BE Computer Engineering (IOE, TU) Engineering Mathematics II (IOE, SH 451) Question Paper 2079 Nepal

(a) Rank of a matrix and echelon form

Definition. The rank of a matrix is the number of non-zero rows in its row-echelon form, equivalently the order of the largest non-vanishing minor. It equals the maximum number of linearly independent rows (or columns).

Reduce

$$A = \begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & 1 & -2 \\ 3 & 6 & 2 & -1 \end{bmatrix}.$$

$R_2 \to R_2 - 2R_1,\; R_3 \to R_3 - 3R_1$:

$$\begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 3 & -8 \\ 0 & 0 & 5 & -10 \end{bmatrix}.$$

$R_3 \to R_3 - \tfrac{5}{3}R_2$:

$$\begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 0 & 3 & -8 \\ 0 & 0 & 0 & \tfrac{10}{3} \end{bmatrix}.$$

There are 3 non-zero rows, so $\boxed{\operatorname{rank}(A) = 3}$.

(b) Consistency of the system

$$x+y+z=6,\quad x+2y+3z=10,\quad x+2y+\lambda z=\mu.$$

Form the augmented matrix and eliminate:

$$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 10 \\ 1 & 2 & \lambda & \mu \end{array}\right] \xrightarrow{R_2-R_1,\,R_3-R_1} \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 4 \\ 0 & 1 & \lambda-1 & \mu-6 \end{array}\right]$$ $$\xrightarrow{R_3-R_2} \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & \lambda-3 & \mu-10 \end{array}\right].$$

The last row gives $(\lambda-3)z = \mu-10$.

• (i) No solution: $\lambda - 3 = 0$ and $\mu - 10 \ne 0$, i.e. $\lambda = 3,\ \mu \ne 10$ (rank of coefficient matrix $=2$, augmented $=3$).
• (ii) Unique solution: $\lambda - 3 \ne 0$, i.e. $\lambda \ne 3$ (any $\mu$); then rank $=3=$ number of unknowns.
• (iii) Infinitely many solutions: $\lambda = 3$ and $\mu = 10$ (rank $=2 <3$).

Solving the infinite case ($\lambda=3,\mu=10$). Let $z = t$ (free). From $R_2$: $y + 2t = 4 \Rightarrow y = 4 - 2t$. From $R_1$: $x + y + z = 6 \Rightarrow x = 6 - (4-2t) - t = 2 + t$.

$$\boxed{x = 2 + t,\quad y = 4 - 2t,\quad z = t,\quad t \in \mathbb{R}.}$$

(a) Evaluate $\displaystyle\int_0^1 \int_{x^2}^{x} (x^2+y^2)\,dy\,dx$

Region: $0\le x\le 1$, with $y$ from the parabola $y=x^2$ up to the line $y=x$. The two curves meet at $(0,0)$ and $(1,1)$; the region is the lens between line and parabola in the first quadrant.

Inner integral:

$$\int_{x^2}^{x}(x^2+y^2)\,dy = \Big[x^2 y + \tfrac{y^3}{3}\Big]_{x^2}^{x} = \Big(x^3 + \tfrac{x^3}{3}\Big) - \Big(x^4 + \tfrac{x^6}{3}\Big) = \tfrac{4}{3}x^3 - x^4 - \tfrac{1}{3}x^6.$$

Outer integral:

$$\int_0^1\Big(\tfrac{4}{3}x^3 - x^4 - \tfrac{1}{3}x^6\Big)dx = \tfrac{4}{3}\cdot\tfrac14 - \tfrac15 - \tfrac13\cdot\tfrac17 = \tfrac13 - \tfrac15 - \tfrac{1}{21}.$$

LCD $=105$: $\tfrac{35 - 21 - 5}{105} = \dfrac{9}{105} = \boxed{\dfrac{3}{35}}.$

(b) Change the order of integration in $\displaystyle\int_0^a \int_{x^2/a}^{2a-x} xy\,dy\,dx$

The region is bounded by the parabola $y = x^2/a$ (i.e. $x=\sqrt{ay}$) and the line $y = 2a - x$ (i.e. $x = 2a - y$), with $0\le x\le a$. At $x=a$ both give $y=a$, so the curves meet at $(a,a)$; the parabola meets $x=0$ at $y=0$ and the line meets $x=0$ at $y=2a$.

Splitting on $y$: for $0\le y\le a$, $x$ runs $0 \to \sqrt{ay}$; for $a\le y\le 2a$, $x$ runs $0 \to 2a - y$.

$$I = \int_0^a \int_0^{\sqrt{ay}} xy\,dx\,dy + \int_a^{2a}\int_0^{2a-y} xy\,dx\,dy.$$

First part: $\int_0^{\sqrt{ay}} x\,dx = \tfrac{ay}{2}$, so $\int_0^a y\cdot\tfrac{ay}{2}\,dy = \tfrac{a}{2}\cdot\tfrac{a^3}{3} = \tfrac{a^4}{6}.$

Second part: $\int_0^{2a-y} x\,dx = \tfrac{(2a-y)^2}{2}$, so $\tfrac12\int_a^{2a} y(2a-y)^2\,dy.$ Let me expand $y(4a^2 - 4ay + y^2)=4a^2 y -4a y^2 + y^3$. Then

$$\int_a^{2a}(4a^2 y - 4a y^2 + y^3)\,dy = \Big[2a^2 y^2 - \tfrac{4a}{3}y^3 + \tfrac{y^4}{4}\Big]_a^{2a}.$$

At $2a$: $8a^4 - \tfrac{32}{3}a^4 + 4a^4 = (12 - \tfrac{32}{3})a^4 = \tfrac{4}{3}a^4.$ At $a$: $2a^4 - \tfrac{4}{3}a^4 + \tfrac14 a^4 = \tfrac{11}{12}a^4.$ Difference $= \tfrac{4}{3}a^4 - \tfrac{11}{12}a^4 = \tfrac{5}{12}a^4.$ Half of it: $\tfrac{5}{24}a^4.$

$$I = \tfrac{a^4}{6} + \tfrac{5a^4}{24} = \tfrac{4a^4 + 5a^4}{24} = \boxed{\dfrac{3a^4}{8}}.$$

(c) Volume bounded by $x^2+y^2=4$, $z=0$, $z=3-x$

Over the disk $R: x^2+y^2\le 4$ the height is $(3-x)$ (which is $\ge 1>0$ on the disk):

$$V = \iint_R (3-x)\,dA = 3\cdot(\text{area of disk}) - \iint_R x\,dA.$$

The area $=\pi(2)^2 = 4\pi$ and $\iint_R x\,dA = 0$ by symmetry.

$$\boxed{V = 3(4\pi) = 12\pi \text{ cubic units.}}$$

(a) $(D^2 - 4D + 4)y = e^{2x} + x^3$

CF: auxiliary equation $m^2 - 4m + 4 = (m-2)^2 = 0 \Rightarrow m = 2,2$.

$$y_c = (C_1 + C_2 x)e^{2x}.$$

PI for $e^{2x}$: since $2$ is a double root, multiply by $x^2$:

$$y_{p1} = \frac{x^2}{2!}e^{2x} = \frac{x^2}{2}e^{2x}.$$

PI for $x^3$: $\dfrac{1}{(D-2)^2}x^3 = \dfrac{1}{4}\Big(1 - D\Big)^{-2}x^3$ with $\dfrac{1}{4}\big(1 + D + \tfrac34 D^2 + \tfrac12 D^3 + \dots\big)x^3$ where I expand $(1-\tfrac{D}{2})^{-2}=1 + D + \tfrac34 D^2 + \tfrac12 D^3+\cdots$.

$$y_{p2} = \tfrac14\big(x^3 + 3x^2 + \tfrac34\cdot 6x + \tfrac12\cdot 6\big) = \tfrac14\big(x^3 + 3x^2 + \tfrac{9}{2}x + 3\big).$$

General solution:

$$\boxed{y = (C_1 + C_2 x)e^{2x} + \tfrac{x^2}{2}e^{2x} + \tfrac14\Big(x^3 + 3x^2 + \tfrac92 x + 3\Big).}$$

(b) Variation of parameters: $y'' + y = \sec x$

CF: $y_c = C_1\cos x + C_2\sin x$, so $y_1=\cos x,\ y_2=\sin x$, Wronskian $W = \cos x\cos x - (-\sin x)\sin x...$ actually $W = y_1 y_2' - y_1' y_2 = \cos^2 x + \sin^2 x = 1.$

With $f(x)=\sec x$:

$$u_1 = -\int \frac{y_2 f}{W}dx = -\int \sin x\sec x\,dx = -\int \tan x\,dx = \ln|\cos x|,$$ $$u_2 = \int \frac{y_1 f}{W}dx = \int \cos x\sec x\,dx = \int 1\,dx = x.$$ $$y_p = u_1 y_1 + u_2 y_2 = \cos x\,\ln|\cos x| + x\sin x.$$ $$\boxed{y = C_1\cos x + C_2\sin x + \cos x\,\ln|\cos x| + x\sin x.}$$

(c) Cauchy–Euler: $x^2 y'' - 3x y' + 4y = x^2\ln x$

Put $x = e^t$, $t=\ln x$, $\theta = \dfrac{d}{dt}$. Then $x y' = \theta y$ and $x^2 y'' = \theta(\theta-1)y$. The equation becomes

$$\theta(\theta-1)y - 3\theta y + 4y = e^{2t}\,t \;\Rightarrow\; (\theta^2 - 4\theta + 4)y = t\,e^{2t}.$$

CF: $(\theta-2)^2=0 \Rightarrow y_c = (A + Bt)e^{2t} = (A + B\ln x)x^2.$

PI: $\dfrac{1}{(\theta-2)^2}t e^{2t} = e^{2t}\dfrac{1}{\theta^2}t = e^{2t}\dfrac{t^3}{6}.$ In terms of $x$: $\dfrac{x^2 (\ln x)^3}{6}.$

$$\boxed{y = (A + B\ln x)x^2 + \frac{x^2(\ln x)^3}{6}.}$$

(a) Definitions and $\nabla\times(\nabla\phi)=\vec 0$

• Gradient of a scalar field $\phi$: $\nabla\phi = \dfrac{\partial\phi}{\partial x}\hat i + \dfrac{\partial\phi}{\partial y}\hat j + \dfrac{\partial\phi}{\partial z}\hat k$ — a vector giving the direction and rate of greatest increase of $\phi$.
• Divergence of $\vec F = F_1\hat i + F_2\hat j + F_3\hat k$: $\nabla\cdot\vec F = \dfrac{\partial F_1}{\partial x} + \dfrac{\partial F_2}{\partial y} + \dfrac{\partial F_3}{\partial z}$ — a scalar measuring net outflow per unit volume.
• Curl: $\nabla\times\vec F = \Big(\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z}\Big)\hat i + \Big(\dfrac{\partial F_1}{\partial z}-\dfrac{\partial F_3}{\partial x}\Big)\hat j + \Big(\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y}\Big)\hat k$ — measures local rotation.

Proof $\nabla\times(\nabla\phi)=\vec 0$. With $\vec F = \nabla\phi$, the $\hat i$-component of the curl is

$$\frac{\partial}{\partial y}\Big(\frac{\partial\phi}{\partial z}\Big) - \frac{\partial}{\partial z}\Big(\frac{\partial\phi}{\partial y}\Big) = \phi_{zy} - \phi_{yz} = 0$$

since mixed partials are equal for a twice-differentiable $\phi$. Likewise the $\hat j$ and $\hat k$ components vanish. Hence $\nabla\times(\nabla\phi)=\vec 0.$

(b) $\vec F=(x^2-yz)\hat i+(y^2-zx)\hat j+(z^2-xy)\hat k$

Irrotational check: $\nabla\times\vec F$. $\hat i$: $\partial_y(z^2-xy)-\partial_z(y^2-zx) = -x-(-x)=0.$ $\hat j$: $\partial_z(x^2-yz)-\partial_x(z^2-xy) = -y-(-y)=0.$ $\hat k$: $\partial_x(y^2-zx)-\partial_y(x^2-yz) = -z-(-z)=0.$ So $\nabla\times\vec F=\vec 0$ — irrotational.

Potential $\phi$: $\phi_x = x^2-yz \Rightarrow \phi = \tfrac{x^3}{3} - xyz + g(y,z).$ Then $\phi_y = -xz + g_y = y^2 - zx \Rightarrow g_y = y^2 \Rightarrow g = \tfrac{y^3}{3} + h(z).$ Then $\phi_z = -xy + h'(z) = z^2 - xy \Rightarrow h'=z^2 \Rightarrow h=\tfrac{z^3}{3}.$

$$\boxed{\phi = \tfrac13(x^3 + y^3 + z^3) - xyz + C.}$$

(c) Green's theorem for $\oint_C[(xy+y^2)dx + x^2 dy]$, $C$: $y=x^2$ to $y=x$

Here $M = xy+y^2,\ N = x^2$, region $R$: between $y=x^2$ (lower) and $y=x$ (upper), $0\le x\le 1$, traversed counter-clockwise.

Double integral side: $N_x - M_y = 2x - (x+2y) = x - 2y.$

$$\iint_R (x-2y)\,dA = \int_0^1\int_{x^2}^{x}(x-2y)\,dy\,dx.$$

Inner: $[xy - y^2]_{x^2}^{x} = (x^2 - x^2) - (x^3 - x^4) = x^4 - x^3.$ Outer: $\int_0^1(x^4 - x^3)dx = \tfrac15 - \tfrac14 = -\tfrac{1}{20}.$

Line integral side: Along $C_1: y=x^2$, $x:0\to1$, $dy=2x\,dx$: $M\,dx + N\,dy = (x\cdot x^2 + x^4)dx + x^2(2x\,dx) = (x^3 + x^4 + 2x^3)dx = (3x^3 + x^4)dx.$ $\int_0^1(3x^3+x^4)dx = \tfrac34 + \tfrac15 = \tfrac{19}{20}.$ Along $C_2: y=x$ back, $x:1\to0$, $dy=dx$: $M\,dx + N\,dy = (x\cdot x + x^2)dx + x^2 dx = (x^2 + x^2 + x^2)dx = 3x^2 dx.$ $\int_1^0 3x^2 dx = -1.$ Total line integral $= \tfrac{19}{20} - 1 = -\tfrac{1}{20}.$

Both sides equal $-\dfrac{1}{20}$, so Green's theorem is verified.

Eigenvalues and eigenvectors of $A=\begin{bmatrix}2&1&1\\1&2&1\\0&0&1\end{bmatrix}$

Characteristic equation $\det(A-\lambda I)=0$. Expanding along the third row:

$$\det\begin{bmatrix}2-\lambda&1&1\\1&2-\lambda&1\\0&0&1-\lambda\end{bmatrix} = (1-\lambda)\big[(2-\lambda)^2 - 1\big].$$

$(2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda-1)(\lambda-3).$ So

$$(1-\lambda)(\lambda-1)(\lambda-3)=0 \Rightarrow \boxed{\lambda = 1,\,1,\,3.}$$

Eigenvector for $\lambda=3$: $(A-3I)x=0$: $-x_1+x_2+x_3=0,\ x_1 - x_2 + x_3=0,\ -2x_3=0.$ So $x_3=0,\ x_1=x_2$. Eigenvector $\mathbf{v_1}=(1,1,0)^T.$

Eigenvectors for $\lambda=1$: $(A-I)x=0$: $x_1 + x_2 + x_3 = 0$ (rows 1 and 2 identical), third row $0=0$. One independent equation $\Rightarrow$ two free parameters. Choosing $x_3=0,x_2=1$ gives $(-1,1,0)^T$; choosing $x_2=0,x_3=1$ gives $(-1,0,1)^T$. So eigenvectors $\mathbf{v_2}=(-1,1,0)^T,\ \mathbf{v_3}=(-1,0,1)^T.$

Cayley–Hamilton verification

The characteristic polynomial is $(\lambda-1)^2(\lambda-3) = \lambda^3 - 5\lambda^2 + 7\lambda - 3.$ We must show $A^3 - 5A^2 + 7A - 3I = 0.$

$$A^2 = \begin{bmatrix}5&4&4\\4&5&4\\0&0&1\end{bmatrix},\qquad A^3 = A^2 A = \begin{bmatrix}14&13&13\\13&14&13\\0&0&1\end{bmatrix}.$$

Then $A^3 - 5A^2 + 7A - 3I$, computed entrywise:

• $(1,1):14 - 25 + 14 - 3 = 0$; $(1,2):13-20+7-0=0$; $(1,3):13-20+7-0=0.$
• $(2,1):13-20+7-0=0$; $(2,2):14-25+14-3=0$; $(2,3):13-20+7-0=0.$
• Row 3: $0,0,\ 1-5+7-3=0.$

All entries are zero, so $A^3 - 5A^2 + 7A - 3I = 0$, verifying the Cayley–Hamilton theorem.

(a) Scalar triple product $[\vec a\ \vec b\ \vec c]$

$\vec a=(2,-1,1),\ \vec b=(1,2,-3),\ \vec c=(3,-4,5).$

$$[\vec a\ \vec b\ \vec c] = \begin{vmatrix}2&-1&1\\1&2&-3\\3&-4&5\end{vmatrix}.$$

Expand along row 1:

$$2(2\cdot5 - (-3)(-4)) - (-1)(1\cdot5 - (-3)(3)) + 1(1\cdot(-4) - 2\cdot3)$$ $$= 2(10-12) + 1(5+9) + (-4-6) = 2(-2) + 14 - 10 = -4 + 14 - 10 = 0.$$

Since $[\vec a\ \vec b\ \vec c] = 0$, the three vectors are coplanar (they lie in one plane / are linearly dependent).

(b) Diagonals of a rhombus bisect each other at right angles

Let the rhombus be $ABCD$ with $\vec{AB} = \vec a$ and $\vec{AD} = \vec b$, where $|\vec a| = |\vec b|$ (all sides equal). Then $\vec{AC} = \vec a + \vec b$ (one diagonal) and $\vec{BD} = \vec b - \vec a$ (the other diagonal).

Bisection: In a parallelogram (which a rhombus is) the diagonals bisect each other — the midpoint of $AC$ is $\tfrac12(\vec a+\vec b)$ and the midpoint of $BD$ (from $A$) is $\vec a + \tfrac12(\vec b-\vec a)=\tfrac12(\vec a+\vec b)$; same point.

Right angles: Compute the dot product of the diagonals:

$$\vec{AC}\cdot\vec{BD} = (\vec a+\vec b)\cdot(\vec b-\vec a) = \vec a\cdot\vec b - |\vec a|^2 + |\vec b|^2 - \vec b\cdot\vec a = |\vec b|^2 - |\vec a|^2.$$

Since $|\vec a| = |\vec b|$, this is $0$. Hence the diagonals are perpendicular.

Therefore the diagonals of a rhombus bisect each other at right angles. $\blacksquare$

(a) De Moivre's theorem and $\cos4\theta,\sin4\theta$

Statement. For any integer $n$, $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta.$

With $n=4$, expand the left side by the binomial theorem (write $c=\cos\theta,\ s=\sin\theta$):

$$(c+is)^4 = c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4$$ $$= c^4 + 4ic^3 s - 6c^2 s^2 - 4i c s^3 + s^4.$$

Equating real and imaginary parts with $\cos4\theta + i\sin4\theta$:

$$\boxed{\cos4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta,}$$ $$\boxed{\sin4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta.}$$

(b) Cube roots of $1+i$

Write $1+i$ in polar form: $r = \sqrt{1^2+1^2} = \sqrt2$, argument $\theta = \dfrac{\pi}{4}$. So

$$1+i = \sqrt2\Big(\cos\tfrac\pi4 + i\sin\tfrac\pi4\Big).$$

By De Moivre, the three cube roots are

$$(1+i)^{1/3} = 2^{1/6}\Big[\cos\frac{\tfrac\pi4 + 2k\pi}{3} + i\sin\frac{\tfrac\pi4 + 2k\pi}{3}\Big],\quad k=0,1,2.$$

The arguments are $\dfrac{\pi}{12},\ \dfrac{\pi}{12}+\dfrac{2\pi}{3}=\dfrac{9\pi}{12}=\dfrac{3\pi}{4},\ \dfrac{\pi}{12}+\dfrac{4\pi}{3}=\dfrac{17\pi}{12}.$ Thus

$$z_0 = 2^{1/6}\big(\cos\tfrac{\pi}{12} + i\sin\tfrac{\pi}{12}\big),\ z_1 = 2^{1/6}\big(\cos\tfrac{3\pi}{4} + i\sin\tfrac{3\pi}{4}\big),\ z_2 = 2^{1/6}\big(\cos\tfrac{17\pi}{12} + i\sin\tfrac{17\pi}{12}\big).$$

Argand diagram (described): the three roots all lie on a circle of radius $2^{1/6}\approx 1.122$ centred at the origin, spaced $120^\circ$ apart, at angles $15^\circ,\ 135^\circ,\ 255^\circ$ — forming the vertices of an equilateral triangle.

(a) Ratio test and $\sum n!/n^n$

D'Alembert's ratio test. For a series $\sum u_n$ of positive terms, let $L = \lim_{n\to\infty}\dfrac{u_{n+1}}{u_n}$. Then the series converges if $L<1$, diverges if $L>1$, and the test is inconclusive if $L=1$.

For $u_n = \dfrac{n!}{n^n}$:

$$\frac{u_{n+1}}{u_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} = \frac{(n+1)\,n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \Big(\frac{n}{n+1}\Big)^n = \frac{1}{\big(1+\tfrac1n\big)^n}.$$

As $n\to\infty$, $\big(1+\tfrac1n\big)^n \to e$, so $L = \dfrac1e \approx 0.368 < 1.$

Since $L<1$, the series $\displaystyle\sum \frac{n!}{n^n}$ converges.

(b) $\sum 1/n^p$ by the integral test

Integral test. If $f(x)=\dfrac{1}{x^p}$ is positive, continuous and decreasing for $x\ge1$ (true for $p>0$), then $\sum_{n=1}^\infty \dfrac{1}{n^p}$ and $\int_1^\infty \dfrac{dx}{x^p}$ converge or diverge together.

Case $p\ne1$:

$$\int_1^\infty x^{-p}\,dx = \Big[\frac{x^{1-p}}{1-p}\Big]_1^\infty.$$

• If $p>1$: $1-p<0$, $x^{1-p}\to0$, integral $=\dfrac{1}{p-1}$ (finite) $\Rightarrow$ converges.
• If $p<1$: $1-p>0$, $x^{1-p}\to\infty$ $\Rightarrow$ integral diverges $\Rightarrow$ diverges.

Case $p=1$: $\int_1^\infty \dfrac{dx}{x} = [\ln x]_1^\infty = \infty$ $\Rightarrow$ diverges (harmonic series).

Conclusion (p-series): $\displaystyle\sum_{n=1}^\infty \frac{1}{n^p}$ converges if and only if $\boxed{p>1}$; it diverges for $p\le1$.

(a) Plane through $(1,1,0),(-2,2,-1),(1,2,1)$

Let $P=(1,1,0)$. Two in-plane vectors:

$$\vec{u} = (-2,2,-1)-(1,1,0) = (-3,1,-1),\quad \vec{v} = (1,2,1)-(1,1,0) = (0,1,1).$$

Normal $\vec n = \vec u\times\vec v$:

$$\vec n = \begin{vmatrix}\hat i&\hat j&\hat k\\-3&1&-1\\0&1&1\end{vmatrix} = \hat i(1\cdot1 -(-1)\cdot1) - \hat j((-3)\cdot1 - (-1)\cdot0) + \hat k((-3)\cdot1 - 1\cdot0)$$ $$= \hat i(1+1) - \hat j(-3) + \hat k(-3) = (2,\,3,\,-3).$$

Plane: $2(x-1) + 3(y-1) - 3(z-0) = 0$, i.e.

$$\boxed{2x + 3y - 3z = 5.}$$

(Check: $(1,2,1):2+6-3=5$ ✓.)

(b) Shortest distance between the two lines

Line 1: point $A=(1,2,3)$, direction $\vec d_1=(2,3,4)$. Line 2: point $B=(2,4,5)$, direction $\vec d_2=(3,4,5)$.

$$\vec d_1\times\vec d_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\2&3&4\\3&4&5\end{vmatrix} = \hat i(15-16) - \hat j(10-12) + \hat k(8-9) = (-1,\,2,\,-1).$$

$|\vec d_1\times\vec d_2| = \sqrt{1+4+1} = \sqrt6.$

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

$$(\vec d_1\times\vec d_2)\cdot\vec{AB} = (-1)(1) + 2(2) + (-1)(2) = -1+4-2 = 1.$$

Shortest distance:

$$d = \frac{|(\vec d_1\times\vec d_2)\cdot\vec{AB}|}{|\vec d_1\times\vec d_2|} = \frac{1}{\sqrt6} = \boxed{\frac{1}{\sqrt6}\approx 0.408 \text{ units.}}$$