01. 공간벡터와 그 연산Vectors in Space and Their Operations

Solution. $\overrightarrow{PQ} = Q-P = (-5-(-3),\, 2-4,\, 2-1) = (-2,-2,1)$.

Hence $|\overrightarrow{PQ}| = \sqrt{(-2)^2+(-2)^2+1^2} = \sqrt{9} = 3$.

Solution. Since $\mathbf{F}$ has magnitude $20$ and makes an angle of $45^\circ$ with the positive $x$-axis, $\mathbf{F} = 20(\cos45^\circ, \sin45^\circ) = 20\left(\dfrac{\sqrt2}{2}, \dfrac{\sqrt2}{2}\right) = (10\sqrt2, 10\sqrt2)$.

Solution. The centroid of a triangle is the average of the position vectors of its three vertices, so

$$\overrightarrow{OP} = \frac{1}{3}\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right) = \frac{1}{3}\big((1,-1,2)+(2,1,3)+(-1,2,-1)\big) = \frac{1}{3}(2,2,4) = \left(\frac{2}{3},\frac{2}{3},\frac{4}{3}\right).$$

Solution. $\mathbf{u}=(-1,3,1)$ and $\mathbf{v}=(4,7,0)$.

(1) $2\mathbf{u}+3\mathbf{v} = (-2,6,2)+(12,21,0) = (10,27,2)$.

(2) $\mathbf{u}-\mathbf{v} = (-1-4,\,3-7,\,1-0) = (-5,-4,1)$.

(3) Since $\dfrac{1}{2}\mathbf{u} = \left(-\dfrac12,\dfrac32,\dfrac12\right)$, we have $\left|\dfrac12\mathbf{u}\right| = \dfrac12|\mathbf{u}| = \dfrac12\sqrt{(-1)^2+3^2+1^2} = \dfrac{\sqrt{11}}{2}$.

Solution. $\overrightarrow{P_1P_2} = P_2-P_1 = (3-1,\,2-0,\,0-1) = (2,2,-1)$.

Since $|\overrightarrow{P_1P_2}| = \sqrt{4+4+1} = 3$, the direction is $\dfrac{\overrightarrow{P_1P_2}}{|\overrightarrow{P_1P_2}|} = \left(\dfrac23,\dfrac23,-\dfrac13\right)$.

Solution. Since $\mathbf{v}=3\mathbf{i}-4\mathbf{j}=(3,-4)$, the speed is $|\mathbf{v}| = \sqrt{3^2+(-4)^2} = \sqrt{25} = 5$.

The direction is $\dfrac{\mathbf{v}}{|\mathbf{v}|} = \left(\dfrac35,-\dfrac45\right)$.

Solution. Since $\mathbf{v}=2\mathbf{i}+2\mathbf{j}-\mathbf{k}=(2,2,-1)$ and $|\mathbf{v}| = \sqrt{4+4+1} = 3$, the direction of $\mathbf{v}$ is $\dfrac{\mathbf{v}}{|\mathbf{v}|} = \left(\dfrac23,\dfrac23,-\dfrac13\right)$.

Since $\mathbf{F}$ has magnitude $6$ in this direction,

$$\mathbf{F} = 6\cdot\frac{\mathbf{v}}{|\mathbf{v}|} = 6\left(\frac23,\frac23,-\frac13\right) = (4,4,-2).$$

Solution. (1) $(1,-2,-1)\cdot(-6,2,-3) = 1\cdot(-6)+(-2)\cdot2+(-1)\cdot(-3) = -6-4+3 = -7$.

(2) $\left(\dfrac12\mathbf{i}+3\mathbf{j}+\mathbf{k}\right)\cdot(4\mathbf{i}-\mathbf{j}+2\mathbf{k}) = \dfrac12\cdot4 + 3\cdot(-1) + 1\cdot2 = 2-3+2 = 1$.

Solution. If $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$, then $\cos\theta = \dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$.

(1) $\mathbf{u}\cdot\mathbf{v} = 1\cdot6+(-2)\cdot3+(-2)\cdot2 = 6-6-4 = -4$, and $|\mathbf{u}| = \sqrt{1+4+4}=3$, $|\mathbf{v}| = \sqrt{36+9+4}=7$, so $\cos\theta = -\dfrac{4}{21}$, i.e., $\theta = \cos^{-1}\left(-\dfrac{4}{21}\right)$.

(2) $\mathbf{u}\cdot\mathbf{v} = 3\cdot0+(-2)\cdot2+1\cdot4 = -4+4 = 0$, so $\cos\theta=0$, i.e., $\theta = \dfrac{\pi}{2}$.

Solution. $\angle ABC$ is the angle between $\overrightarrow{BA}$ and $\overrightarrow{BC}$.

$\overrightarrow{BA} = A-B = (1,-1,0)$, $\overrightarrow{BC} = C-B = (0,-1,1)$, and $\overrightarrow{BA}\cdot\overrightarrow{BC} = 0+1+0=1$, $|\overrightarrow{BA}|=|\overrightarrow{BC}|=\sqrt2$.

Hence $\cos(\angle ABC) = \dfrac{1}{\sqrt2\cdot\sqrt2} = \dfrac12$, so $\angle ABC = \dfrac{\pi}{3}$.

Proof.

(1) True. Since $\mathbf{u}\cdot\mathbf{u} = |\mathbf{u}|^2$, we have $\mathbf{u}\cdot\mathbf{u}=0 \Leftrightarrow |\mathbf{u}|^2=0 \Leftrightarrow |\mathbf{u}|=0 \Leftrightarrow \mathbf{u}=\mathbf{0}$. Hence $\mathbf{u}\cdot\mathbf{u}=0$ is a necessary and sufficient condition for $\mathbf{u}=\mathbf{0}$.

(2) False. The equality $\mathbf{u}\cdot\mathbf{v}_1=\mathbf{u}\cdot\mathbf{v}_2$ is equivalent to $\mathbf{u}\cdot(\mathbf{v}_1-\mathbf{v}_2)=0$, i.e., $\mathbf{v}_1-\mathbf{v}_2$ is perpendicular to $\mathbf{u}$, so we can find $\mathbf{v}_1\neq\mathbf{v}_2$ for which the equality still holds. For instance, let $\mathbf{u}=(1,0,0)$, $\mathbf{v}_1=(0,1,0)$, $\mathbf{v}_2=(0,1,1)$. Then $\mathbf{u}\neq\mathbf{0}$ and $\mathbf{u}\cdot\mathbf{v}_1 = 0 = \mathbf{u}\cdot\mathbf{v}_2$, but $\mathbf{v}_1\neq\mathbf{v}_2$. Hence the statement is false.

Solution. $|\mathbf{v}-2\mathbf{w}|^2 = |\mathbf{v}|^2 - 4\,\mathbf{v}\cdot\mathbf{w} + 4|\mathbf{w}|^2$, and $\mathbf{v}\cdot\mathbf{w} = |\mathbf{v}||\mathbf{w}|\cos\dfrac{\pi}{3} = 2\cdot3\cdot\dfrac12 = 3$.

Hence $|\mathbf{v}-2\mathbf{w}|^2 = 4 - 4\cdot3 + 4\cdot9 = 4-12+36 = 28$, so $|\mathbf{v}-2\mathbf{w}| = \sqrt{28} = 2\sqrt7$.

Proof. Assume $\mathbf{u}\neq\mathbf{0}$, $\mathbf{v}\neq\mathbf{0}$, and let $\mathbf{a}=|\mathbf{v}|\mathbf{u}$, $\mathbf{b}=|\mathbf{u}|\mathbf{v}$. Then $\mathbf{w}=\mathbf{a}+\mathbf{b}$, and

$$|\mathbf{a}| = |\mathbf{v}||\mathbf{u}| = |\mathbf{u}||\mathbf{v}| = |\mathbf{b}|,$$

so $\mathbf{a}$ and $\mathbf{b}$ have equal magnitude. Since $\mathbf{a}$ is a positive multiple of $\mathbf{u}$ and $\mathbf{b}$ is a positive multiple of $\mathbf{v}$, the angle between $\mathbf{a}$ and $\mathbf{b}$ equals the angle between $\mathbf{u}$ and $\mathbf{v}$.

Let $\theta_1$ be the angle between $\mathbf{w}=\mathbf{a}+\mathbf{b}$ and $\mathbf{a}$, and $\theta_2$ the angle between $\mathbf{w}$ and $\mathbf{b}$. Writing $k=|\mathbf{a}|=|\mathbf{b}|$,

$$\cos\theta_1 = \frac{\mathbf{w}\cdot\mathbf{a}}{|\mathbf{w}||\mathbf{a}|} = \frac{(\mathbf{a}+\mathbf{b})\cdot\mathbf{a}}{|\mathbf{w}|k} = \frac{k^2+\mathbf{a}\cdot\mathbf{b}}{|\mathbf{w}|k}, \qquad \cos\theta_2 = \frac{\mathbf{w}\cdot\mathbf{b}}{|\mathbf{w}||\mathbf{b}|} = \frac{(\mathbf{a}+\mathbf{b})\cdot\mathbf{b}}{|\mathbf{w}|k} = \frac{\mathbf{a}\cdot\mathbf{b}+k^2}{|\mathbf{w}|k}.$$

Hence $\cos\theta_1=\cos\theta_2$, and since $\theta_1,\theta_2\in[0,\pi]$, we get $\theta_1=\theta_2$.

Thus $\mathbf{w}$ bisects the angle between $\mathbf{a}$ and $\mathbf{b}$, hence it bisects the angle between $\mathbf{u}$ and $\mathbf{v}$. (Here we treat the case $\mathbf{w}\neq\mathbf{0}$, i.e., $\mathbf{u}$ and $\mathbf{v}$ are not exactly opposite in direction.)

Solution. The work done by the force is $W = \mathbf{F}\cdot\mathbf{d} = |\mathbf{F}||\mathbf{d}|\cos\theta$.

Since $|\mathbf{F}|=40$, $|\mathbf{d}|=3$, $\theta=60^\circ$,

$$W = 40\cdot3\cdot\cos60^\circ = 120\cdot\frac12 = 60\ \text{J}.$$

Solution. The projection is $\operatorname{proj}_{\mathbf{v}}\mathbf{u} = \dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|^2}\mathbf{v}$, and the scalar component in the direction of $\mathbf{v}$ is $\dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|}$.

(1) $\mathbf{F}=(5,2)$, $\mathbf{v}=(1,-3)$. Since $\mathbf{F}\cdot\mathbf{v} = 5-6=-1$ and $|\mathbf{v}|^2=1+9=10$,

$$\operatorname{proj}_{\mathbf{v}}\mathbf{F} = \frac{-1}{10}(1,-3) = \left(-\frac1{10},\frac3{10}\right),$$

and the scalar component is $\dfrac{\mathbf{F}\cdot\mathbf{v}}{|\mathbf{v}|} = \dfrac{-1}{\sqrt{10}} = -\dfrac{\sqrt{10}}{10}$.

(2) $\mathbf{u}=(6,3,2)$, $\mathbf{v}=(1,-2,-2)$. Since $\mathbf{u}\cdot\mathbf{v} = 6-6-4=-4$ and $|\mathbf{v}|^2=1+4+4=9$,

$$\operatorname{proj}_{\mathbf{v}}\mathbf{u} = \frac{-4}{9}(1,-2,-2) = \left(-\frac49,\frac89,\frac89\right),$$

and the scalar component is $\dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|} = \dfrac{-4}{3} = -\dfrac43$.

Proof. Let $c = \dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|^2}$, so that $\operatorname{proj}_{\mathbf{v}}\mathbf{u} = c\mathbf{v}$.

(1) $(\mathbf{u}-\operatorname{proj}_{\mathbf{v}}\mathbf{u})\cdot\operatorname{proj}_{\mathbf{v}}\mathbf{u} = (\mathbf{u}-c\mathbf{v})\cdot(c\mathbf{v}) = c(\mathbf{u}\cdot\mathbf{v}) - c^2|\mathbf{v}|^2$. Substituting $c = \dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|^2}$,

$$c(\mathbf{u}\cdot\mathbf{v}) - c^2|\mathbf{v}|^2 = \frac{(\mathbf{u}\cdot\mathbf{v})^2}{|\mathbf{v}|^2} - \frac{(\mathbf{u}\cdot\mathbf{v})^2}{|\mathbf{v}|^2} = 0,$$

so $\mathbf{u}-\operatorname{proj}_{\mathbf{v}}\mathbf{u}$ is perpendicular to $\operatorname{proj}_{\mathbf{v}}\mathbf{u}$.

(2) If $\mathbf{u}$ and $\mathbf{v}$ are perpendicular, then $\mathbf{u}\cdot\mathbf{v}=0$, so $c = \dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|^2} = 0$. Hence $\operatorname{proj}_{\mathbf{v}}\mathbf{u} = c\mathbf{v} = \mathbf{0}$.

Solution. $\mathbf{u}=(2,1,1)$, $\mathbf{v}=(-4,3,1)$.

$$\mathbf{u}\times\mathbf{v} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&1&1\\-4&3&1\end{vmatrix} = (1\cdot1-1\cdot3)\mathbf{i} - (2\cdot1-1\cdot(-4))\mathbf{j} + (2\cdot3-1\cdot(-4))\mathbf{k} = -2\mathbf{i}-6\mathbf{j}+10\mathbf{k} = (-2,-6,10).$$

By anticommutativity of the cross product, $\mathbf{v}\times\mathbf{u} = -(\mathbf{u}\times\mathbf{v}) = (2,6,-10)$.

Solution. $\overrightarrow{PQ} = Q-P = (1,2,-1)$, $\overrightarrow{PR} = R-P = (-2,2,2)$.

(1) The vector $\overrightarrow{PQ}\times\overrightarrow{PR}$ is perpendicular to the plane through $P,Q,R$.

$$\overrightarrow{PQ}\times\overrightarrow{PR} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&-1\\-2&2&2\end{vmatrix} = (2\cdot2-(-1)\cdot2)\mathbf{i}-(1\cdot2-(-1)\cdot(-2))\mathbf{j}+(1\cdot2-2\cdot(-2))\mathbf{k} = (6,0,6),$$

and $|\overrightarrow{PQ}\times\overrightarrow{PR}| = \sqrt{36+0+36}=6\sqrt2$, so the required unit vector is

$$\frac{\overrightarrow{PQ}\times\overrightarrow{PR}}{|\overrightarrow{PQ}\times\overrightarrow{PR}|} = \frac{(6,0,6)}{6\sqrt2} = \left(\frac{\sqrt2}{2},0,\frac{\sqrt2}{2}\right).$$

(2) The area of triangle $PQR$ is $\dfrac12|\overrightarrow{PQ}\times\overrightarrow{PR}| = \dfrac12\cdot6\sqrt2 = 3\sqrt2$.

Solution. In a parallelogram, the diagonals bisect each other, so we compare the midpoints of $\overline{AB}$ and $\overline{CD}$.

The midpoint of $A,B$ is $\left(1,\dfrac72,\dfrac12\right)$, and the midpoint of $C,D$ is also $\left(1,\dfrac72,\dfrac12\right)$, so they coincide. Hence $A,B$ form one diagonal and $C,D$ form the other, so the vertices of the parallelogram are arranged in the order $A,C,B,D$.

That is, the two sides adjacent to $A$ are $\overrightarrow{AC}$ and $\overrightarrow{AD}$. Since $\overrightarrow{AC} = (1,4,0)$ and $\overrightarrow{AD} = (-1,3,3)$,

$$\overrightarrow{AC}\times\overrightarrow{AD} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&4&0\\-1&3&3\end{vmatrix} = (4\cdot3-0\cdot3)\mathbf{i}-(1\cdot3-0\cdot(-1))\mathbf{j}+(1\cdot3-4\cdot(-1))\mathbf{k} = (12,-3,7),$$

so the area of the parallelogram is $|\overrightarrow{AC}\times\overrightarrow{AD}| = \sqrt{144+9+49} = \sqrt{202}$.

Proof.

(1) True. Since $\mathbf{u}\times\mathbf{v}$ is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$, we have $(\mathbf{u}\times\mathbf{v})\cdot\mathbf{u} = 0$ and $(\mathbf{u}\times\mathbf{v})\cdot\mathbf{v}=0$. By commutativity of the dot product, $\mathbf{v}\cdot(\mathbf{u}\times\mathbf{v}) = (\mathbf{u}\times\mathbf{v})\cdot\mathbf{v} = 0$. Since both sides equal $0$, $(\mathbf{u}\times\mathbf{v})\cdot\mathbf{u} = \mathbf{v}\cdot(\mathbf{u}\times\mathbf{v})$ holds.

(2) False. Since $\mathbf{u}\times\mathbf{v}_1 = \mathbf{u}\times\mathbf{v}_2 \Leftrightarrow \mathbf{u}\times(\mathbf{v}_1-\mathbf{v}_2) = \mathbf{0}$, the equality holds whenever $\mathbf{v}_1-\mathbf{v}_2$ is parallel to $\mathbf{u}$. For instance, let $\mathbf{u}=(1,0,0)$, $\mathbf{v}_1=(0,1,0)$, $\mathbf{v}_2=\mathbf{v}_1+\mathbf{u}=(1,1,0)$. Then

$$\mathbf{u}\times\mathbf{v}_1 = (1,0,0)\times(0,1,0) = (0,0,1), \qquad \mathbf{u}\times\mathbf{v}_2 = (1,0,0)\times(1,1,0) = (0,0,1),$$

which are equal, but $\mathbf{u}\neq\mathbf{0}$ and $\mathbf{v}_1=(0,1,0)\neq(1,1,0)=\mathbf{v}_2$. Hence the statement is false.

(3) True. Let $\mathbf{d}=\mathbf{v}_1-\mathbf{v}_2$. By hypothesis $\mathbf{u}\cdot\mathbf{d}=0$ and $\mathbf{u}\times\mathbf{d}=\mathbf{0}$. Since $\mathbf{u}\times\mathbf{d}=\mathbf{0}$, either $\mathbf{d}=\mathbf{0}$ or $\mathbf{d}$ is parallel to $\mathbf{u}$. If $\mathbf{d}$ is parallel to $\mathbf{u}$, say $\mathbf{d}=\lambda\mathbf{u}$ with $\lambda\neq0$, then $\mathbf{u}\cdot\mathbf{d} = \lambda|\mathbf{u}|^2 = 0$; since $\mathbf{u}\neq\mathbf{0}$, $|\mathbf{u}|^2\neq0$, forcing $\lambda=0$, a contradiction. Hence $\mathbf{d}=\mathbf{0}$, i.e., $\mathbf{v}_1=\mathbf{v}_2$.

Solution. The volume of the parallelepiped is the absolute value of the scalar triple product, $|\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})|$.

$$\mathbf{v}\times\mathbf{w} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-2&0&3\\0&7&-4\end{vmatrix} = (0\cdot(-4)-3\cdot7)\mathbf{i}-((-2)\cdot(-4)-3\cdot0)\mathbf{j}+((-2)\cdot7-0\cdot0)\mathbf{k} = (-21,-8,-14),$$

so

$$\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w}) = 1\cdot(-21)+2\cdot(-8)+(-1)\cdot(-14) = -21-16+14 = -23.$$

Hence the volume is $|-23|=23$.

Solution. The volume of a tetrahedron is $\dfrac16$ of the absolute value of the scalar triple product of the three edge vectors from one vertex.

$\overrightarrow{AB}=(3,-2,-2)$, $\overrightarrow{AC}=(2,-5,-1)$, $\overrightarrow{AD}=(-1,-1,-4)$, and

$$\overrightarrow{AC}\times\overrightarrow{AD} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-5&-1\\-1&-1&-4\end{vmatrix} = ((-5)(-4)-(-1)(-1))\mathbf{i}-((2)(-4)-(-1)(-1))\mathbf{j}+((2)(-1)-(-5)(-1))\mathbf{k} = (19,9,-7),$$

so

$$\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD}) = 3\cdot19+(-2)\cdot9+(-2)\cdot(-7) = 57-18+14 = 53.$$

Hence the volume of the tetrahedron is $\dfrac{1}{6}|53| = \dfrac{53}{6}$.

Solution. Since $A$ is a vertex of the parallelepiped and $B,C,D$ are the three vertices adjacent to it, the volume is $|\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD})|$.

$\overrightarrow{AB}=(1,2,0)$, $\overrightarrow{AC}=(0,-3,2)$, $\overrightarrow{AD}=(3,-4,5)$, and

$$\overrightarrow{AC}\times\overrightarrow{AD} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&-3&2\\3&-4&5\end{vmatrix} = ((-3)(5)-(2)(-4))\mathbf{i}-((0)(5)-(2)(3))\mathbf{j}+((0)(-4)-(-3)(3))\mathbf{k} = (-7,6,9),$$

so

$$\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD}) = 1\cdot(-7)+2\cdot6+0\cdot9 = -7+12 = 5.$$

Hence the volume of the parallelepiped is $|5|=5$.

Proof.

(1) Three vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ with initial point at the origin lie in a single plane exactly when they are linearly dependent, i.e., when the parallelepiped they determine has volume $0$. This volume equals $|(\mathbf{u}\times\mathbf{v})\cdot\mathbf{w}|$, so the vectors lie in a single plane if and only if $(\mathbf{u}\times\mathbf{v})\cdot\mathbf{w}=0$.

More precisely, if $\mathbf{u}\times\mathbf{v}\neq\mathbf{0}$, then $\mathbf{u}$ and $\mathbf{v}$ determine a plane $\Pi$ through the origin with normal vector $\mathbf{n}=\mathbf{u}\times\mathbf{v}$, and $\mathbf{w}$ lies in the same plane as $\mathbf{u},\mathbf{v}$ exactly when $\mathbf{w}$ is a linear combination of $\mathbf{u},\mathbf{v}$, i.e., when $\mathbf{w}\perp\mathbf{n}$, i.e., when $(\mathbf{u}\times\mathbf{v})\cdot\mathbf{w}=0$. If $\mathbf{u}\times\mathbf{v}=\mathbf{0}$, then $\mathbf{u}$ and $\mathbf{v}$ are parallel, so $\mathbf{u},\mathbf{v},\mathbf{w}$ always lie in the plane spanned by the origin, the common direction of $\mathbf{u}$ (or $\mathbf{v}$), and $\mathbf{w}$; in this case $(\mathbf{u}\times\mathbf{v})\cdot\mathbf{w}=\mathbf{0}\cdot\mathbf{w}=0$ as well, so the equality holds. Hence the necessary and sufficient condition holds in either case.

(2) Let $\overrightarrow{AB}=(-2,-1,3)$, $\overrightarrow{AC}=(-1,1,0)$, $\overrightarrow{AD}=(1,-3,2)$. By part (1), whether the four points lie in a single plane can be determined by checking whether $(\overrightarrow{AC}\times\overrightarrow{AD})\cdot\overrightarrow{AB}=0$.

$$\overrightarrow{AC}\times\overrightarrow{AD} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&1&0\\1&-3&2\end{vmatrix} = (1\cdot2-0\cdot(-3))\mathbf{i}-((-1)\cdot2-0\cdot1)\mathbf{j}+((-1)(-3)-1\cdot1)\mathbf{k} = (2,2,2),$$

so

$$\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD}) = (-2)\cdot2+(-1)\cdot2+3\cdot2 = -4-2+6 = 0.$$

Since the scalar triple product is $0$, the four points $A,B,C,D$ lie in a single plane.