1. Meaning of scalar product of three vectors

For three vectors (\vec a,\vec b,\vec c), the scalar triple product is defined as
[ [\vec a\ \vec b\ \vec c]=\vec a\cdot(\vec b\times\vec c). ] It is a scalar quantity. In component form, [ [\vec a\ \vec b\ \vec c]= \begin{vmatrix} a_1&a_2&a_3\ b_1&b_2&b_3\ c_1&c_2&c_3 \end{vmatrix}. ] It is also called the box product. Its absolute value gives the volume of the parallelepiped whose coterminous edges are (\vec a,\vec b,\vec c).

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2. Scalar triple product gives the volume of the parallelepiped

Let the parallelepiped have coterminous edges (\vec a,\vec b,\vec c). Take the parallelogram formed by (\vec b) and (\vec c) as the base.

Area of the base is
[ A=|\vec b\times\vec c|. ] Let
[ \hat n=\frac{\vec b\times\vec c}{|\vec b\times\vec c|} ] be the unit vector normal to the base.

The height (h) of the parallelepiped is the perpendicular distance of the terminal point of (\vec a) from the base. Hence (h) is the component of (\vec a) along (\hat n): [ h=|\vec a\cdot \hat n| =\left|\vec a\cdot \frac{\vec b\times\vec c}{|\vec b\times\vec c|}\right| =\frac{|\vec a\cdot(\vec b\times\vec c)|}{|\vec b\times\vec c|}. ] Therefore, [ V=A h =|\vec b\times\vec c|\cdot \frac{|\vec a\cdot(\vec b\times\vec c)|}{|\vec b\times\vec c|} =|\vec a\cdot(\vec b\times\vec c)|. ] Thus, [ V=|[\vec a\ \vec b\ \vec c]|. ] So the scalar triple product gives the signed volume of the parallelepiped, and its absolute value gives the actual volume. If (\vec a,\vec b,\vec c) form a right-handed system, then ([\vec a\ \vec b\ \vec c]>0); otherwise it is negative.

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3. Scalar triple product of three coplanar vectors is zero

If (\vec a,\vec b,\vec c) are coplanar, then (\vec b) and (\vec c) lie in the same plane. Hence
[ \vec b\times\vec c ] is perpendicular to that plane. Since (\vec a) also lies in the same plane, (\vec a) is perpendicular to (\vec b\times\vec c). Therefore, [ \vec a\cdot(\vec b\times\vec c)=0. ] Hence, [ [\vec a\ \vec b\ \vec c]=0. ] Equivalently, the volume of the parallelepiped formed by three coplanar vectors is zero. Conversely, if ([\vec a\ \vec b\ \vec c]=0), then the volume is zero, so the three vectors must be coplanar. Also, if any two vectors are parallel or collinear, then (\vec b\times\vec c=0), and the scalar triple product is again zero.
[ \boxed{[\vec a\ \vec b\ \vec c]=0 \quad \text{for coplanar vectors}} ]

Given $3x^2 - 6x + 4y^2 + 14 = 0$.

Complete the square in $x$: $3(x^2 - 2x) + 4y^2 + 14 = 0$ $3\left[(x-1)^2 - 1\right] + 4y^2 + 14 = 0$ $3(x-1)^2 + 4y^2 + 11 = 0$

Shift the origin to $(1, 0)$, i.e. put $x = X + 1,\ y = Y$. The equation becomes

$3X^2 + 4Y^2 + 11 = 0,$

which has no linear terms.

First alternative: Tangents at the ends of a focal chord of an ellipse meet on the directrix.

Let the ellipse be [ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. ] Its focus is (S(ae,0)), and the corresponding directrix is [ x=\frac{a}{e}. ]

Let (P(x_1,y_1)) and (Q(x_2,y_2)) be the ends of a focal chord, so (P,Q,S) are collinear. Let the tangents at (P) and (Q) meet at (T(h,k)).

The tangent at (P(x_1,y_1)) is [ \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1. ] Since (T(h,k)) lies on it, [ \frac{hx_1}{a^2}+\frac{ky_1}{b^2}=1. \quad (1) ]

Similarly, the tangent at (Q(x_2,y_2)) gives [ \frac{hx_2}{a^2}+\frac{ky_2}{b^2}=1. \quad (2) ]

From (1) and (2), both (P) and (Q) satisfy the linear equation [ \frac{hx}{a^2}+\frac{ky}{b^2}=1. ] Hence this is the equation of the chord (PQ).

But (PQ) is a focal chord, so it passes through the focus (S(ae,0)). Substituting (x=ae,\ y=0) in the chord equation, [ \frac{h(ae)}{a^2}+\frac{k(0)}{b^2}=1 ] [ \frac{he}{a}=1 ] [ h=\frac{a}{e}. ]

Thus the point of intersection (T(h,k)) lies on [ x=\frac{a}{e}, ] which is the directrix of the ellipse. Hence the tangents at the ends of a focal chord intersect on the directrix.

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Second alternative: Tangent to the hyperbola [ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 ] at ((\alpha,\beta)).

Differentiate the hyperbola implicitly: [ \frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0. ] So [ \frac{dy}{dx}=\frac{b^2x}{a^2y}. ] At ((\alpha,\beta)), [ m=\frac{b^2\alpha}{a^2\beta}. ]

Therefore the tangent is [ y-\beta=\frac{b^2\alpha}{a^2\beta}(x-\alpha). ] Simplifying, [ a^2\beta y-a^2\beta^2=b^2\alpha x-b^2\alpha^2 ] [ b^2\alpha x-a^2\beta y=b^2\alpha^2-a^2\beta^2. ]

Since ((\alpha,\beta)) lies on the hyperbola, [ \frac{\alpha^2}{a^2}-\frac{\beta^2}{b^2}=1 ] [ b^2\alpha^2-a^2\beta^2=a^2b^2. ]

Therefore, [ b^2\alpha x-a^2\beta y=a^2b^2. ] Dividing by (a^2b^2), [ \boxed{\frac{\alpha x}{a^2}-\frac{\beta y}{b^2}=1}. ]

So the required tangent is [ \boxed{\frac{\alpha x}{a^2}-\frac{\beta y}{b^2}=1}. ]

Chord joining two points of a conic. Let the points have vectorial angles $\alpha$ and $\beta$ on $\dfrac{l}{r} = 1 + e\cos\theta$. The chord is a straight line not through the pole, so we seek it in the form $\dfrac{l}{r} = e\cos\theta + \sec\delta\,\cos(\theta-\mu)$ (the term $e\cos\theta$ makes it a line in the same family as the conic). Choose $\mu = \dfrac{\alpha+\beta}{2}$ and $\delta = \dfrac{\alpha-\beta}{2}$. Then at $\theta=\alpha$ and $\theta=\beta$, $\sec\delta\cos(\alpha-\mu) = \sec\delta\cos\delta = 1,\qquad \sec\delta\cos(\beta-\mu)=\sec\delta\cos(-\delta)=1,$ so both points satisfy it, hence this is the chord:

$\frac{l}{r} = e\cos\theta + \sec\frac{\alpha-\beta}{2}\cos\!\left(\theta - \frac{\alpha+\beta}{2}\right)$

Here $\alpha = \dfrac{\pi}{3}$, $\beta = \dfrac{\pi}{2}$: $\ \dfrac{\alpha-\beta}{2} = -\dfrac{\pi}{12}$, $\ \dfrac{\alpha+\beta}{2} = \dfrac{5\pi}{12}$. The required chord is

$\frac{l}{r} = e\cos\theta + \sec\frac{\pi}{12}\cos\!\left(\theta - \frac{5\pi}{12}\right)$

or equivalently $\;l\cos\dfrac{\pi}{12} = r\left[\cos\!\left(\theta - \dfrac{5\pi}{12}\right) + e\cos\dfrac{\pi}{12}\cos\theta\right]$.

Check at $\theta=\pi/2$: right side $= 0 + \sec\frac{\pi}{12}\cos\frac{\pi}{12} = 1 = l/r$, as required.

Let the required plane pass through the intersection of
$2x+y+z-5=0,\qquad 3x-2y+4z-5=0.$ Its equation can be written as
$(2x+y+z-5)+\lambda(3x-2y+4z-5)=0.$ So its normal vector is
$\vec n=(2+3\lambda,\;1-2\lambda,\;1+4\lambda).$

The required plane is normal to the plane
$x+y+5z=0,$ whose normal vector is ((1,1,5)). Hence the two normals are perpendicular: $\vec n\cdot(1,1,5)=0.$ Thus, $(2+3\lambda)+(1-2\lambda)+5(1+4\lambda)=0$ $8+21\lambda=0$ $\lambda=-\frac{8}{21}.$

Substituting (\lambda=-\frac{8}{21}) into the plane equation gives
$\left(2-\frac{24}{21}\right)x+\left(1+\frac{16}{21}\right)y+\left(1-\frac{32}{21}\right)z-5\left(1-\frac{8}{21}\right)=0.$ Simplifying, $\frac{18}{21}x+\frac{37}{21}y-\frac{11}{21}z-\frac{65}{21}=0.$ Multiplying by (21),

$\boxed{18x+37y-11z-65=0}$

This is the required equation of the plane.

The lines are

$L_1:\frac{x}{2}=\frac{y}{-3}=\frac{z}{1},\qquad L_2:\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}.$

Direction vectors:

$\vec d_1=(2,-3,1),\qquad \vec d_2=(3,-5,2).$

Points on the lines:

$A=(0,0,0),\qquad B=(2,1,-2).$

The shortest distance between skew lines is

$d=\frac{|(\vec{AB})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Now,

$$\vec d_1\times \vec d_2= \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 2 & -3 & 1\\ 3 & -5 & 2 \end{vmatrix} =(-1,-1,-1).$$

Also,

$\vec{AB}=B-A=(2,1,-2).$

So,

$$(\vec{AB})\cdot(\vec d_1\times \vec d_2) =(2,1,-2)\cdot(-1,-1,-1) =-2-1+2=-1.$$

Hence,

$$d=\frac{|-1|}{\sqrt{(-1)^2+(-1)^2+(-1)^2}} =\frac{1}{\sqrt3} =\frac{\sqrt3}{3}.$$

$\boxed{\text{Shortest distance}=\frac{1}{\sqrt3}\text{ units}=\frac{\sqrt3}{3}\text{ units}}$

Right circular cylinder. The axis $x = 2y = -z$ is $\dfrac{x}{2}=\dfrac{y}{1}=\dfrac{z}{-2}$: it passes through the origin with direction ratios $(2, 1, -2)$, and $\sqrt{4+1+4}=3$.

A point $P(x,y,z)$ lies on the cylinder when its distance from the axis equals 4. Distance$^2$ = $OP^2$ − (projection of $OP$ on the axis)$^2$: $x^2+y^2+z^2 - \frac{(2x+y-2z)^2}{9} = 16$ $9(x^2+y^2+z^2) - (2x+y-2z)^2 = 144$

Expanding $(2x+y-2z)^2 = 4x^2+y^2+4z^2+4xy-4yz-8zx$:

$5x^2 + 8y^2 + 5z^2 - 4xy + 4yz + 8zx - 144 = 0$

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OR

Reciprocal cones. Two cones with a common vertex are reciprocal if each is the locus of the normals, through the vertex, to the tangent planes of the other.

Cone $S$: $ayz+bzx+cxy=0$, i.e. $a_{11}=a_{22}=a_{33}=0$, $f=\tfrac a2,\ g=\tfrac b2,\ h=\tfrac c2$.

The plane $lx+my+nz=0$ touches the cone $ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0$ iff $A l^2+Bm^2+Cn^2+2Fmn+2Gnl+2Hlm=0$, where $A=bc-f^2$ etc. are the cofactors. Here the coefficients are zero on the diagonal, so $A=-\tfrac{a^2}{4},\ B=-\tfrac{b^2}{4},\ C=-\tfrac{c^2}{4},\quad F=gh-af=\tfrac{bc}{4},\ G=\tfrac{ca}{4},\ H=\tfrac{ab}{4}.$ The normal to the plane has direction $(l,m,n)$, so replacing $(l,m,n)$ by $(x,y,z)$ gives the reciprocal cone: $-\tfrac{1}{4}(a^2x^2+b^2y^2+c^2z^2) + \tfrac{1}{2}(bc\,yz+ca\,zx+ab\,xy)=0$ $a^2x^2+b^2y^2+c^2z^2-2bc\,yz-2ca\,zx-2ab\,xy=0. \quad (1)$

Now rationalise $\sqrt{ax}+\sqrt{by}+\sqrt{cz}=0$: $\sqrt{ax}+\sqrt{by}=-\sqrt{cz}$. Squaring, $ax+by+2\sqrt{abxy}=cz$, so $ax+by-cz=-2\sqrt{abxy}$. Squaring again, $a^2x^2+b^2y^2+c^2z^2+2abxy-2bcyz-2cazx = 4abxy$ $a^2x^2+b^2y^2+c^2z^2-2abxy-2bcyz-2cazx=0,$ which is exactly (1). Hence the two cones are reciprocal. $\blacksquare$

Tangent planes. Write the conicoid as $\dfrac{3}{17}x^2-\dfrac{6}{17}y^2+\dfrac{9}{17}z^2=1$, so $a=\dfrac{3}{17},\ b=-\dfrac{6}{17},\ c=\dfrac{9}{17}$.

A plane $lx+my+nz=p$ touches $ax^2+by^2+cz^2=1$ iff $p^2=\frac{l^2}{a}+\frac{m^2}{b}+\frac{n^2}{c}.$ For planes parallel to $x+4y-2z=0$, take $(l,m,n)=(1,4,-2)$: $p^2 = 17\left(\frac{1}{3}-\frac{16}{6}+\frac{4}{9}\right)=17\left(\frac{3-24+4}{9}\right)=-\frac{289}{9}<0.$

So $p$ is not real: this conicoid has no real tangent plane parallel to $x+4y-2z=0$ (the planes $x+4y-2z=\pm\frac{17}{3}i$ are imaginary), and hence no real points of contact. (Tangent planes would be real only if the constant on the right were large enough relative to the signs of the coefficients; as printed, none exist.)

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OR

Locus of mid-points of parallel chords. Let the chord have mid-point $(\alpha,\beta,\gamma)$ and direction ratios $l,m,n$. The chord of $ax^2+by^2+cz^2=1$ with this mid-point has equation $T=S_1$: $a\alpha x+b\beta y+c\gamma z = a\alpha^2+b\beta^2+c\gamma^2 .$ This chord lies in that plane and has direction $(l,m,n)$, so it is perpendicular to the plane's normal: $a\alpha\,l+b\beta\,m+c\gamma\,n=0.$ Replacing $(\alpha,\beta,\gamma)$ by $(x,y,z)$, the locus is the diametral plane

$alx+bmy+cnz=0.$

Solution (first alternative):

For a reciprocal system, [ \vec{a}\cdot\vec{u}=\vec{b}\cdot\vec{v}=\vec{c}\cdot\vec{w}=1, ] and all mixed dot products such as (\vec{a}\cdot\vec{v},\ \vec{a}\cdot\vec{w},\ \vec{b}\cdot\vec{u}) etc. are zero. Also, [ \vec{u}=\frac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]},\qquad \vec{v}=\frac{\vec{c}\times\vec{a}}{[\vec{a}\ \vec{b}\ \vec{c}]},\qquad \vec{w}=\frac{\vec{a}\times\vec{b}}{[\vec{a}\ \vec{b}\ \vec{c}]}. ]

Now, [ \vec{v}\times\vec{w} =\frac{(\vec{c}\times\vec{a})\times(\vec{a}\times\vec{b})}{[\vec{a}\ \vec{b}\ \vec{c}]^2}. ] Using the vector triple product, [ (\vec{c}\times\vec{a})\times(\vec{a}\times\vec{b}) =((\vec{c}\times\vec{a})\cdot\vec{b})\vec{a} -((\vec{c}\times\vec{a})\cdot\vec{a})\vec{b}. ] But ((\vec{c}\times\vec{a})\cdot\vec{a}=0), and [ (\vec{c}\times\vec{a})\cdot\vec{b}=[\vec{c}\ \vec{a}\ \vec{b}] =[\vec{a}\ \vec{b}\ \vec{c}]. ] Hence, [ (\vec{c}\times\vec{a})\times(\vec{a}\times\vec{b}) =[\vec{a}\ \vec{b}\ \vec{c}]\vec{a}. ] Therefore, [ \vec{v}\times\vec{w} =\frac{\vec{a}}{[\vec{a}\ \vec{b}\ \vec{c}]}. ]

Also, [ [\vec{u}\ \vec{v}\ \vec{w}] =\vec{u}\cdot(\vec{v}\times\vec{w}) =\frac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]} \cdot \frac{\vec{a}}{[\vec{a}\ \vec{b}\ \vec{c}]} =\frac{[\vec{b}\ \vec{c}\ \vec{a}]}{[\vec{a}\ \vec{b}\ \vec{c}]^2} =\frac{1}{[\vec{a}\ \vec{b}\ \vec{c}]}. ]

Thus, [ \frac{\vec{v}\times\vec{w}}{[\vec{u}\ \vec{v}\ \vec{w}]}

\frac{\vec{a}/[\vec{a}\ \vec{b}\ \vec{c}]}{1/[\vec{a}\ \vec{b}\ \vec{c}]} =\vec{a}. ] Hence, [ \boxed{\vec{a}=\frac{\vec{v}\times\vec{w}}{[\vec{u}\ \vec{v}\ \vec{w}]}}. ]

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OR

Let [ \vec{A}=\vec{b}\times\vec{c},\qquad \vec{B}=\vec{c}\times\vec{a},\qquad \vec{C}=\vec{a}\times\vec{b}. ] The vectors (\vec{A},\vec{B},\vec{C}) are coplanar if and only if [ [\vec{A}\ \vec{B}\ \vec{C}]=0. ] Now, [ [\vec{A}\ \vec{B}\ \vec{C}] =(\vec{b}\times\vec{c})\cdot\big((\vec{c}\times\vec{a})\times(\vec{a}\times\vec{b})\big). ] Using the identity proved above, [ (\vec{c}\times\vec{a})\times(\vec{a}\times\vec{b}) =[\vec{a}\ \vec{b}\ \vec{c}]\vec{a}. ] So, [ [\vec{A}\ \vec{B}\ \vec{C}] =(\vec{b}\times\vec{c})\cdot([\vec{a}\ \vec{b}\ \vec{c}]\vec{a}) =[\vec{a}\ \vec{b}\ \vec{c}],(\vec{b}\times\vec{c})\cdot\vec{a} =[\vec{a}\ \vec{b}\ \vec{c}]^2. ] Therefore, [ [\vec{A}\ \vec{B}\ \vec{C}]=0 \iff [\vec{a}\ \vec{b}\ \vec{c}]=0. ] Hence (\vec{b}\times\vec{c},\ \vec{c}\times\vec{a},\ \vec{a}\times\vec{b}) are coplanar if and only if (\vec{a},\vec{b},\vec{c}) are coplanar.

Given, [ \vec r = e^{bt}\vec{c_1} + e^{-bt}\vec{c_2} ] where (\vec{c_1}) and (\vec{c_2}) are constant vectors.

Differentiate with respect to (t): [ \frac{d\vec r}{dt} = \frac{d}{dt}\left(e^{bt}\vec{c_1}\right)

• \frac{d}{dt}\left(e^{-bt}\vec{c_2}\right) ] Since (\vec{c_1}) and (\vec{c_2}) are constants, [ \frac{d\vec r}{dt} = b e^{bt}\vec{c_1} - b e^{-bt}\vec{c_2} ]

Again differentiate with respect to (t): [ \frac{d^2\vec r}{dt^2} = \frac{d}{dt}\left(b e^{bt}\vec{c_1}\right)

• \frac{d}{dt}\left(b e^{-bt}\vec{c_2}\right) ] [ = b^2 e^{bt}\vec{c_1} + b^2 e^{-bt}\vec{c_2} ] [ = b^2\left(e^{bt}\vec{c_1} + e^{-bt}\vec{c_2}\right) ] [ \therefore \frac{d^2\vec r}{dt^2} = b^2 \vec r ]

Hence proved.

Given: Vectors $\vec{a}$ and $\vec{b}$ are irrotational, i.e.

$\nabla \times \vec{a} = 0 \quad \text{and} \quad \nabla \times \vec{b} = 0$

To show: $\vec{a} \times \vec{b}$ is solenoidal, i.e. $\nabla \cdot (\vec{a} \times \vec{b}) = 0$.

Proof: Using the vector identity for the divergence of a cross product,

$\nabla \cdot (\vec{a} \times \vec{b}) = \vec{b} \cdot (\nabla \times \vec{a}) - \vec{a} \cdot (\nabla \times \vec{b})$

Since $\vec{a}$ and $\vec{b}$ are irrotational, $\nabla \times \vec{a} = 0$ and $\nabla \times \vec{b} = 0$. Substituting,

$\nabla \cdot (\vec{a} \times \vec{b}) = \vec{b} \cdot 0 - \vec{a} \cdot 0 = 0$

Result: Since $\nabla \cdot (\vec{a} \times \vec{b}) = 0$, the vector $\vec{a} \times \vec{b}$ is solenoidal. Hence proved.

Let the given circle be the intersection of

[ S: x^2+y^2+z^2=9 ]

and the plane

[ P: x+2y+3z=3. ]

Any sphere passing through this circle is of the form

[ S+\lambda P=0, ]

i.e.

[ x^2+y^2+z^2-9+\lambda(x+2y+3z-3)=0. ]

So the required sphere is

[ x^2+y^2+z^2+\lambda x+2\lambda y+3\lambda z-(9+3\lambda)=0. ]

Its centre is

[ C\left(-\frac{\lambda}{2},-\lambda,-\frac{3\lambda}{2}\right) ]

and radius (r) is given by

[ r^2=\left(\frac{\lambda}{2}\right)^2+\lambda^2+\left(\frac{3\lambda}{2}\right)^2+9+3\lambda. ]

Hence

[ r^2=\frac{7}{2}\lambda^2+3\lambda+9. ]

The sphere touches the plane

[ 4x+3y=15, ]

so the perpendicular distance from the centre to the plane equals the radius.

Distance from (C) to (4x+3y-15=0) is

[ d=\frac{\left|4\left(-\frac{\lambda}{2}\right)+3(-\lambda)-15\right|}{\sqrt{4^2+3^2}} ]

[ =\frac{|-2\lambda-3\lambda-15|}{5} =\frac{|{-5\lambda-15}|}{5} =|\lambda+3|. ]

For tangency,

[ d=r. ]

Therefore,

[ (\lambda+3)^2=\frac{7}{2}\lambda^2+3\lambda+9. ]

Expanding,

[ \lambda^2+6\lambda+9=\frac{7}{2}\lambda^2+3\lambda+9 ]

[ 0=\frac{5}{2}\lambda^2-3\lambda ]

[ \lambda\left(\frac{5}{2}\lambda-3\right)=0. ]

Thus,

[ \lambda=0 \quad \text{or} \quad \lambda=\frac{6}{5}. ]

Case 1: (\lambda=0)

The sphere is

[ x^2+y^2+z^2=9. ]

Case 2: (\lambda=\frac{6}{5})

The sphere is

[ x^2+y^2+z^2+\frac{6}{5}x+\frac{12}{5}y+\frac{18}{5}z-\left(9+\frac{18}{5}\right)=0. ]

Multiplying by (5),

[ 5x^2+5y^2+5z^2+6x+12y+18z-63=0. ]

Hence the required spheres are

[ \boxed{x^2+y^2+z^2=9} ]

and

[ \boxed{5x^2+5y^2+5z^2+6x+12y+18z-63=0}. ]

Reciprocal system of vectors. Let $\vec a,\vec b,\vec c$ be three non-coplanar vectors, so $[\vec a\ \vec b\ \vec c]=\vec a\cdot(\vec b\times\vec c)\neq 0$. The vectors

$\vec a'=\frac{\vec b\times\vec c}{[\vec a\ \vec b\ \vec c]},\qquad \vec b'=\frac{\vec c\times\vec a}{[\vec a\ \vec b\ \vec c]},\qquad \vec c'=\frac{\vec a\times\vec b}{[\vec a\ \vec b\ \vec c]}$

form the reciprocal system to $\vec a,\vec b,\vec c$. (The statement is read with $\vec a',\vec b',\vec c'$ as the reciprocal vectors.)

(i) $\vec a'\cdot\vec b=\vec b'\cdot\vec c=\vec c'\cdot\vec a=0$.

$\vec a'$ is parallel to $\vec b\times\vec c$, which is perpendicular to both $\vec b$ and $\vec c$. Hence $\vec a'\cdot\vec b=\vec a'\cdot\vec c=0$. In the same way $\vec b'\perp\vec c,\vec a$ and $\vec c'\perp\vec a,\vec b$, so $\vec b'\cdot\vec c=\vec c'\cdot\vec a=0$.

(ii) $\vec a'\cdot\vec a=\vec b'\cdot\vec b=\vec c'\cdot\vec c=1$.

$\vec a'\cdot\vec a=\frac{(\vec b\times\vec c)\cdot\vec a}{[\vec a\ \vec b\ \vec c]}=\frac{[\vec a\ \vec b\ \vec c]}{[\vec a\ \vec b\ \vec c]}=1$

Since $[\vec b\ \vec c\ \vec a]=[\vec c\ \vec a\ \vec b]=[\vec a\ \vec b\ \vec c]$, the same calculation gives $\vec b'\cdot\vec b=\dfrac{(\vec c\times\vec a)\cdot\vec b}{[\vec a\ \vec b\ \vec c]}=1$ and $\vec c'\cdot\vec c=\dfrac{(\vec a\times\vec b)\cdot\vec c}{[\vec a\ \vec b\ \vec c]}=1$.

(iii) $[\vec a'\ \vec b'\ \vec c']\,[\vec a\ \vec b\ \vec c]=1$.

Write $V=[\vec a\ \vec b\ \vec c]$. Then $[\vec a'\ \vec b'\ \vec c']=\frac{1}{V^3}(\vec b\times\vec c)\cdot\big[(\vec c\times\vec a)\times(\vec a\times\vec b)\big].$

Using $(\vec p\times\vec q)\times(\vec r\times\vec s)=[\vec p\ \vec q\ \vec s]\,\vec r-[\vec p\ \vec q\ \vec r]\,\vec s$ with $\vec p=\vec c,\vec q=\vec a,\vec r=\vec a,\vec s=\vec b$: $(\vec c\times\vec a)\times(\vec a\times\vec b)=[\vec c\ \vec a\ \vec b]\,\vec a-[\vec c\ \vec a\ \vec a]\,\vec b=V\,\vec a,$ because $[\vec c\ \vec a\ \vec a]=0$. Therefore $[\vec a'\ \vec b'\ \vec c']=\frac{1}{V^3}\,V\,(\vec b\times\vec c)\cdot\vec a=\frac{V^2}{V^3}=\frac1V,$ and so $[\vec a'\ \vec b'\ \vec c']\,[\vec a\ \vec b\ \vec c]=\dfrac1V\cdot V=1$.

The given curve is a circle with centre $(h,-k)$ and radius $r$.

Shifting the origin to $(h-r,\,k)$ without rotating the axes gives the transformation $x=X+(h-r),\qquad y=Y+k.$

Substituting, $(X+h-r-h)^2+(Y+k+k)^2=r^2$

$\boxed{(X-r)^2+(Y+2k)^2=r^2}$

In the new axes the centre is $(r,-2k)$ and the radius is still $r$.

Let $P(h,k)$ be the pole of a chord. The chord (polar of $P$) is $\frac{hx}{a^2}+\frac{ky}{b^2}=1. \qquad (1)$

The normal at the point $(a\cos\theta,\,b\sin\theta)$ of the ellipse is $\frac{ax}{\cos\theta}-\frac{by}{\sin\theta}=a^2-b^2 .$ Dividing by $a^2-b^2$, $\frac{a}{(a^2-b^2)\cos\theta}\,x-\frac{b}{(a^2-b^2)\sin\theta}\,y=1. \qquad (2)$

If the chord is a normal chord, (1) and (2) are the same line, so the coefficients are equal: $\frac{h}{a^2}=\frac{a}{(a^2-b^2)\cos\theta},\qquad \frac{k}{b^2}=\frac{-b}{(a^2-b^2)\sin\theta}.$ Hence $\cos\theta=\frac{a^3}{(a^2-b^2)h},\qquad \sin\theta=\frac{-b^3}{(a^2-b^2)k}.$

Using $\cos^2\theta+\sin^2\theta=1$: $\frac{a^6}{h^2}+\frac{b^6}{k^2}=(a^2-b^2)^2.$

Replacing $(h,k)$ by $(x,y)$, the locus of the poles of normal chords is $\boxed{\frac{a^6}{x^2}+\frac{b^6}{y^2}=(a^2-b^2)^2}$ This is the standard form of the required curve (the printed question abbreviates $a^6,b^6$ as $a^2,b^2$).

The line is the intersection of the planes $x+2y+3z-6=0$ and $3x+4y+5z-2=0$.

Direction ratios. The line is perpendicular to both normals $(1,2,3)$ and $(3,4,5)$, so its direction is their cross product: $\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&2&3\\3&4&5\end{vmatrix}=(10-12)\,\mathbf i-(5-9)\,\mathbf j+(4-6)\,\mathbf k=(-2,\,4,\,-2)\propto(1,-2,1).$

A point on the line. Put $z=0$: $x+2y=6$ and $3x+4y=2$. Then $3(6-2y)+4y=2$ gives $y=8$ and $x=-10$. The point is $(-10,8,0)$. Check: $-10+16-6=0$ and $-30+32-2=0$.

Symmetrical form: $\boxed{\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z}{1}}$

Plane section of a sphere. The curve in which a plane cuts a sphere is a circle, called the plane section of the sphere. Its equation is given by the equations of the sphere and the plane taken together. If the plane passes through the centre the circle is a great circle, its radius equals that of the sphere. Otherwise it is a small circle.

Sphere through the circle and the point $(2,2,3)$. (The question asks for the sphere through the circle $x^2+y^2+z^2=9,\ x+3y+4z=2$ and the point $(2,2,3)$.)

Every sphere through the circle has the form $x^2+y^2+z^2-9+\lambda(x+3y+4z-2)=0.$ It passes through $(2,2,3)$, so $(4+4+9-9)+\lambda(2+6+12-2)=0$, giving $8+18\lambda=0$ and $\lambda=-\dfrac49$.

Therefore $9(x^2+y^2+z^2-9)-4(x+3y+4z-2)=0$, i.e. $\boxed{9x^2+9y^2+9z^2-4x-12y-16z-73=0}$

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OR

Cone. A cone is the surface generated by a straight line (generator) that passes through a fixed point (the vertex) and meets a given curve (the guiding curve).

Make the plane $x-y+2z=7$ homogeneous by writing $\dfrac{x-y+2z}{7}=1$ and substitute in the sphere-type surface $x^2+y^2+2z^2=1$: $x^2+y^2+2z^2=\left(\frac{x-y+2z}{7}\right)^2$ $49(x^2+y^2+2z^2)=x^2+y^2+4z^2-2xy+4xz-4yz$

$\boxed{24x^2+24y^2+47z^2+xy-2xz+2yz=0}$

This is homogeneous of degree 2, so it represents a cone with vertex at the origin.

Let the focus (S) be the pole (O), and let the directrix be the line (x=d), perpendicular to the polar axis.

Let (P(r,\theta)) be any point on the conic. Then

[ SP = r ]

and the perpendicular distance from (P) to the directrix is

[ PM = d - r\cos\theta. ]

By the definition of a conic,

[ SP = e,PM, ]

where (e) is the eccentricity. Hence

[ r = e(d-r\cos\theta). ]

Rearranging,

[ r = ed - er\cos\theta ]

[ r(1+e\cos\theta)=ed. ]

Therefore,

[ r=\frac{ed}{1+e\cos\theta}. ]

Let (l=ed), the semi-latus rectum. Then the polar equation of the conic is

[ \boxed{r=\frac{l}{1+e\cos\theta}}. ]

If the semi-latus rectum is taken as (1), i.e. (l=1), this becomes

[ \boxed{r=\frac{1}{1+e\cos\theta}}. ]

This represents an ellipse if (e<1), a parabola if (e=1), and a hyperbola if (e>1).

Definition (1 mark)

The product of three vectors in which two vectors are crossed first and the result is then crossed with the third vector is called a vector triple product.
It is a vector quantity, e.g. [ \vec{a}\times(\vec{b}\times\vec{c}) \quad\text{or}\quad (\vec{a}\times\vec{b})\times\vec{c}. ] It is not associative in general.

Standard vector triple product formula

[ \vec{x}\times(\vec{y}\times\vec{z}) =(\vec{x}\cdot\vec{z})\vec{y}-(\vec{x}\cdot\vec{y})\vec{z}. ]

This formula can be obtained by expanding in components. Hence,

[ \vec{a}\times(\vec{b}\times\vec{c}) =(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}. ]

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(i) Prove that

[ \vec{a}\times(\vec{c}\times\vec{b}) +\vec{c}\times(\vec{b}\times\vec{a}) +\vec{b}\times(\vec{a}\times\vec{c})=0 ]

Using the vector triple product formula,

[ \vec{a}\times(\vec{c}\times\vec{b}) =(\vec{a}\cdot\vec{b})\vec{c}-(\vec{a}\cdot\vec{c})\vec{b} ]

[ \vec{c}\times(\vec{b}\times\vec{a}) =(\vec{c}\cdot\vec{a})\vec{b}-(\vec{c}\cdot\vec{b})\vec{a} =(\vec{a}\cdot\vec{c})\vec{b}-(\vec{b}\cdot\vec{c})\vec{a} ]

[ \vec{b}\times(\vec{a}\times\vec{c}) =(\vec{b}\cdot\vec{c})\vec{a}-(\vec{b}\cdot\vec{a})\vec{c} =(\vec{b}\cdot\vec{c})\vec{a}-(\vec{a}\cdot\vec{b})\vec{c} ]

Adding the three results,

[ \begin{aligned} &(\vec{a}\cdot\vec{b})\vec{c}-(\vec{a}\cdot\vec{c})\vec{b} +(\vec{a}\cdot\vec{c})\vec{b}-(\vec{b}\cdot\vec{c})\vec{a} +(\vec{b}\cdot\vec{c})\vec{a}-(\vec{a}\cdot\vec{b})\vec{c} \ &=0. \end{aligned} ]

Hence,

[ \boxed{ \vec{a}\times(\vec{c}\times\vec{b}) +\vec{c}\times(\vec{b}\times\vec{a}) +\vec{b}\times(\vec{a}\times\vec{c})=0 } ]

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(ii) Prove that

[ \vec{a}\times(\vec{b}\times\vec{a}) =(\vec{a}\cdot\vec{a})\vec{b}-(\vec{a}\cdot\vec{b})\vec{a} ]

Using the vector triple product formula with
[ \vec{x}=\vec{a},\qquad \vec{y}=\vec{b},\qquad \vec{z}=\vec{a}, ]

we get

[ \vec{a}\times(\vec{b}\times\vec{a}) =(\vec{a}\cdot\vec{a})\vec{b}-(\vec{a}\cdot\vec{b})\vec{a}. ]

Hence,

[ \boxed{ \vec{a}\times(\vec{b}\times\vec{a}) =(\vec{a}\cdot\vec{a})\vec{b}-(\vec{a}\cdot\vec{b})\vec{a} } ]

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OR

Prove that the necessary and sufficient condition for the vector function (\vec{a}(t)) to have constant magnitude is

[ \vec{a}\cdot\frac{d\vec{a}}{dt}=0. ]

Let

[ \vec{a}\cdot\vec{a}=|\vec{a}|^2. ]

If (|\vec{a}|) is constant, then (\vec{a}\cdot\vec{a}) is constant. Differentiating with respect to (t),

[ \frac{d}{dt}(\vec{a}\cdot\vec{a})

\vec{a}\cdot\frac{d\vec{a}}{dt} + \frac{d\vec{a}}{dt}\cdot\vec{a}

2\vec{a}\cdot\frac{d\vec{a}}{dt}. ]

Since (\vec{a}\cdot\vec{a}) is constant,

[ 2\vec{a}\cdot\frac{d\vec{a}}{dt}=0 ]

so

[ \vec{a}\cdot\frac{d\vec{a}}{dt}=0. ]

Thus the condition is necessary.

Conversely, suppose

[ \vec{a}\cdot\frac{d\vec{a}}{dt}=0. ]

Then

[ \frac{d}{dt}(\vec{a}\cdot\vec{a})

2\vec{a}\cdot\frac{d\vec{a}}{dt}=0. ]

Therefore (\vec{a}\cdot\vec{a}) is constant. Hence

[ |\vec{a}|=\sqrt{\vec{a}\cdot\vec{a}} ]

is constant. Thus the condition is sufficient.

Therefore, the necessary and sufficient condition for (\vec{a}(t)) to have constant magnitude is

[ \boxed{\vec{a}\cdot\frac{d\vec{a}}{dt}=0}. ]

Option 1

Constant vector: A vector whose magnitude and direction do not change with respect to the coordinate system is called a constant vector. If
[ \vec{a}=a_1\vec{i}+a_2\vec{j}+a_3\vec{k}, ] then (a_1,a_2,a_3) are constants, and its derivative with respect to any scalar variable is zero.

Given, [ \vec{r}=e^{kt}\vec{a}+e^{-kt}\vec{b}, ] where (\vec{a},\vec{b}) are constant vectors.

Differentiating with respect to (t), [ \frac{d\vec{r}}{dt}=k e^{kt}\vec{a}-k e^{-kt}\vec{b}. ]

Again differentiating, [ \frac{d^2\vec{r}}{dt^2}=k^2 e^{kt}\vec{a}+k^2 e^{-kt}\vec{b}. ] [ \frac{d^2\vec{r}}{dt^2}=k^2\left(e^{kt}\vec{a}+e^{-kt}\vec{b}\right). ] Hence, [ \boxed{\frac{d^2\vec{r}}{dt^2}=k^2\vec{r}} ]

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OR

Scalar triple product: For three vectors (\vec{a},\vec{b},\vec{c}), the scalar triple product is defined as
[ \vec{a}\cdot(\vec{b}\times\vec{c}) ] and is denoted by ([\vec{a}\ \vec{b}\ \vec{c}]). It is a scalar quantity. Its absolute value gives the volume of the parallelepiped whose coterminous edges are (\vec{a},\vec{b},\vec{c}).

Let the concurrent edges be [ \vec{a}=2\vec{i}+3\vec{j}+4\vec{k},\qquad \vec{b}=\vec{i}-\vec{j}+\vec{k},\qquad \vec{c}=\vec{i}+\vec{j}. ]

The volume of the parallelepiped is [ V=\left|\vec{a}\cdot(\vec{b}\times\vec{c})\right|. ]

Now, [ \vec{b}\times\vec{c}

\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\ 1 & -1 & 1\ 1 & 1 & 0 \end{vmatrix} ] [ =\vec{i}[(-1)(0)-1(1)]-\vec{j}[(1)(0)-1(1)]+\vec{k}[(1)(1)-(-1)(1)] ] [ =-\vec{i}+\vec{j}+2\vec{k}. ]

Thus, [ \vec{a}\cdot(\vec{b}\times\vec{c}) =(2\vec{i}+3\vec{j}+4\vec{k})\cdot(-\vec{i}+\vec{j}+2\vec{k}) ] [ =-2+3+8=9. ]

Therefore, [ V=|9|=9. ]

[ \boxed{\text{Volume}=9\text{ cubic units}} ]

Solenoidal vector field:
A vector field (\vec F) is called solenoidal if its divergence is zero everywhere, i.e. [ \nabla\cdot \vec F=\operatorname{div}\vec F=0. ] Physically, a solenoidal field represents an incompressible fluid flow, and the net flux through any closed surface is zero.

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Find a unit vector normal to the surface

[ x^2-y^2+z-2=0 ] at the point ((1,-1,2)).

Let [ f(x,y,z)=x^2-y^2+z-2. ] A normal vector to the surface is given by (\nabla f): [ \nabla f=\frac{\partial f}{\partial x}\vec i+\frac{\partial f}{\partial y}\vec j+\frac{\partial f}{\partial z}\vec k ] [ =2x\vec i-2y\vec j+\vec k. ] At ((1,-1,2)), [ \nabla f=2(1)\vec i-2(-1)\vec j+\vec k =2\vec i+2\vec j+\vec k. ] Its magnitude is [ |\nabla f|=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3. ] Hence a unit vector normal to the surface is [ \boxed{\hat n=\pm \frac{2\vec i+2\vec j+\vec k}{3}} ] or, taking the positive sign, [ \boxed{\hat n=\frac{2}{3}\vec i+\frac{2}{3}\vec j+\frac{1}{3}\vec k}. ]

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OR

Given [ \vec F=(x+y+1)\vec i+\vec j-(x+y)\vec k. ] Let [ P=x+y+1,\qquad Q=1,\qquad R=-(x+y). ] Then [ \operatorname{curl}\vec F=\nabla\times \vec F

\begin{vmatrix} \vec i & \vec j & \vec k\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\ P & Q & R \end{vmatrix}. ] Now, [ \frac{\partial R}{\partial y}=-1,\qquad \frac{\partial Q}{\partial z}=0, ] [ \frac{\partial P}{\partial z}=0,\qquad \frac{\partial R}{\partial x}=-1, ] [ \frac{\partial Q}{\partial x}=0,\qquad \frac{\partial P}{\partial y}=1. ] Therefore, [ \operatorname{curl}\vec F

\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)\vec i + \left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)\vec j + \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\vec k ] [ =(-1-0)\vec i+(0+1)\vec j+(0-1)\vec k ] [ =-\vec i+\vec j-\vec k. ] Now, [ \vec F\cdot \operatorname{curl}\vec F

\big[(x+y+1)\vec i+\vec j-(x+y)\vec k\big]\cdot(-\vec i+\vec j-\vec k) ] [ =-(x+y+1)+1+(x+y) ] [ =-x-y-1+1+x+y=0. ] Hence, [ \boxed{\vec F\cdot \operatorname{curl}\vec F=0}. ] Proved.

The given line is the intersection of the planes
[ 5x+6y-18=0,\qquad 3y-z=0. ] Hence any plane containing this line can be written as [ 5x+6y-18+k(3y-z)=0 ] i.e. [ 5x+(6+3k)y-kz-18=0. \quad (1) ]

For a plane (lx+my+nz+d=0) to touch the ellipsoid [ 5x^2+3y^2+z^2=36, ] the tangency condition is [ d^2=36\left(\frac{l^2}{5}+\frac{m^2}{3}+n^2\right). ]

Comparing with (1), [ l=5,\quad m=6+3k,\quad n=-k,\quad d=-18. ] So, [ (-18)^2=36\left(\frac{25}{5}+\frac{(6+3k)^2}{3}+k^2\right). ] Thus, [ 324=36\left(5+3(k+2)^2+k^2\right). ] Dividing by (36), [ 9=5+3(k+2)^2+k^2. ] Hence, [ 4=3(k+2)^2+k^2 ] [ 4=4k^2+12k+12 ] [ k^2+3k+2=0 ] [ (k+1)(k+2)=0. ] Therefore, [ k=-1,,-2. ]

For (k=-1), the plane is [ 5x+3y+z-18=0. ]

For (k=-2), the plane is [ 5x+2z-18=0. ]

Therefore, the required equations of the two planes are [ \boxed{5x+3y+z=18} ] and [ \boxed{5x+2z=18}. ]

Given conic: [ x^2-4xy+y^2-2x-20y-11=0 ]

Compare with [ ax^2+2hxy+by^2+2gx+2fy+c=0 ] [ a=1,\quad h=-2,\quad b=1,\quad g=-1,\quad f=-10,\quad c=-11 ]

Since [ h^2-ab=(-2)^2-(1)(1)=3>0, ] the conic is a hyperbola, so it has two asymptotes.

Let the centre be ((\alpha,\beta)). Then [ a\alpha+h\beta+g=0,\qquad h\alpha+b\beta+f=0 ] [ \alpha-2\beta-1=0,\qquad -2\alpha+\beta-10=0 ] Solving, [ \alpha=-7,\qquad \beta=-4 ] So the centre is [ (-7,-4) ]

The directions of the asymptotes are given by the quadratic part: [ x^2-4xy+y^2=0 ] Put (y=mx): [ 1-4m+m^2=0 ] [ m^2-4m+1=0 ] [ m=\frac{4\pm\sqrt{16-4}}{2}=2\pm\sqrt3 ]

Hence the asymptotes through ((-7,-4)) are [ y+4=(2+\sqrt3)(x+7) ] and [ y+4=(2-\sqrt3)(x+7) ]

Therefore, [ \boxed{y-(2+\sqrt3)x-(10+7\sqrt3)=0} ] and [ \boxed{y-(2-\sqrt3)x-(10-7\sqrt3)=0} ]

Combined equation of the asymptotes: [ \boxed{x^2-4xy+y^2-2x-20y-47=0} ]