$$ \small 11.\ 다음\ 중\ 두\ 벡터가\ 이루는\ 각\ \theta가\ 직각인\ 것을\ 찾고\\ \small 직각이\ 아니면\ 그때의\ cos\theta\ 값을\ 구하라.$$
(1) A=(1, 2, 1), B=(-2, -5, -4)
(2) A=(2, 1, 0), B=(-2, 4, 0)
(3) A=(1, 2, 3), B=(-3, 0, 1)
(4) A=(3, 0, 1), B=(-9, 0, -3)
※ 직각 조건 : AxB=0
$$ \small ※\ cos\theta = \frac{A\cdot B}{|A|\cdot |B|}$$
(1)
$$ \small \begin{align}A\cdot B&=(1,\ 2,\ 1)\cdot (-2,\ -5,\ -4)\\&=(1\cdot (-2)+2\cdot (-5)+1\cdot (-4))\\&=-2+(-10)+(-4)\\&=-16 \end{align}$$
$$ \small \begin{align}cos\theta &=\frac{A\cdot B}{|A|\cdot |B|}\\&=-\frac{16}{\sqrt{(1,\ 2,\ 1)^{2}}\cdot \sqrt{(-2,\ -5,\ -4)^{2}}}\\&=-\frac{16}{\sqrt{1+4+1}\cdot \sqrt{4+25+16}}\\&=-\frac{16}{\sqrt{6} \cdot \sqrt{45}}\\&=-\frac{16}{\sqrt{6} \cdot 3\sqrt{5}}\\&=-\frac{16}{3\sqrt{30}} \end{align}$$
(2)
$$ \small \begin{align}A\cdot B&=(2,\ 1,\ 0)\cdot (-2,\ 4,\ 0)\\&=2\cdot (-2)+1\cdot 4+0\\&=-4+4+0\\&=0 \end{align}$$
(3)
$$ \small \begin{align}A\cdot B&=(1,\ 2,\ 3)\cdot (-3,\ 0,\ 1)\\&=1\cdot (-3)+2\cdot 0+3\cdot 1\\&=-3+0+3\\&=0 \end{align}$$
(4)
$$ \small \begin{align}A\cdot B&=(3,\ 0,\ 1)\cdot (-9,\ 0,\ -3)\\&=3\cdot (-9)+0\cdot 0+1\cdot (-3)\\&=-27+0+(-3)\\&=-30 \end{align} $$
$$ \small \begin{align}cos\theta &=\frac{A\cdot B}{|A|\cdot |B|}\\&=-\frac{30}{\sqrt{(3,\ 0,\ 1)^{2}}\cdot \sqrt{(-9,\ 0,\ -3)^{2}}}\\&=-\frac{30}{\sqrt{9+0+1}\cdot \sqrt{81+0+9}}\\&=-\frac{30}{\sqrt{10}\cdot \sqrt{90}}\\&=-\frac{30}{\sqrt{10}\cdot 3\sqrt{10}}\\&=-\frac{30}{30}\\&=-1 \end{align}$$
12. [정리 7.6]을 이용하여 A=(4, 3, -2), B=(1, 2, 5)가 수직임을 보여라.
$$ \small [정리7.6]\\ \small R^{2}\ 공간이나\ R^{3}\ 공간에서\ O벡터가\ 아닌\\ \small 두\ 벡터\ A,\ B가\ 수직인\ 것은\ A\cdot B=0인\ 것과\ 동치이다.$$
$$ \small \begin{align}A\cdot B&=(4,\ 3,\ -2)\cdot (1,\ 2,\ 5)\\&=4\cdot 1+3\cdot 2+(-2)\cdot 5\\&=4+6+(-10)\\&=0 \end{align}$$
A·B는 0이므로, 수직이다.
13. 다음을 구하라
$$ \small (1)\ 점\ (1,\ 1,\ 1)에서\ 직선\ \frac{x-2}{-1}=\frac{y-2}{-2}=\frac{z-2}{-3}까지\ 거리를\ 구하라.$$
$$ \small (2)\ 점\ (1,\ 0,\ 0)에서\ 직선\ \frac{x+1}{2}=\frac{y}{2}=\frac{z}{2}까지\ 거리를\ 구하라.$$
점에서 직선까지의 거리는 수직선의 길이,
수직선과 만나는 점을 H=(a, b, c)라고 하고 구한다.
(1)
$$ \small 직선의\ 방정식\ :\ \frac{a-2}{-1}=\frac{b-2}{-2}=\frac{c-2}{-3}$$
이때,
$$ \small \frac{a-2}{-1}=\frac{b-2}{-2}=\frac{c-2}{-3}=k $$
로 놓으면,
$$ \small a=-k+2, b=-2k+2, c=-3k+2$$
$$ \small 따라서,\ H=(-k+2,\ -2k+2,\ -3k+2)이다.$$
$$ \small 또한, \overrightarrow{AH}와\\ \small 직선의\ 방향벡터\ U=(-1, -2, -3)은\ 서로\ 수직이다.$$
$$ \small \begin{align}0&=\overrightarrow{AH}\cdot U\\&=((-k+2)-1,\ (-2k+2)-1,\ (-3k+2)-1)\cdot (-1,\ -2,\ -3)\\&=(-k+1,\ -2k+1,\ -3k+1)\cdot (-1,\ -2,\ -3)\\&=(k-1)+(4k-2)+(9k-3)\\&=14k-6\\&=7k-3\end{align}$$
$$ \small k=\frac{3}{7}일\ 때,\ 두\ 벡터가\ 수직이고,\\ \small H는(\frac{11}{7},\ \frac{8}{7},\ \frac{5}{7})이 된다.$$
점 A와 H의 거리를 구하게 되면
$$ \small \begin{align}AH&=\sqrt{(1-\frac{11}{7})^{2}+(1-\frac{8}{7})^{2}+(1-\frac{5}{7})^2}\\&=\sqrt{(-\frac{4}{7})^{2}+(-\frac{1}{7})^{2}+(\frac{2}{7})^{2}}\\&=\sqrt{(\frac{16}{49})+(\frac{1}{49})+(\frac{4}{49})}\\&=\sqrt{\frac{21}{49}}\\&=\frac{\sqrt{21}}{7} \end{align}$$
(2)
위의 방법과 동일하게 적용한다.
$$ \small 직선의\ 방정식\ :\ \frac{a+1}{2}=\frac{y}{2}=\frac{z}{2}$$
$$ \small \frac{a+1}{2}=\frac{y}{2}=\frac{z}{2}=k$$
$$ \small a=2k-1, b=2k, c=2k$$
H=(2k-1, 2k, 2k)
$$ \small 직선의\ 방향벡터\ U=(2, 2, 2)$$
$$ \small \begin{align}0=\overrightarrow{AH}\cdot U&=((2k-1)-1,\ (2k)-0,\ (2k)-0)\cdot (2,\ 2,\ 2)\\&=(2k-2,\ 2k,\ 2k)\cdot (2,\ 2,\ 2)\\&=(4k-4)+(4k)+(4k)\\&=12k-4\\&=3k-1\end{align}$$
$$ \small H=(-\frac{1}{3},\ \frac{2}{3},\ \frac{2}{3}) $$
$$ \small \begin{align}AH&=\sqrt{(1-(-\frac{1}{3}))^{2}+(0-\frac{2}{3})^{2}+(0-\frac{2}{3})^{2}}\\&=\sqrt{(\frac{4}{3})^{2}+(-\frac{2}{3})^{2}+(-\frac{2}{3})^{2}}\\&=\sqrt{(\frac{16}{9})+(\frac{4}{9})+(\frac{4}{9})}\\&=\sqrt{\frac{24}{9}}\\&=\frac{2\sqrt{6}}{3}\end{align}$$
14. 다음 두 벡터 A, B에 수직인 벡터를 구하라.
(1) A=(-1, -2, -3), B=(3, 1, 2)
(2) A=(1, 4, -5), B=(3, 3, 2)
$$ \small [정의\ 7.16]\\ \small 평행하지\ 않는\ 두\ 벡터\\ \small A=(a_{1},\ a_{2},\ a_{3}),\ B=(b_{1},\ b_{2},\ b_{3})\\ \small 모두에\ 수직인\ 벡터\\ \small C=(a_{2}b_{3}-a_{3}b_{2},\ a_{3}b_{1}-a_{1}b_{3},\ a_{1}b_{2}-a_{2}b_{1})$$
(1)
$$ \small \begin{align}C&=(a_{2}b_{3}-a_{3}b_{2},\ a_{3}b_{1}-a_{1}b_{3},\ a_{1}b_{2}-a_{2}b_{1})\\&=((-2)\cdot 2-(-3)\cdot 1,\ (-3)\cdot 3-(-1)\cdot 2,\ (-1)\cdot 1-(-2)\cdot 3)\\&=((-4)-(-3),\ (-9)-(-2),\ (-1)-(-6))\\&=(-1,\ -7,\ 5)\end{align}$$
(2)
$$ \small \begin{align}C&=(a_{2}b_{3}-a_{3}b_{2},\ a_{3}b_{1}-a_{1}b_{3},\ a_{1}b_{2}-a_{2}b_{1})\\&=(4\cdot 2-(-5)\cdot 3,\ (-5)\cdot 3-1\cdot 2,\ 1\cdot 3-4\cdot 3)\\&=(8-(-15),\ (-15)-2,\ 3-12)\\&=(23,\ -17,\ -9)\end{align}$$
15. A=(1, 3, 0), B=(0, 1, -3), C=(3, 0, 1)일 때 다음을 계산하라.
(1) (-A)x(3B) (2) (AxB)xC
(3) (A+B)xC (4) (AxC)·(BxC)
$$ \small [벡터의\ 합,\ 내적,\ 외적의\ 표현]\\ \small 벡터\ 합\ :\ A+B\\ \small 내적\ :\ A\cdot B\\ \small 외적\ :\ A\times B$$
(1)
$$ \small \begin{align}(-A)\times (3B)&=(-1)\times 3\times(A\times B)\\&=(-3)\times (A\times B)\\&=(-3)\times (a_{2}b_{3}-a_{3}b_{2},\ a_{3}b_{1}-a_{1}b_{3},\ a_{1}b_{2}-a_{2}b_{1})\\&=(-3)\times (3\cdot (-3)-0\cdot 1,\ 0\cdot 0-1\cdot (-3),\ 1\cdot 1-3\cdot 0)\\&=(-3)\times ((-9)-0,\ 0-(-3),\ 1-0)\\&=(-3)times (-9,\ 3,\ 1)\\&=(27,\ -9,\ -3) \end{align}$$
(2)
$$ \small \begin{align}(A\times B)\times C&=(-9,\ 3,\ 1)\times (3,\ 0,\ 1)\\&=(3\cdot 1-1\cdot 0,\ 1\cdot 3-(-9)\cdot 1,\ (-9)\cdot 0-3\cdot 3)\\&=(3-0,\ 3-(-9),\ 0-9)\\&=(3,\ 12,\ -9) \end{align}$$
(3)
$$ \small \begin{align}(A+B)\times C&=((1,\ 3,\ 0)+(0,\ 1,\ -3))\times (3,\ 0,\ 1)\\&=(1,\ 4,\ -3)\times (3,\ 0,\ 1)\\&=(4\cdot 1-(-3)\cdot 0,\ (-3)\cdot 3-1\cdot 1,\ 1\cdot 0-4\cdot 3)\\&=(4-0,\ (-9)-1,\ 0-12)\\&=(4,\ -10,\ -12)\end{align}$$
(4)
$$ \small \begin{align}(A\times C)\cdot (B\times C)&=(a_{2}c_{3}-a_{3}c_{2},\ a_{3}c_{1}-a_{1}c_{3},\ a_{1}c_{2}-a_{2}c_{1})\cdot (b_{2}c_{3}-b_{3}c_{2},\ b_{3}c_{1}-b_{1}c_{3},\ b_{1}c_{2}-b_{2}c_{1})\\&=(3\cdot 1-0\cdot 0,\ 0\cdot 3-1\cdot 1,\ 1\cdot 0-3\cdot 3)\cdot (1\cdot 1-(-3)\cdot 0,\ (-3)\cdot 3-0\cdot 1,\ 0\cdot 0-1\cdot 3)\\&=(3-0,\ 0-1,\ 0-9)\cdot (1-0,\ (-9)-0,\ 0-3)\\&=(3,\ -1,\ -9)\cdot (1,\ -9,\ -3)\\&=3\cdot 1+(-1)\cdot (-9)+(-9)\cdot (-3)\\&=3+9+27\\&=39\end{align}$$
16. 벡터 A=(1, -1, -1), B=(2, -2, 2)에 의해 만들어지는 평행사변형의 넓이를 구하라.
벡터로 만들어지는 평행사변형 넓이(=외적의 절대값) : |AxB|
$$ \small \begin{align}|A\times B|&=\sqrt{(a_{2}b_{3}-a_{3}b_{2})^{2}+(a_{3}b_{1}-a_{1}b_{3})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}\\&=\sqrt{((-1)\cdot 2-(-1)\cdot (-2))^{2}+((-1)\cdot 2-1\cdot 2)^{2}+(1\cdot (-2)-(-1)\cdot 2)^{2}}\\&=\sqrt{((-2)-2)^{2}+((-2)-2)^{2}+((-2)-(-2))^{2}}\\&=\sqrt{(-4)^{2}+(-4)^{2}+(0)^{2}}\\&=\sqrt{16+16+0}\\&=\sqrt{32}\\&=4\sqrt{2}\end{align}$$
17. 문제 9번에서 A와 B의 사이각θ를 기하학적 방법으로 구하라.
$$ \small 평행사변형\ 넓이의\ 기하학적\ 방법\ :\ |A\times B| = |A||B|sin\theta$$
$$ \small \begin{align}|A\times B|&=|A||B|sin\theta\\&=\sqrt{1^{2}+(-1)^{2}+(-1)^{2}}\cdot \sqrt{2^{2}+(-2)^{2}+2^{2}}sin\theta\\&=\sqrt{1+1+1}\cdot \sqrt{4+4+4}sin\theta \\&=\sqrt{3}\cdot \sqrt{12}sin\theta\\&=\sqrt{36}sin\theta \\&=6sin\theta \end{align}$$
$$ \small 6sin\theta = 4\sqrt{2}$$
$$\begin{align}sin\theta&=\frac{4\sqrt{2}}{6}\\&=\frac{2\sqrt{2}}{3} \end{align}$$
18. A=(3, 0, 0), B=(0, 2, 0), C=(0, 0, 2)가 이루는 평행육면체의 부피를 구하라.
$$ \small 평행육면체의\ 부피\ :\ (A\times B)\cdot C=\begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}$$
$$ \small \begin{align}(A\times B)\cdot C&=\begin{vmatrix}3&0&0\\0&2&0\\0&0&2\end{vmatrix}\\&=|3\cdot 2\cdot 2|\\&=|12|\\&=12\end{align}$$
19. 다음 주어진 세 점을 지나는 평면의 방정식을 구하라.
(1) (1, 1, -1), (1, -1, 1), (1, 1, 1)
$$ \small (2)\ (1,\ 2,\ 1),\ (-1,\ 3,\ 0),\ (0,\ 1,\ -\frac{1}{2}) $$
평면에 대한 수직인 벡터를 찾고, 두 벡터의 외적을 구하자.
(1)
$$ \small \begin{align}\overrightarrow{AB}&=(1,\ -1,\ 1)-(1,\ 1,\ -1)\\&=(0,\ -2,\ 2)\end{align}$$
$$ \small \begin{align}\overrightarrow{AC}&=(1,\ 1,\ 1)-(1,\ 1,\ -1)\\&=(0,\ 0,\ 2)\end{align}$$
$$ \small \begin{align}\overrightarrow{AB}\times \overrightarrow{AC}&=(a_{2}b-{3}-a_{3}b_{2},\ a_{3}b_{1}-a_{1}b_{3},\ a_{1}b_{2}-a_{2}b_{1})\\&=((-2)\cdot 2-2\cdot 0,\ 3\cdot 0-0\cdot 2,\ 0\cdot 0-(-2)\cdot 0)\\&=((-4)-0,\ 0-0,\ 0-0)\\&=(-4,\ 0,\ 0)\\&=(1,\ 0,\ 0)\end{align}$$
$$ \small \begin{align}a(x-x_{1})+b(y-y_{1})+c(z-z_{1})&=1(x-1)+0(y-1)+0(z-(-1)\\&=x-1+0+0\\&=x-1\end{align}$$
x-1=0
(2)
$$ \small \begin{align}\overrightarrow{AB}&=(-1,\ 3,\ 0)-(1,\ 2,\ 1)\\&=(-2,\ 1,\ -1)\end{align}$$
$$ \small \begin{align}\overrightarrow{AC}&=(0,\ 1,\ -\frac{1}{2})-(1,\ 2,\ 1)\\&=(-1,\ -1,\ -\frac{3}{2})\end{align}$$
$$ \small \begin{align}\overrightarrow{AB}\times \overrightarrow{AC}&=(a_{2}b-{3}-a_{3}b_{2},\ a_{3}b_{1}-a_{1}b_{3},\ a_{1}b_{2}-a_{2}b_{1})\\&=(1\cdot (-\frac{3}{2})-(-1)\cdot (-1),\ (-1)\cdot (-1)-(-2)\cdot (-\frac{3}{2}),\ (-2)\cdot (-1)-1\cdot (-1))\\&=(-\frac{3}{2}-1,\ 1-3,\ 2-(-1))\\&=(-\frac{5}{2},\ -2,\ 3)\end{align}$$
$$ \small \begin{align}a(x-x_{1})+b(y-y_{1})+c(z-z_{1})&=(-\frac{5}{2})\cdot (x-1)+(-2)\cdot (y-2)+3\cdot (z-1)\\&=-\frac{5}{2}x+\frac{5}{2}-2y+4+3z-3\\&=-5x+5-4y+8+6z-6\\&=-5x-4y+6z+7\end{align}$$
$$ \small -5x-4y+6z+7=0 $$