1. Meaning of the scalar product of three vectors [1]

By the scalar product of three vectors we mean the scalar triple product. For three vectors (\vec a,\vec b,\vec c), it 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} ]

---

2. Scalar triple product gives the volume of the parallelepiped [4]

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

The area of the base is

[ A=|\vec b\times\vec c| ]

The vector (\vec b\times\vec c) is perpendicular to the base. Let (\phi) be the angle between (\vec a) and the normal (\vec b\times\vec c). Then the perpendicular height of the parallelepiped is

[ h=|\vec a||\cos\phi| ]

Therefore, the volume (V) of the parallelepiped is

[ V=A h ]

[ V=|\vec b\times\vec c|,|\vec a||\cos\phi| ]

[ V=|\vec a\cdot(\vec b\times\vec c)| ]

Hence,

[ V=|[\vec a\ \vec b\ \vec c]| ]

Thus the scalar triple product gives the volume of the parallelepiped determined by the three vectors. If (\vec a,\vec b,\vec c) form a right-handed system, then

[ [\vec a\ \vec b\ \vec c]>0 ]

and the scalar triple product is exactly the volume. If they form a left-handed system, it is negative, so its absolute value gives the volume.

---

3. Scalar triple product of three coplanar vectors is zero [2]

If (\vec a,\vec b,\vec c) are coplanar, then (\vec a) lies in the plane determined by (\vec b) and (\vec c). But (\vec b\times\vec c) is perpendicular to that plane. Hence (\vec a) is perpendicular to (\vec b\times\vec c).

Therefore,

[ \vec a\cdot(\vec b\times\vec c)=0 ]

So,

[ [\vec a\ \vec b\ \vec c]=0 ]

Alternatively, since the vectors are coplanar, we can write

[ \vec c=\lambda \vec a+\mu \vec b ]

Then

[ [\vec a\ \vec b\ \vec c]

\vec a\cdot{\vec b\times(\lambda \vec a+\mu \vec b)} ]

[ =\lambda \vec a\cdot(\vec b\times\vec a)+\mu \vec a\cdot(\vec b\times\vec b) ]

But

[ \vec b\times\vec a=-(\vec a\times\vec b) ]

and

[ \vec b\times\vec b=0 ]

Hence,

[ [\vec a\ \vec b\ \vec c]=0 ]

Thus, the scalar triple product of three coplanar vectors is always zero. Conversely, if the scalar triple product of three vectors is zero, then the vectors are coplanar or linearly dependent.

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

Shift the origin to $(h, k)$, i.e. put $x = X + h,\ y = Y + k$:

$3(X+h)^2 - 6(X+h) + 4(Y+k)^2 + 14 = 0$

$3X^2 + 4Y^2 + (6h - 6)X + 8kY + (3h^2 - 6h + 4k^2 + 14) = 0$

For the linear terms to vanish, the coefficients of $X$ and $Y$ must be zero:

$6h - 6 = 0 \Rightarrow h = 1, \qquad 8k = 0 \Rightarrow k = 0$

The constant term becomes $3(1)^2 - 6(1) + 0 + 14 = 11$.

Hence, with the origin shifted to $(1, 0)$, the equation becomes

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

First alternative: proof for the ellipse

Let the ellipse be

[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b. ]

Its focus is (S=(ae,0)) and the corresponding directrix is

[ x=\frac{a}{e}, ]

where

[ e^2=1-\frac{b^2}{a^2}. ]

Let a focal chord through (S) meet the ellipse at

[ P=(a\cos\alpha,;b\sin\alpha),\qquad Q=(a\cos\beta,;b\sin\beta). ]

Since (P,S,Q) are collinear,

[ \frac{b\sin\alpha}{a\cos\alpha-ae}

\frac{b\sin\beta}{a\cos\beta-ae}. ]

Simplifying,

[ \sin\alpha(\cos\beta-e)=\sin\beta(\cos\alpha-e), ]

so

[ \sin\alpha\cos\beta-\sin\beta\cos\alpha

e(\sin\alpha-\sin\beta). ]

Hence

[ \sin(\alpha-\beta)=e(\sin\alpha-\sin\beta). \quad (1) ]

Now the tangents at (P) and (Q) are

[ \frac{x\cos\alpha}{a}+\frac{y\sin\alpha}{b}=1, ]

[ \frac{x\cos\beta}{a}+\frac{y\sin\beta}{b}=1. ]

Let their point of intersection be (T=(X,Y)). Solving the two equations by Cramer’s rule gives

[ X=\frac{a(\sin\beta-\sin\alpha)}{\sin(\beta-\alpha)}. ]

But

[ \sin(\beta-\alpha)=-\sin(\alpha-\beta) =e(\sin\beta-\sin\alpha) ]

using (1). Therefore,

[ X=\frac{a(\sin\beta-\sin\alpha)}{e(\sin\beta-\sin\alpha)} =\frac{a}{e}. ]

Thus the point of intersection (T) lies on the line

[ 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.

[ \boxed{\text{Proved.}} ]

---

OR alternative: tangent to the hyperbola

For the hyperbola

[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, ]

the tangent at the point ((\alpha,\beta)) on it is

[ \boxed{\frac{x\alpha}{a^2}-\frac{y\beta}{b^2}=1}. ]

This follows by differentiating the hyperbola equation and using the point-slope form of the tangent.

Let the chord joining the points with vectorial angles
[ \theta=\frac{\pi}{3},\qquad \theta=\frac{\pi}{2} ] be [ \frac{1}{r}=A\cos\theta+B\sin\theta. ]

For the conic [ \frac{1}{r}=1+e\cos\theta, ] at (\theta=\frac{\pi}{3}), [ \frac{1}{r}=1+e\cos\frac{\pi}{3}=1+\frac e2. ] At (\theta=\frac{\pi}{2}), [ \frac{1}{r}=1+e\cos\frac{\pi}{2}=1. ]

Substituting in the chord equation:

At (\theta=\frac{\pi}{2}), [ 1=A(0)+B(1)\implies B=1. ]

At (\theta=\frac{\pi}{3}), [ 1+\frac e2=A\left(\frac12\right)+B\left(\frac{\sqrt3}{2}\right). ] Using (B=1), [ 1+\frac e2=\frac A2+\frac{\sqrt3}{2} ] [ A+\sqrt3=2+e ] [ A=2+e-\sqrt3. ]

Hence the required chord is [ \boxed{\frac{1}{r}=(2+e-\sqrt3)\cos\theta+\sin\theta} ] or equivalently, [ \boxed{r\left[(2+e-\sqrt3)\cos\theta+\sin\theta\right]=1}. ]

In Cartesian form, since (x=r\cos\theta,\ y=r\sin\theta), the chord is [ \boxed{(2+e-\sqrt3)x+y=1}. ]

Let the required plane pass through the intersection of
[ 2x+y+z=5,\qquad 3x-2y+4z=5. ] So its equation can be taken as [ (2x+y+z-5)+\lambda(3x-2y+4z-5)=0. ] 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, [ \vec n\cdot(1,1,5)=0. ] [ (2+3\lambda)+(1-2\lambda)+5(1+4\lambda)=0 ] [ 8+21\lambda=0 ] [ \lambda=-\frac{8}{21}. ]

Thus the required plane is [ (2x+y+z-5)-\frac{8}{21}(3x-2y+4z-5)=0. ] Multiplying by (21): [ 21(2x+y+z-5)-8(3x-2y+4z-5)=0 ] [ 42x+21y+21z-105-24x+16y-32z+40=0 ] [ 18x+37y-11z-65=0. ]

[ \boxed{18x+37y-11z=65} ]

This is the required equation of the plane.

Let the two lines be

$$L_1:\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$$

so a point on (L_1) is (A=(0,0,0)) and direction vector

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

Also,

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

so a point on (L_2) is (B=(2,1,-2)) and direction vector

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

The shortest distance between skew lines is

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

Now,

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

Also,

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

So,

$$|\vec d_1\times \vec d_2|=\sqrt{(-1)^2+(-1)^2+(-1)^2}=\sqrt3.$$

And,

$$(2,1,-2)\cdot(-1,-1,-1) =-2-1+2=-1.$$

Hence,

$$d=\frac{|-1|}{\sqrt3} =\frac1{\sqrt3} =\frac{\sqrt3}{3}.$$

Therefore, the shortest distance between the given lines is

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

Choosing the first alternative:

The axis is the line
[ x=2y=-z. ] Let the common value be (t). Then [ x=t,\qquad y=\frac t2,\qquad z=-t. ] So the direction ratios of the axis are [ (2,1,-2). ] The axis passes through the origin.

For any point (P(x,y,z)), the distance from the axis is (4). Hence [ d^2=4^2=16. ] Using the formula for perpendicular distance from a point to a line through the origin, [ d^2=x^2+y^2+z^2-\frac{(2x+y-2z)^2}{2^2+1^2+(-2)^2}. ] Since [ 2^2+1^2+(-2)^2=9, ] we get [ x^2+y^2+z^2-\frac{(2x+y-2z)^2}{9}=16. ] Multiplying by (9), [ 9(x^2+y^2+z^2)-(2x+y-2z)^2=144. ] Now, [ (2x+y-2z)^2=4x^2+y^2+4z^2+4xy-8xz-4yz. ] Therefore, [ 9x^2+9y^2+9z^2-\left(4x^2+y^2+4z^2+4xy-8xz-4yz\right)=144. ] Simplifying, [ \boxed{5x^2+8y^2+5z^2-4xy+8xz+4yz=144}. ]

So the required equation of the right circular cylinder is
[ \boxed{5x^2+8y^2+5z^2-4xy+8xz+4yz=144}. ]

Tangent planes parallel to $x + 4y - 2z = 0$

Let the conicoid be $3x^2 - 6y^2 + 9z^2 = 17$. A plane $lx + my + nz = p$ touches $ax^2+by^2+cz^2=1$ if $\dfrac{l^2}{a}+\dfrac{m^2}{b}+\dfrac{n^2}{c}=p^2$. Here $a=\dfrac{3}{17},\ b=-\dfrac{6}{17},\ c=\dfrac{9}{17}$.

A plane parallel to the given plane is $x + 4y - 2z = p$, with $l=1,\ m=4,\ n=-2$. The condition of tangency gives

$p^2 = \frac{1}{3/17} + \frac{16}{-6/17} + \frac{4}{9/17} = \frac{17}{3} - \frac{136}{3} + \frac{68}{9} = \frac{51 - 408 + 68}{9} = -\frac{289}{9}$

Since $p^2<0$, no real plane qualifies. To find the points of contact, use the tangent plane at $(x_1,y_1,z_1)$: $3x_1x - 6y_1y + 9z_1z = 17$, parallel to $x+4y-2z=p$, so

$\frac{3x_1}{1} = \frac{-6y_1}{4} = \frac{9z_1}{-2} = \lambda \Rightarrow x_1 = \frac{\lambda}{3},\quad y_1 = -\frac{2\lambda}{3},\quad z_1 = -\frac{2\lambda}{9}$

The point lies on the conicoid:

$3\cdot\frac{\lambda^2}{9} - 6\cdot\frac{4\lambda^2}{9} + 9\cdot\frac{4\lambda^2}{81} = 17$

$\frac{\lambda^2}{3} - \frac{8\lambda^2}{3} + \frac{4\lambda^2}{9} = 17 \Rightarrow \frac{3\lambda^2 - 24\lambda^2 + 4\lambda^2}{9} = 17 \Rightarrow -\frac{17\lambda^2}{9}=17$

so $\lambda^2 = -9$, which has no real solution, agreeing with $p^2<0$.

Conclusion: for this conicoid, with the given coefficients, no real tangent plane is parallel to $x+4y-2z=0$; the planes are imaginary, $x+4y-2z=\pm\dfrac{17i}{3}$, touching at the imaginary points $\left(\pm i,\ \mp 2i,\ \mp\dfrac{2i}{3}\right)$.

OR: Locus of mid-points of chords parallel to $\dfrac{x}{l}=\dfrac{y}{m}=\dfrac{z}{n}$

Let $(\alpha,\beta,\gamma)$ be the mid-point of a chord of $ax^2+by^2+cz^2=1$ with direction ratios $l,m,n$. The chord is

$x = \alpha + lr,\quad y = \beta + mr,\quad z = \gamma + nr.$

Substituting in the conicoid:

$(al^2+bm^2+cn^2)r^2 + 2(a\alpha l+b\beta m+c\gamma n)r + (a\alpha^2+b\beta^2+c\gamma^2-1)=0$

Since $(\alpha,\beta,\gamma)$ is the mid-point, the two roots of $r$ are equal and opposite, so the coefficient of $r$ is zero:

$a\alpha l + b\beta m + c\gamma n = 0$

Hence the locus of $(\alpha,\beta,\gamma)$ is

$\boxed{alx + bmy + cnz = 0}$

which is a plane through the centre (a diametral plane) of the conicoid.

First part

Let
[ \Delta=[\vec a\ \vec b\ \vec c]=\vec a\cdot(\vec b\times\vec c)\neq 0. ] Since (\vec a,\vec b,\vec c) and (\vec u,\vec v,\vec w) are reciprocal systems, [ \vec u=\frac{\vec b\times\vec c}{\Delta},\qquad \vec v=\frac{\vec c\times\vec a}{\Delta},\qquad \vec w=\frac{\vec a\times\vec b}{\Delta}. ]

Now, [ \vec v\times\vec w =\frac{1}{\Delta^2}\left[(\vec c\times\vec a)\times(\vec a\times\vec b)\right]. ] Using the identity [ (\vec x\times\vec y)\times(\vec z\times\vec w) =[\vec x\ \vec z\ \vec w]\vec y-[\vec y\ \vec z\ \vec w]\vec x, ] we get [ (\vec c\times\vec a)\times(\vec a\times\vec b) =[\vec c\ \vec a\ \vec b]\vec a-[\vec a\ \vec a\ \vec b]\vec c =\Delta \vec a. ] Hence [ \vec v\times\vec w=\frac{\Delta \vec a}{\Delta^2}=\frac{\vec a}{\Delta}. ]

Also, [ [\vec u\ \vec v\ \vec w] =\vec u\cdot(\vec v\times\vec w) =\frac{\vec b\times\vec c}{\Delta}\cdot\frac{\vec a}{\Delta} =\frac{(\vec b\times\vec c)\cdot\vec a}{\Delta^2} =\frac{\Delta}{\Delta^2} =\frac1\Delta. ] Therefore, [ \vec v\times\vec w=\frac{\vec a}{\Delta} =\vec a,[\vec u\ \vec v\ \vec w]. ] So, [ \boxed{\vec a=\frac{\vec v\times\vec w}{[\vec u\ \vec v\ \vec w]}}. ]

---

OR

Let [ \vec X=\vec b\times\vec c,\qquad \vec Y=\vec c\times\vec a,\qquad \vec Z=\vec a\times\vec b. ] The vectors (\vec X,\vec Y,\vec Z) are coplanar iff [ [\vec X\ \vec Y\ \vec Z]=0. ]

Now, [ [\vec X\ \vec Y\ \vec Z] =\vec X\cdot(\vec Y\times\vec Z) =(\vec b\times\vec c)\cdot\left[(\vec c\times\vec a)\times(\vec a\times\vec b)\right]. ] Using the same vector identity, [ (\vec c\times\vec a)\times(\vec a\times\vec b) =[\vec c\ \vec a\ \vec b]\vec a =\Delta \vec a. ] Therefore, [ [\vec X\ \vec Y\ \vec Z] =(\vec b\times\vec c)\cdot(\Delta \vec a) =\Delta[(\vec b\times\vec c)\cdot\vec a] =\Delta\cdot\Delta =\Delta^2. ] Thus, [ [\vec b\times\vec c,\ \vec c\times\vec a,\ \vec a\times\vec b] =[\vec a\ \vec b\ \vec c]^2. ]

Hence, [ [\vec b\times\vec c,\ \vec c\times\vec a,\ \vec a\times\vec b]=0 \iff [\vec a\ \vec b\ \vec c]^2=0 \iff [\vec a\ \vec b\ \vec c]=0. ] Therefore, (\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.
[ \boxed{\text{Proved.}} ]

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 (\vec r) 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,\vec c_2) are constant,

[ \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} = b^2 e^{bt}\vec c_1 + b^2 e^{-bt}\vec c_2 ]

Taking (b^2) common,

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

But
[ e^{bt}\vec c_1+e^{-bt}\vec c_2=\vec r ]

Therefore,

[ \boxed{\frac{d^2\vec r}{dt^2}=b^2\vec r} ]

Hence proved.

Given: (\vec{a}) and (\vec{b}) are irrotational, i.e.

$\nabla \times \vec{a} = 0, \qquad \nabla \times \vec{b} = 0$

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

Proof: Using the vector identity

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

Substituting the given conditions (\nabla \times \vec{a} = 0) and (\nabla \times \vec{b} = 0),

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

Since the divergence of (\vec{a} \times \vec{b}) is zero, the vector (\vec{a} \times \vec{b}) is solenoidal. (\blacksquare)

Taking the given circle as the intersection of
[ x^2+y^2+z^2=9,\qquad x+2y+3z=3, ] any sphere passing through this circle is of the form [ x^2+y^2+z^2-9+\lambda(x+2y+3z-3)=0. ] So [ 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). ] Its radius squared is [ r^2=\left(\frac{\lambda}{2}\right)^2+\lambda^2+\left(\frac{3\lambda}{2}\right)^2+9+3\lambda =\frac72\lambda^2+3\lambda+9. ]

The given plane is [ 4x+3y=15 \quad\Rightarrow\quad 4x+3y-15=0. ] Distance from the centre to this plane is [ d=\frac{\left|4\left(-\frac{\lambda}{2}\right)+3(-\lambda)-15\right|}{\sqrt{4^2+3^2}} =\frac{|-5\lambda-15|}{5}=|\lambda+3|. ]

Since the sphere touches the plane, [ d=r. ] Hence [ (\lambda+3)^2=\frac72\lambda^2+3\lambda+9. ] [ \lambda^2+6\lambda+9=\frac72\lambda^2+3\lambda+9 ] [ 0=\frac52\lambda^2-3\lambda =\lambda\left(\frac52\lambda-3\right). ] Therefore, [ \lambda=0 \quad\text{or}\quad \lambda=\frac65. ]

For (\lambda=0), the sphere is [ \boxed{x^2+y^2+z^2=9}. ]

For (\lambda=\frac65), [ x^2+y^2+z^2+\frac65x+\frac{12}{5}y+\frac{18}{5}z-\frac{63}{5}=0, ] or multiplying by (5), [ \boxed{5(x^2+y^2+z^2)+6x+12y+18z-63=0}. ]

Therefore, the required spheres are [ \boxed{x^2+y^2+z^2=9} ] and [ \boxed{5(x^2+y^2+z^2)+6x+12y+18z-63=0}. ]

Reciprocal system of vectors

Let $\vec{a},\ \vec{b},\ \vec{c}$ be three non-coplanar (non-coplanar means $[\vec{a}\,\vec{b}\,\vec{c}]\neq 0$) vectors. The reciprocal system of vectors is the set of three vectors $\vec{a}'$, $\vec{b}'$, $\vec{c}'$ defined by

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

where $[\vec{a}\,\vec{b}\,\vec{c}]=\vec{a}\cdot(\vec{b}\times\vec{c})$ is the scalar triple product. The reciprocal vectors are also written as $\vec{a}'=\vec{a}^{-1}$, etc. They satisfy the following properties.

---

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

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

The scalar triple product $(\vec{b}\times\vec{c})\cdot\vec{b}=[\vec{b}\,\vec{c}\,\vec{b}]=0$, since two of its vectors are equal. Hence

$$\vec{a}'\cdot\vec{b}=0$$

Similarly,

$$\vec{b}'\cdot\vec{c}=\frac{(\vec{c}\times\vec{a})\cdot\vec{c}}{[\vec{a}\,\vec{b}\,\vec{c}]}=\frac{[\vec{c}\,\vec{a}\,\vec{c}]}{[\vec{a}\,\vec{b}\,\vec{c}]}=0$$ $$\vec{c}'\cdot\vec{a}=\frac{(\vec{a}\times\vec{b})\cdot\vec{a}}{[\vec{a}\,\vec{b}\,\vec{c}]}=\frac{[\vec{a}\,\vec{b}\,\vec{a}]}{[\vec{a}\,\vec{b}\,\vec{c}]}=0$$

Thus $\vec{a}'\cdot\vec{b}=\vec{b}'\cdot\vec{c}=\vec{c}'\cdot\vec{a}=0$. $\blacksquare$

---

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

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

Similarly,

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

Thus $\vec{a}'\cdot\vec{a}=\vec{b}'\cdot\vec{b}=\vec{c}'\cdot\vec{c}=1$. $\blacksquare$

---

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

$$[\vec{a}'\,\vec{b}'\,\vec{c}'] =\vec{a}'\cdot(\vec{b}'\times\vec{c}') =\frac{\vec{b}\times\vec{c}}{[\vec{a}\,\vec{b}\,\vec{c}]}\cdot\left(\frac{\vec{c}\times\vec{a}}{[\vec{a}\,\vec{b}\,\vec{c}]}\times\frac{\vec{a}\times\vec{b}}{[\vec{a}\,\vec{b}\,\vec{c}]}\right)$$ $$=\frac{1}{[\vec{a}\,\vec{b}\,\vec{c}]^{3}}\ (\vec{b}\times\vec{c})\cdot\big[(\vec{c}\times\vec{a})\times(\vec{a}\times\vec{b})\big]$$

Using the vector triple product identity $\vec{p}\times(\vec{q}\times\vec{r})=(\vec{p}\cdot\vec{r})\vec{q}-(\vec{p}\cdot\vec{q})\vec{r}$ with $\vec{p}=\vec{c}\times\vec{a}$, $\vec{q}=\vec{a}$, $\vec{r}=\vec{b}$:

$$(\vec{c}\times\vec{a})\times(\vec{a}\times\vec{b}) =(\vec{c}\times\vec{a})\cdot\vec{b}\ \vec{a}-(\vec{c}\times\vec{a})\cdot\vec{a}\ \vec{b} =[\vec{a}\,\vec{b}\,\vec{c}]\,\vec{a}-0 =[\vec{a}\,\vec{b}\,\vec{c}]\,\vec{a}$$

Therefore,

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

Hence

$$[\vec{a}'\,\vec{b}'\,\vec{c}'][\vec{a}\,\vec{b}\,\vec{c}]=1 \qquad \blacksquare$$

Given: $(x-h)^2+(y+k)^2=r^2$, a circle with centre $(h,-k)$ and radius $r$.

Shifting the origin (axes kept parallel). For the new origin $O'(h-r,\,k)$ the transformation is

$x = X + (h-r), \qquad y = Y + k$

where $(X,Y)$ are the new coordinates.

Substituting:

$x-h = X + h - r - h = X - r, \qquad y+k = Y + k + k = Y + 2k$

$\therefore\ (X-r)^2 + (Y+2k)^2 = r^2$

Result: the equation becomes $(X-r)^2+(Y+2k)^2=r^2$, which is $X^2+Y^2-2rX+4kY+4k^2=0$ on expanding.

Check: the old centre $(h,-k)$ has new coordinates $X=h-(h-r)=r$ and $Y=-k-k=-2k$, i.e. centre $(r,-2k)$ with the same radius $r$, as the new equation shows.

Setting up. Let the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and let $P(h,k)$ be the pole of a chord. The chord is the polar of $P$:

$\frac{hx}{a^2}+\frac{ky}{b^2}=1 \qquad (1)$

Normal at a point. The normal at $\alpha=(a\cos\alpha,\,b\sin\alpha)$ is

$\frac{ax}{\cos\alpha}-\frac{by}{\sin\alpha}=a^2-b^2$

Dividing by $a^2-b^2$ to put it in the form "$\ldots=1$":

$\frac{a\sec\alpha}{a^2-b^2}\,x-\frac{b\,\csc\alpha}{a^2-b^2}\,y=1 \qquad (2)$

Condition for a normal chord. The chord (1) is a normal chord when it coincides with the normal (2) at some point $\alpha$. Comparing coefficients:

$\frac{h}{a^2}=\frac{a\sec\alpha}{a^2-b^2},\qquad \frac{k}{b^2}=-\frac{b\csc\alpha}{a^2-b^2}$

$\Rightarrow\ \cos\alpha=\frac{a^3}{h\,(a^2-b^2)},\qquad \sin\alpha=-\frac{b^3}{k\,(a^2-b^2)}$

Eliminating $\alpha$ using $\cos^2\alpha+\sin^2\alpha=1$:

$\frac{a^6}{h^2(a^2-b^2)^2}+\frac{b^6}{k^2(a^2-b^2)^2}=1$

$\frac{a^6}{h^2}+\frac{b^6}{k^2}=(a^2-b^2)^2$

Locus. Replacing $(h,k)$ by $(x,y)$, the locus of the poles of normal chords is

$\frac{a^6}{x^2}+\frac{b^6}{y^2}=(a^2-b^2)^2$

This is the standard form of the required curve; the printed statement of the question writes it with $a^2,b^2$ in the numerators, but the derivation above gives the powers $a^6,b^6$.

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

Direction ratios. The direction is perpendicular to both normals $(1,2,3)$ and $(3,4,5)$:

$\vec d=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&3\\3&4&5\end{vmatrix}=(10-12,\;9-5,\;4-6)=(-2,\,4,\,-2)\propto(1,-2,1)$

A point on the line. Put $z=0$:

$x+2y=6,\qquad 3x+4y=2$

Substituting $x=6-2y$ in the second gives $18-2y=2$, so $y=8$ and $x=-10$. The point is $(-10,\,8,\,0)$.

Symmetrical form:

$\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z-0}{1}$

Check: $(1,-2,1)\cdot(1,2,3)=1-4+3=0$ and $(1,-2,1)\cdot(3,4,5)=3-8+5=0$.

First alternative

Plane section of a sphere: The curve obtained by intersecting a sphere with a plane is a circle. If the plane passes through the centre of the sphere, it is a great circle; otherwise it is a small circle.

The given circle is the intersection of
[ x^2+y^2+z^2=9,\qquad x+3y+4z=2. ] Let the required sphere through this circle be [ 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 ] [ 8+18\lambda=0 \Rightarrow \lambda=-\frac49. ] Hence the required sphere is [ x^2+y^2+z^2-9-\frac49(x+3y+4z-2)=0 ] or [ \boxed{9x^2+9y^2+9z^2-4x-12y-16z-73=0.} ]

---

OR alternative

Cone: A cone is the surface generated by a straight line (generator) passing through a fixed point (vertex) and intersecting a given curve (guiding curve).

Let the vertex be the origin. A point ((x,y,z)) lies on the cone if the line through the origin and ((x,y,z)) meets the guiding curve [ x^2+y^2+2z^2=1,\qquad x-y+2z=7. ] Put ((X,Y,Z)=t(x,y,z)). Then [ t^2(x^2+y^2+2z^2)=1,\qquad t(x-y+2z)=7. ] Eliminating (t), [ 49(x^2+y^2+2z^2)=(x-y+2z)^2. ] Therefore the required cone is [ \boxed{49(x^2+y^2+2z^2)-(x-y+2z)^2=0} ] or [ \boxed{24x^2+24y^2+47z^2+xy-2xz+2yz=0.} ]

Let the focus (S) be the pole (O), and let the directrix be the line
[ x=d \qquad (d>0) ] perpendicular to the polar axis (OX).

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 focus-directrix definition of a conic,

[ SP=e,PM ]

so

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

Rearranging,

[ r=ed-er\cos\theta ]

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

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

Now choose the directrix so that (ed=1), i.e. (d=\frac1e). Then

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

This is the required polar equation of a conic, where (e) is the eccentricity.

• If (e<1), it represents an ellipse.
• If (e=1), it represents a parabola.
• If (e>1), it represents a hyperbola.

Vector triple product

If (\vec{a},\vec{b},\vec{c}) are three vectors, then an expression of the form
[ \vec{a}\times(\vec{b}\times\vec{c}) ] is called the vector triple product of (\vec{a},\vec{b},\vec{c}). It is a vector quantity. The standard identity is [ \vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}. ]

---

(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 identity, [ \vec{x}\times(\vec{y}\times\vec{z})=(\vec{x}\cdot\vec{z})\vec{y}-(\vec{x}\cdot\vec{y})\vec{z}, ] we get

[ \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 these 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} ]

---

(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 identity, [ \vec{x}\times(\vec{y}\times\vec{z})=(\vec{x}\cdot\vec{z})\vec{y}-(\vec{x}\cdot\vec{y})\vec{z}, ] put [ \vec{x}=\vec{a},\qquad \vec{y}=\vec{b},\qquad \vec{z}=\vec{a}. ] Then [ \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}} ]

---

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}=\vec{a}(t). ] Then [ |\vec{a}|^2=\vec{a}\cdot\vec{a}. ]

Necessary condition

Suppose (|\vec{a}|) is constant. Then (|\vec{a}|^2=\vec{a}\cdot\vec{a}) is also constant. Differentiating with respect to (t), [ \frac{d}{dt}(\vec{a}\cdot\vec{a})=0. ] But [ \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}. ] Therefore, [ 2\vec{a}\cdot\frac{d\vec{a}}{dt}=0 ] so that [ \vec{a}\cdot\frac{d\vec{a}}{dt}=0. ]

Sufficient condition

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. ] Hence (\vec{a}\cdot\vec{a}=|\vec{a}|^2) is constant. Therefore (|\vec{a}|) is constant.

Thus, the necessary and sufficient condition for (\vec{a}(t)) to have constant magnitude is [ \boxed{\vec{a}\cdot\frac{d\vec{a}}{dt}=0}. ]

Solution (First alternative):

Constant vector: A vector whose magnitude and direction remain unchanged as the variable changes is called a constant vector. If
[ \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}, ] then (\vec{a}) is constant when (a_1,a_2,a_3) are constants. Hence [ \frac{d\vec{a}}{dt}=\vec{0}. ]

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

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

Differentiate again: [ \frac{d^2\vec{r}}{dt^2}=k^2 e^{kt}\vec{a}+k^2 e^{-kt}\vec{b} ] [ =k^2\left(e^{kt}\vec{a}+e^{-kt}\vec{b}\right) ] [ =k^2\vec{r}. ]

Hence, [ \boxed{\frac{d^2\vec{r}}{dt^2}=k^2\vec{r}} ]

---

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}). ] It is a scalar quantity. Its absolute value gives the volume of the parallelepiped whose concurrent edges are represented by (\vec{a},\vec{b},\vec{c}).

Given: [ \vec{a}=2\hat{i}+3\hat{j}+4\hat{k},\qquad \vec{b}=\hat{i}-\hat{j}+\hat{k},\qquad \vec{c}=\hat{i}+\hat{j}. ]

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

First compute (\vec{b}\times\vec{c}): [ \vec{b}\times\vec{c}

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

Now, [ \vec{a}\cdot(\vec{b}\times\vec{c})

(2\hat{i}+3\hat{j}+4\hat{k})\cdot(-\hat{i}+\hat{j}+2\hat{k}) ] [ =2(-1)+3(1)+4(2) ] [ =-2+3+8=9. ]

Therefore, [ V=|9|=9\text{ cubic units}. ]

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

Solenoidal vector field:
A vector field $\vec{F}$ is said to be solenoidal if its divergence is zero everywhere, i.e. [ \nabla \cdot \vec{F}=0. ] Physically, a solenoidal field has no source or sink and represents an incompressible fluid flow.

---

Find a unit vector normal to the surface

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

Let [ \phi=x^2-y^2+z-2. ] The normal vector to the surface is given by $\nabla \phi$: [ \nabla \phi=\frac{\partial \phi}{\partial x}\vec{i}+\frac{\partial \phi}{\partial y}\vec{j}+\frac{\partial \phi}{\partial z}\vec{k} ] [ =2x\vec{i}-2y\vec{j}+\vec{k}. ] At the point $(1,-1,2)$, [ \nabla \phi=2(1)\vec{i}-2(-1)\vec{j}+\vec{k} ] [ =2\vec{i}+2\vec{j}+\vec{k}. ] Its magnitude is [ |\nabla \phi|=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3. ] Hence the required unit normal vector is [ \hat{n}=\frac{\nabla \phi}{|\nabla \phi|} =\frac{2\vec{i}+2\vec{j}+\vec{k}}{3} ] [ \boxed{\hat{n}=\frac{2}{3}\vec{i}+\frac{2}{3}\vec{j}+\frac{1}{3}\vec{k}} ]

The opposite direction $-\hat{n}$ is also a unit normal.

---

OR

If [ \vec{F}=(x+y+1)\vec{i}+\vec{j}-(x+y)\vec{k}, ] show that [ \vec{F}\cdot \operatorname{curl}\vec{F}=0. ]

Let [ \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}, ] where [ P=x+y+1,\qquad Q=1,\qquad R=-(x+y). ]

Now, [ \operatorname{curl}\vec{F}=\nabla\times\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}. ]

Compute each term: [ \frac{\partial R}{\partial y}=-1,\qquad \frac{\partial Q}{\partial z}=0 ] so the $\vec{i}$-component is $-1$.

[ \frac{\partial P}{\partial z}=0,\qquad \frac{\partial R}{\partial x}=-1 ] so the $\vec{j}$-component is [ 0-(-1)=1. ]

[ \frac{\partial Q}{\partial x}=0,\qquad \frac{\partial P}{\partial y}=1 ] so the $\vec{k}$-component is [ 0-1=-1. ]

Therefore, [ \operatorname{curl}\vec{F}=-\vec{i}+\vec{j}-\vec{k}. ]

Now, [ \vec{F}\cdot \operatorname{curl}\vec{F} =(x+y+1)(-1)+(1)(1)+-(x+y) ] [ =-(x+y+1)+1+(x+y) ] [ =-x-y-1+1+x+y=0. ]

Hence, [ \boxed{\vec{F}\cdot \operatorname{curl}\vec{F}=0} ] proved.

Any plane containing the line
[ 5x+6y-18=0,\qquad 3y-z=0 ] is of the form [ (5x+6y-18)+t(3y-z)=0. ] So its equation is [ 5x+(6+3t)y-tz-18=0. ] Thus the normal vector is [ \vec n=(5,;6+3t,;-t). ]

For the ellipsoid [ 5x^2+3y^2+z^2=36, ] the condition for the plane (\vec n\cdot \vec r=18) to be tangent is [ 18^2=36\left(\frac{5^2}{5}+\frac{(6+3t)^2}{3}+t^2\right). ] Hence [ 324=36\left[5+3(t+2)^2+t^2\right]. ] Dividing by (36), [ 9=5+3(t+2)^2+t^2 ] [ 4=3(t^2+4t+4)+t^2 ] [ 4=4t^2+12t+12 ] [ t^2+3t+2=0 ] [ (t+1)(t+2)=0. ] So [ t=-1 \quad \text{or} \quad t=-2. ]

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

For the conic [ x^2-4xy+y^2-2x-20y-11=0 ] compare with [ ax^2+2hxy+by^2+2gx+2fy+c=0. ] Here, [ a=1,\quad h=-2,\quad b=1,\quad g=-1,\quad f=-10,\quad c=-11. ]

The centre ((x_0,y_0)) is found from [ ax_0+hy_0+g=0,\qquad hx_0+by_0+f=0. ] So, [ x_0-2y_0-1=0, ] [ -2x_0+y_0-10=0. ] Solving, [ x_0=-7,\qquad y_0=-4. ] Thus the centre is ((-7,-4)).

The asymptotes have the same quadratic part and pass through the centre. Hence their slopes (m) satisfy [ m^2-4m+1=0. ] [ m=\frac{4\pm\sqrt{16-4}}{2} =\frac{4\pm2\sqrt3}{2} =2\pm\sqrt3. ]

Therefore the asymptotes are [ \boxed{y+4=(2+\sqrt3)(x+7)} ] and [ \boxed{y+4=(2-\sqrt3)(x+7)}. ]

Their combined equation is [ \boxed{x^2-4xy+y^2-2x-20y-47=0}. ]