Analytic Geometry in Space: Very Difficult Problems with Solutions

Solution:
The line [tex]g_1[/tex] passes through [tex]P_1(3, -1, 0)[/tex] with direction vector [tex]\vec{p}_1=(1, 2, 4)[/tex], and [tex]g_2[/tex] has direction vector [tex]\vec{p}_2=(3, 0, -5)[/tex]. The point [tex]A[/tex] is not on [tex]g_1[/tex] (from [tex]3+\lambda=1[/tex] we get [tex]\lambda=-2[/tex], but then [tex]y=-5\ne1[/tex]), so [tex]A[/tex] and [tex]g_1[/tex] determine a plane [tex]\varepsilon[/tex], and every line through [tex]A[/tex] that meets [tex]g_1[/tex] lies in [tex]\varepsilon[/tex]. A normal vector of [tex]\varepsilon[/tex]:
[tex]\vec{p}_1×\overrightarrow{P_1A}=(1, 2, 4)×(-2, 2, 1)=(-6, -9, 6)\parallel(2, 3, -2)[/tex]
[tex]\varepsilon:\ 2(x-1)+3(y-1)-2(z-1)=0\iff 2x+3y-2z-3=0[/tex]
The required line meets [tex]g_2[/tex] at a point of [tex]\varepsilon[/tex]. Since [tex](2, 3, -2)⋅(3, 0, -5)=16\ne0[/tex], the line [tex]g_2[/tex] meets [tex]\varepsilon[/tex] at exactly one point:
[tex]2(-2+3\mu)+3\times (-1)-2(4-5\mu)-3=16\mu-18=0\ \Rightarrow\ \mu=\frac{9}{8},\quad P\left(\frac{11}{8}, -1, -\frac{13}{8}\right)[/tex]
So the required line must be [tex]AP[/tex], with direction vector
[tex]8\,\overrightarrow{AP}=8\left(\frac{3}{8}, -2, -\frac{21}{8}\right)=(3, -16, -21)[/tex]
It lies in [tex]\varepsilon[/tex] and is not parallel to [tex]g_1[/tex], so it really meets [tex]g_1[/tex]: from [tex]1+3s=3+\lambda[/tex], [tex]1-16s=-1+2\lambda[/tex], [tex]1-21s=4\lambda[/tex] we get [tex]s=\frac{3}{11}[/tex], [tex]\lambda=-\frac{13}{11}[/tex]. Therefore the unique such line is
[tex]\frac{x-1}{3}=\frac{y-1}{-16}=\frac{z-1}{-21}[/tex]

Solution:
Let [tex]A'[/tex] be the mirror image of [tex]A[/tex] in [tex]\alpha[/tex]. By the law of reflection, the reflected ray lies on the line through [tex]A'[/tex] and the reflection point [tex]P[/tex], and [tex]AP=A'P[/tex]. The normal vector of [tex]\alpha[/tex] is [tex]\vec{n}=(3, 2, 2)[/tex]; the perpendicular from [tex]A[/tex] to [tex]\alpha[/tex] consists of the points [tex]A+\lambda\vec{n}[/tex], and its foot corresponds to
[tex]3(1+3\lambda)+2\times 2\lambda+2(-3+2\lambda)-65=17\lambda-68=0\ \Rightarrow\ \lambda=4,\qquad A'=A+8\vec{n}=(25, 16, 13)[/tex]
Let [tex]Q[/tex] be the point where the reflected ray meets [tex]a[/tex]. Then [tex]PQ\perp a[/tex], and since [tex]A'[/tex], [tex]P[/tex], [tex]Q[/tex] are collinear, [tex]A'Q\perp a[/tex], i.e. [tex]Q[/tex] is the foot of the perpendicular from [tex]A'[/tex] to [tex]a[/tex]. With [tex]Q(4+t, -2-2t, 1+2t)[/tex] and the direction vector [tex]\vec{d}=(1, -2, 2)[/tex] of [tex]a[/tex]:
[tex]\overrightarrow{A'Q}⋅\vec{d}=(t-21)-2(-2t-18)+2(2t-12)=9t-9=0\ \Rightarrow\ t=1,\quad Q(5, -4, 3)[/tex]
The left side of the equation of [tex]\alpha[/tex] equals [tex]-68[/tex] at [tex]A[/tex], [tex]-52[/tex] at [tex]Q[/tex] and [tex]68[/tex] at [tex]A'[/tex]. So [tex]A[/tex] and [tex]Q[/tex] are on the same side of [tex]\alpha[/tex], and the segment [tex]A'Q[/tex] crosses [tex]\alpha[/tex] at the reflection point [tex]P\left(\frac{41}{3}, \frac{14}{3}, \frac{22}{3}\right)[/tex]; the path [tex]A\to P\to Q[/tex] is realized. Therefore its length is
[tex]AP+PQ=A'P+PQ=A'Q=\left|(-20, -20, -10)\right|=\sqrt{400+400+100}=30[/tex]
(Indeed, [tex]AP=17[/tex] and [tex]PQ=13[/tex].)

Solution:
Putting [tex]z=t[/tex], we get parametric equations of [tex]a[/tex]: it passes through [tex]P_0(2, 3, 0)[/tex] and has direction vector [tex]\vec{e}=(1, 1, 1)[/tex]:
[tex]a:\ x=2+t,\quad y=3+t,\quad z=t[/tex]
Let [tex]\sigma[/tex] be the orthogonal symmetry with respect to [tex]a[/tex] (the rotation by [tex]180^\circ[/tex] about [tex]a[/tex]) and [tex]M'=\sigma(M)[/tex]. The symmetry fixes [tex]P[/tex] and maps the line [tex]MP[/tex] onto the line [tex]M'P[/tex]. By the given law, the reflected ray is the image of the continuation of [tex]MP[/tex] beyond [tex]P[/tex], so it lies on the line [tex]M'P[/tex], on the side of [tex]P[/tex] opposite to [tex]M'[/tex]. Hence [tex]P[/tex] is the common point of the line [tex]M'N[/tex] and [tex]a[/tex], and [tex]P[/tex] lies between [tex]M'[/tex] and [tex]N[/tex]. The foot [tex]H[/tex] of the perpendicular from [tex]M[/tex] to [tex]a[/tex]:
[tex]t=\frac{\overrightarrow{P_0M}⋅\vec{e}}{|\vec{e}|^2}=\frac{(0, -6, -6)⋅(1, 1, 1)}{3}=-4,\quad H(-2, -1, -4),\quad M'=2H-M=(-6, 1, -2)[/tex]
Since [tex]\overrightarrow{M'N}=(14, 2, 2)[/tex], the points of the line [tex]M'N[/tex] are [tex](-6+7w, 1+w, -2+w)[/tex] ([tex]M'[/tex] for [tex]w=0[/tex], [tex]N[/tex] for [tex]w=2[/tex]). The second equation of [tex]a[/tex] holds for every [tex]w[/tex]: [tex]-(1+w)+(-2+w)+3=0[/tex]. The first one gives
[tex](-6+7w)-(-2+w)-2=6w-6=0\ \Rightarrow\ w=1,\quad P(1, 2, -1)[/tex]
Since [tex]0\lt 1\lt 2[/tex], the point [tex]P[/tex] lies between [tex]M'[/tex] and [tex]N[/tex], as required. Check: the symmetry maps the incident direction [tex]\vec{u}=\overrightarrow{MP}=(-1, 5, 5)[/tex] (with [tex]\vec{u}⋅\vec{e}=9[/tex]) to
[tex]2\,\frac{\vec{u}⋅\vec{e}}{|\vec{e}|^2}\,\vec{e}-\vec{u}=6\,(1, 1, 1)-(-1, 5, 5)=(7, 1, 1)=\overrightarrow{PN}[/tex]

Solution:
Take the (non-orthogonal) basis [tex]\vec{a}=\overrightarrow{AB}[/tex], [tex]\vec{b}=\overrightarrow{AD}[/tex], [tex]\vec{c}=\overrightarrow{AA_1}[/tex]. From the given lengths and the angles of [tex]60^\circ[/tex] we get
[tex]\vec{a}^2=9,\quad \vec{b}^2=1,\quad \vec{c}^2=4[/tex]
[tex]\vec{a}⋅\vec{b}=3\times 1\times \frac12=\frac32,\quad \vec{a}⋅\vec{c}=3\times 2\times \frac12=3,\quad \vec{b}⋅\vec{c}=1\times 2\times \frac12=1[/tex]
Then [tex]\overrightarrow{AC}=\vec{a}+\vec{b}[/tex], [tex]\overrightarrow{AB_1}=\vec{a}+\vec{c}[/tex], and the line [tex]CB_1[/tex] has direction vector [tex]\overrightarrow{CB_1}=\vec{c}-\vec{b}[/tex]. The lines are skew, because the scalar triple product [tex]\left(\vec{a},\,\vec{c}-\vec{b},\,\overrightarrow{AC}\right)=(\vec{a},\vec{c},\vec{b})[/tex] is non-zero. The shortest segment is their common perpendicular [tex]PQ[/tex], where [tex]P\in AB[/tex] and [tex]Q\in CB_1[/tex]:
[tex]\overrightarrow{AP}=s\,\vec{a},\quad \overrightarrow{AQ}=\vec{a}+\vec{b}+t(\vec{c}-\vec{b}),\quad \overrightarrow{PQ}=(1-s)\vec{a}+(1-t)\vec{b}+t\vec{c}[/tex]
We need [tex]\overrightarrow{PQ}⋅\vec{a}=0[/tex] and [tex]\overrightarrow{PQ}⋅(\vec{c}-\vec{b})=0[/tex]. Using [tex]\vec{a}⋅(\vec{c}-\vec{b})=\frac32[/tex], [tex]\vec{b}⋅(\vec{c}-\vec{b})=0[/tex] and [tex]\vec{c}⋅(\vec{c}-\vec{b})=3[/tex], we get
[tex]\begin{cases}9(1-s)+\frac32(1-t)+3t=0\\ \frac32(1-s)+3t=0\end{cases}\ \Rightarrow\ s=1+2t,\quad \frac32-\frac{33}{2}t=0\ \Rightarrow\ t=\frac1{11},\quad s=\frac{13}{11}[/tex]
Hence [tex]\overrightarrow{PQ}=\frac{1}{11}\left(-2\vec{a}+10\vec{b}+\vec{c}\right)[/tex], and squaring with the values above gives
[tex]PQ^2=\frac{4\times 9+100\times 1+4+2(-2)(10)\times \frac32+2(-2)(1)\times 3+2(10)(1)\times 1}{121}=\frac{36+100+4-60-12+20}{121}=\frac{8}{11}[/tex]
[tex]PQ=\sqrt{\frac{8}{11}}=\frac{2\sqrt{22}}{11}[/tex]

Solution:
The dot products are
[tex]\vec{a}⋅\vec{b}=1\times 2\times \frac12=1,\quad \vec{b}⋅\vec{c}=2\times \sqrt2\times \frac12=\sqrt2,\quad \vec{a}⋅\vec{c}=0[/tex]
By the double cross product formula [tex](\vec{u}×\vec{v})×\vec{w}=(\vec{u}⋅\vec{w})\vec{v}-(\vec{v}⋅\vec{w})\vec{u}[/tex] we get
[tex]\overrightarrow{OB}=(\vec{a}⋅\vec{c})\vec{b}-(\vec{b}⋅\vec{c})\vec{a}=-\sqrt2\,\vec{a}[/tex]
[tex]\overrightarrow{OC}=\big(\vec{a}⋅(\vec{b}×\vec{c})\big)\vec{b}-\big(\vec{b}⋅(\vec{b}×\vec{c})\big)\vec{a}=(\vec{a},\vec{b},\vec{c})\,\vec{b}[/tex]
Since [tex]\overrightarrow{OB}×\overrightarrow{OC}=-\sqrt2\,(\vec{a},\vec{b},\vec{c})\,\vec{a}×\vec{b}[/tex], the scalar triple product of the edge vectors is
[tex](\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC})=(\vec{a}×\vec{b})⋅\big(\overrightarrow{OB}×\overrightarrow{OC}\big)=-\sqrt2\,(\vec{a},\vec{b},\vec{c})\,|\vec{a}×\vec{b}|^2[/tex]
By Lagrange's identity
[tex]|\vec{a}×\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}⋅\vec{b})^2=4-1=3[/tex]
The square of the scalar triple product equals the Gram determinant:
[tex](\vec{a},\vec{b},\vec{c})^2=\begin{vmatrix}1&1&0\\1&4&\sqrt2\\0&\sqrt2&2\end{vmatrix}=1\times (8-2)-1\times (2-0)=4\ \Rightarrow\ |(\vec{a},\vec{b},\vec{c})|=2[/tex]
In particular, such vectors exist and are not coplanar. Therefore
[tex]V=\frac16\left|(\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC})\right|=\frac16\times \sqrt2\times 2\times 3=\sqrt2[/tex]

Solution:
Take the dot product of the equation with [tex]\vec{b}[/tex]. Since [tex]\vec{x}×\vec{b}\perp\vec{b}[/tex],
[tex]\vec{x}⋅\vec{b}=\vec{a}⋅\vec{b}+(\vec{x}×\vec{b})⋅\vec{b}=\vec{a}⋅\vec{b}[/tex]
Now take the cross product of the equation with [tex]\vec{b}[/tex] and apply the double cross product formula:
[tex]\vec{x}×\vec{b}=\vec{a}×\vec{b}+(\vec{x}×\vec{b})×\vec{b}=\vec{a}×\vec{b}+(\vec{x}⋅\vec{b})\vec{b}-(\vec{b}⋅\vec{b})\vec{x}=\vec{a}×\vec{b}+(\vec{a}⋅\vec{b})\vec{b}-|\vec{b}|^2\vec{x}[/tex]
Substituting this expression into the original equation:
[tex]\vec{x}=\vec{a}+\vec{a}×\vec{b}+(\vec{a}⋅\vec{b})\vec{b}-|\vec{b}|^2\vec{x}\ \Rightarrow\ \left(1+|\vec{b}|^2\right)\vec{x}=\vec{a}+\vec{a}×\vec{b}+(\vec{a}⋅\vec{b})\vec{b}[/tex]
Since [tex]1+|\vec{b}|^2\gt0[/tex], the only possible solution is
[tex]\vec{x}=\frac{\vec{a}+\vec{a}×\vec{b}+(\vec{a}⋅\vec{b})\vec{b}}{1+|\vec{b}|^2}[/tex]
A solution exists: the linear map [tex]\vec{x}\mapsto\vec{x}-\vec{x}×\vec{b}[/tex] is injective (if [tex]\vec{x}=\vec{x}×\vec{b}[/tex], the dot product with [tex]\vec{x}[/tex] gives [tex]|\vec{x}|^2=0[/tex]), hence bijective. So the equation has exactly one solution, namely the vector above (a direct substitution also confirms it).

Solution:
Let [tex]\vec{a}=\overrightarrow{SA}[/tex], [tex]\vec{b}=\overrightarrow{SB}[/tex], [tex]\vec{c}=\overrightarrow{SC}[/tex]. These vectors are linearly independent, because [tex]S[/tex], [tex]A[/tex], [tex]B[/tex], [tex]C[/tex] are not coplanar. From the given ratios
[tex]\overrightarrow{SM}=\frac34\vec{a},\quad \overrightarrow{SN}=\frac12\vec{b},\quad \overrightarrow{SP}=\frac38\vec{c}[/tex]
Since [tex]ABCD[/tex] is a parallelogram, [tex]\overrightarrow{AD}=\overrightarrow{BC}[/tex], so
[tex]\overrightarrow{SD}=\overrightarrow{SA}+\overrightarrow{BC}=\vec{a}+\vec{c}-\vec{b},\qquad \overrightarrow{BD}=\overrightarrow{SD}-\overrightarrow{SB}=\vec{a}-2\vec{b}+\vec{c}[/tex]
The plane [tex](MNP)[/tex] contains the non-collinear vectors
[tex]\overrightarrow{NM}=\frac34\vec{a}-\frac12\vec{b},\qquad \overrightarrow{NP}=\frac38\vec{c}-\frac12\vec{b}[/tex]
We look for [tex]\alpha[/tex], [tex]\beta[/tex] with [tex]\alpha\overrightarrow{NM}+\beta\overrightarrow{NP}=\overrightarrow{BD}[/tex]. The coefficients of [tex]\vec{a}[/tex] and [tex]\vec{c}[/tex] give [tex]\frac34\alpha=1[/tex] and [tex]\frac38\beta=1[/tex], i.e. [tex]\alpha=\frac43[/tex], [tex]\beta=\frac83[/tex], and then the coefficient of [tex]\vec{b}[/tex] also matches:
[tex]-\frac12\times \frac43-\frac12\times \frac83=-2\quad\Rightarrow\quad \overrightarrow{BD}=\frac43\overrightarrow{NM}+\frac83\overrightarrow{NP}[/tex]
Hence [tex]\overrightarrow{BD}[/tex] is parallel to the plane [tex](MNP)[/tex]. It remains to show that [tex]B[/tex] does not lie in this plane. A point [tex]X[/tex] lies in [tex](MNP)[/tex] if and only if [tex]\overrightarrow{SX}=x\overrightarrow{SM}+y\overrightarrow{SN}+z\overrightarrow{SP}[/tex] with [tex]x+y+z=1[/tex]. For [tex]X=B[/tex] the decomposition is unique:
[tex]\overrightarrow{SB}=0\,\overrightarrow{SM}+2\,\overrightarrow{SN}+0\,\overrightarrow{SP},\qquad 0+2+0=2\ne1[/tex]
Therefore the line [tex]BD[/tex] does not lie in the plane [tex](MNP)[/tex] and is parallel to it.

Solution:
The matrix of the quadratic part is
[tex]A=\begin{bmatrix}4&0&2\\0&2&-2\\2&-2&3\end{bmatrix}[/tex]
Its trace is [tex]9[/tex], the sum of its principal minors of order [tex]2[/tex] is [tex]8+8+2=18[/tex] and [tex]\det A=0[/tex], so the characteristic equation is
[tex]\lambda^3-9\lambda^2+18\lambda=\lambda(\lambda-3)(\lambda-6)=0[/tex]
The eigenvalues [tex]3[/tex], [tex]6[/tex], [tex]0[/tex] have the orthonormal eigenvectors (forming a right-handed basis)
[tex]\vec{e}_1=\frac13(-2, -2, 1),\quad \vec{e}_2=\frac13(2, -1, 2),\quad \vec{e}_3=\frac13(-1, 2, 2),\qquad \vec{e}_1×\vec{e}_2=\vec{e}_3[/tex]
The rotation
[tex]x=\frac{-2x'+2y'-z'}{3},\quad y=\frac{-2x'-y'+2z'}{3},\quad z=\frac{x'+2y'+2z'}{3}[/tex]
turns the quadratic part into [tex]3x'^2+6y'^2[/tex] and the linear part [tex]6x+4y+8z[/tex] into [tex]-4x'+8y'+6z'[/tex] (its coefficients are [tex]\vec{e}_i⋅\vec{l}[/tex] with [tex]\vec{l}=(6, 4, 8)[/tex]). Thus
[tex]S:\ 3x'^2+6y'^2-4x'+8y'+6z'+2=0[/tex]
Completing the squares:
[tex]3\left(x'-\frac23\right)^2+6\left(y'+\frac23\right)^2+6z'+2-\frac43-\frac83=0\ \Leftrightarrow\ 3\left(x'-\frac23\right)^2+6\left(y'+\frac23\right)^2=-6\left(z'-\frac13\right)[/tex]
The orientation-preserving change of coordinates (a translation followed by a half-turn about the new [tex]y'[/tex]-axis)
[tex]X=-\left(x'-\frac23\right),\quad Y=y'+\frac23,\quad Z=-\left(z'-\frac13\right)[/tex]
gives [tex]3X^2+6Y^2=6Z[/tex]. Hence [tex]S[/tex] is an elliptic paraboloid with canonical equation
[tex]X^2+2Y^2=2Z[/tex]

Solution:
The affine map [tex]u=\frac{x}{a},\ v=\frac{y}{b},\ w=\frac{z}{c}[/tex] sends the ellipsoid to the unit ball [tex]u^2+v^2+w^2\le1[/tex], planes to planes and [tex]O[/tex] to [tex]O[/tex]. It sends [tex]M[/tex] to [tex]M'\left(\frac{x_0}{a}, \frac{y_0}{b}, \frac{z_0}{c}\right)=M'(u_0, v_0, w_0)[/tex], where [tex]0\lt\rho=|OM'|\lt1[/tex]. All volumes are multiplied by the same factor [tex]\frac{1}{abc}[/tex], so it suffices to solve the problem for the ball. A plane at distance [tex]d[/tex] from [tex]O[/tex] cuts off, on the side not containing [tex]O[/tex], a cap of height [tex]1-d[/tex] with volume
[tex]V(d)=\int_d^1\pi(1-t^2)\,dt=\frac{\pi}{3}(1-d)^2(2+d)[/tex]
Since [tex]V'(d)=-\pi(1-d)(1+d)\lt0[/tex] for [tex]0\le d\lt1[/tex], the volume decreases as [tex]d[/tex] grows, so we need the largest possible [tex]d[/tex]. For a plane through [tex]M'[/tex] we always have [tex]d\le|OM'|=\rho[/tex], with equality only for the plane through [tex]M'[/tex] perpendicular to [tex]OM'[/tex]:
[tex]u_0u+v_0v+w_0w=u_0^2+v_0^2+w_0^2[/tex]
Returning to [tex]x[/tex], [tex]y[/tex], [tex]z[/tex], we get the unique answer
[tex]\frac{x_0x}{a^2}+\frac{y_0y}{b^2}+\frac{z_0z}{c^2}=\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}+\frac{z_0^2}{c^2}[/tex]
Geometrically, this is the plane through [tex]M[/tex] parallel to the tangent plane at the point where the ray [tex]OM[/tex] meets the ellipsoid.

Solution:
Let a plane [tex]\pi[/tex] through [tex]O[/tex] cut the ellipsoid in a circle. The ellipsoid and [tex]\pi[/tex] are both symmetric about [tex]O[/tex], so the section is too; hence the circle is centered at [tex]O[/tex] and, if [tex]r\gt0[/tex] is its radius, it lies on the sphere [tex]x^2+y^2+z^2=r^2[/tex]. Subtracting [tex]\frac{1}{r^2}[/tex] times the equation of the sphere from the equation of the ellipsoid, we see that every point of the circle satisfies
[tex]\left(\frac{1}{a^2}-\frac{1}{r^2}\right)x^2+\left(\frac{1}{b^2}-\frac{1}{r^2}\right)y^2+\left(\frac{1}{c^2}-\frac{1}{r^2}\right)z^2=0\qquad(*)[/tex]
Equation [tex](*)[/tex] is homogeneous, so it holds on every line through [tex]O[/tex] and a point of the circle, i.e. on the whole plane [tex]\pi[/tex]. If [tex]r\ne a[/tex], [tex]r\ne b[/tex] and [tex]r\ne c[/tex], all three coefficients are non-zero and two of them have the same sign; if, say, these are the coefficients of [tex]x^2[/tex] and [tex]y^2[/tex], then [tex]\pi[/tex] and the plane [tex]z=0[/tex] have a common line through [tex]O[/tex], on which [tex](*)[/tex] holds only at [tex]O[/tex], a contradiction (the other cases are the same). If [tex]r=a[/tex], the coefficients of [tex]y^2[/tex] and [tex]z^2[/tex] are positive and [tex](*)[/tex] is only the axis [tex]Ox[/tex]; similarly, [tex]r=c[/tex] gives only the axis [tex]Oz[/tex]. Hence [tex]r=b[/tex], and [tex](*)[/tex] becomes
[tex]\frac{a^2-b^2}{a^2b^2}\,x^2=\frac{b^2-c^2}{b^2c^2}\,z^2\ \Leftrightarrow\ c^2(a^2-b^2)\,x^2=a^2(b^2-c^2)\,z^2\ \Leftrightarrow\ c\sqrt{a^2-b^2}\,x\pm a\sqrt{b^2-c^2}\,z=0[/tex]
Conversely, on each of these two planes [tex](*)[/tex] holds with [tex]r=b[/tex], so such a plane meets the ellipsoid in the same set as the sphere [tex]x^2+y^2+z^2=b^2[/tex], i.e. in a circle of radius [tex]b[/tex] centered at [tex]O[/tex]. Hence there are exactly two such planes (both contain the axis [tex]Oy[/tex]):
[tex]c\sqrt{a^2-b^2}\,x+a\sqrt{b^2-c^2}\,z=0,\qquad c\sqrt{a^2-b^2}\,x-a\sqrt{b^2-c^2}\,z=0[/tex]