)=0\iff x-z-2=0[/tex]
Check: every point [tex](2+s, 2-s, s)[/tex] of [tex]g_2[/tex] satisfies [tex](2+s)-s-2=0[/tex], and [tex]\vec{n}\centerdot\vec{v}_1=-1+0+1=0[/tex]. The point [tex](1, 0, 0)[/tex] of [tex]g_1[/tex] gives [tex]1-0-2\ne0[/tex], so [tex]g_1[/tex] is parallel to [tex]\alpha[/tex] and does not lie in it.

Solution:
If [tex]\vec{a}=\vec{0}[/tex] or [tex]\vec{b}=\vec{0}[/tex], both sides equal [tex]0[/tex]. Otherwise let [tex]\varphi\in\left[0, \pi\right][/tex] be the angle between the vectors. By the definition of the cross product (this also covers collinear vectors, for which [tex]\sin \varphi=0[/tex] and [tex]\vec{a}×\vec{b}=\vec{0}[/tex])
[tex]\left|\vec{a}×\vec{b}\right|=|\vec{a}|\,|\vec{b}|\sin \varphi[/tex]
The scalar square of a vector equals the square of its length, and [tex]0\le\sin ^2\varphi\le 1[/tex], so
[tex](\vec{a}×\vec{b})^2=|\vec{a}|^2|\vec{b}|^2\sin ^2\varphi\le|\vec{a}|^2|\vec{b}|^2=\vec{a}^{\,2}\,\vec{b}^{\,2}[/tex]
More precisely, since [tex]\vec{a}\centerdot\vec{b}=|\vec{a}|\,|\vec{b}|\cos\varphi[/tex] and [tex]\sin ^2\varphi=1-\cos^2\varphi[/tex], we get Lagrange's identity
[tex](\vec{a}×\vec{b})^2=\vec{a}^{\,2}\,\vec{b}^{\,2}-(\vec{a}\centerdot\vec{b})^2\le\vec{a}^{\,2}\,\vec{b}^{\,2}[/tex]
so equality holds exactly when [tex]\vec{a}\centerdot\vec{b}=0[/tex], i.e. when the vectors are perpendicular or one of them is zero.

Solution:
Let [tex]M(x, y, z)[/tex] be a point of the curve. Subtracting the second equation from the first gives
[tex]x^2-y^2=0\iff (x-y)(x+y)=0[/tex]
Hence [tex]x-y=0[/tex] or [tex]x+y=0[/tex], i.e. every point of the curve lies in the plane [tex]\pi_1:\ x-y=0[/tex] or in the plane [tex]\pi_2:\ x+y=0[/tex]. Both planes are really needed: the points of the curve with [tex]x=y[/tex] form the ellipse [tex]x=y,\ x^2+z^2=a^2[/tex] in [tex]\pi_1[/tex], and those with [tex]x=-y[/tex] form the ellipse [tex]x=-y,\ x^2+z^2=a^2[/tex] in [tex]\pi_2[/tex] (all these points satisfy both equations, because [tex]y^2=x^2[/tex]). The normal vectors of the two planes are perpendicular:
[tex]\vec{n}_1=(1, -1, 0),\quad \vec{n}_2=(1, 1, 0),\quad \vec{n}_1\centerdot\vec{n}_2=1-1+0=0[/tex]
Hence [tex]\pi_1\perp\pi_2[/tex], which proves the statement.