Chapter 11: Three Dimensional Geometry (Vectors and 3D Geometry)

$l=\cos 90^{\circ }=0$, $m=\cos 135^{\circ }=-\frac{1}{\sqrt{2}}$, $n=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$.

$0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}$

$l=m=n$ and $l^2+m^2+n^2=1\Rightarrow 3l^2=1\Rightarrow l=\pm \frac{1}{\sqrt{3}}$.

$\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}}$

$\sqrt{(-18{)}^2+12^2+(-4{)}^2}=\sqrt{324+144+16}=\sqrt{484}=22$. Divide each ratio by 22.

$-\frac{9}{11},\frac{6}{11},-\frac{2}{11}$

Let A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7). Direction ratios of AB: $-3,-5,-3$; of AC: $3,5,3$.

These are proportional (AC = −1 × AB), so AB ∥ AC, and they share point A.

A, B, C lie on one line.

A(3, 5, −4), B(−1, 1, 2), C(−5, −5, −2).

AB: ratios $(-4,-4,6)$, length $\sqrt{16+16+36}=2\sqrt{17}$: $-\frac{2}{\sqrt{17}},-\frac{2}{\sqrt{17}},\frac{3}{\sqrt{17}}$.

BC: ratios $(-4,-6,-4)$, length $2\sqrt{17}$: $-\frac{2}{\sqrt{17}},-\frac{3}{\sqrt{17}},-\frac{2}{\sqrt{17}}$.

CA: ratios $(8,10,-2)$, length $\sqrt{168}=2\sqrt{42}$: $\frac{4}{\sqrt{42}},\frac{5}{\sqrt{42}},-\frac{1}{\sqrt{42}}$.

AB: $\left(-\frac{2}{\sqrt{17}},-\frac{2}{\sqrt{17}},\frac{3}{\sqrt{17}}\right)$; BC: $\left(-\frac{2}{\sqrt{17}},-\frac{3}{\sqrt{17}},-\frac{2}{\sqrt{17}}\right)$; CA: $\left(\frac{4}{\sqrt{42}},\frac{5}{\sqrt{42}},-\frac{1}{\sqrt{42}}\right)$

$l_1{l}_2+m_1{m}_2+n_1{n}_2$ for each pair:

1 and 2: $\frac{48-36-12}{169}=0$; 2 and 3: $\frac{12-48+36}{169}=0$; 1 and 3: $\frac{36+12-48}{169}=0$.

Each pair is perpendicular.

Direction ratios: $(2,5,-4)$ and $(3,2,4)$. $2(3)+5(2)+(-4)(4)=6+10-16=0$.

Perpendicular.

Direction ratios: $(-2,-4,-4)$ and $(2,4,4)$: proportional (one is −1 times the other).

Parallel.

$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$, $\vec{b}=3\hat{i}+2\hat{j}-2\hat{k}$.

$\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (3\hat{i}+2\hat{j}-2\hat{k})$; i.e. $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z-3}{-2}$

$\vec{r}=(2\hat{i}-\hat{j}+4\hat{k})+\lambda (\hat{i}+2\hat{j}-\hat{k})$; $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}$

The required line has the same direction ratios 3, 5, 6.

$\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$

Point (5, −4, 6), direction (3, 7, 2).

$\vec{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\lambda (3\hat{i}+7\hat{j}+2\hat{k})$

(i) ${\vec{b}}_1=(3,2,6)$, $|{\vec{b}}_1|=7$; ${\vec{b}}_2=(1,2,2)$, $|{\vec{b}}_2|=3$. $\cos \theta =\frac{3+4+12}{21}=\frac{19}{21}$.

(ii) ${\vec{b}}_1=(1,-1,-2)$, $|{\vec{b}}_1|=\sqrt{6}$; ${\vec{b}}_2=(3,-5,-4)$, $|{\vec{b}}_2|=5\sqrt{2}$. $\cos \theta =\frac{3+5+8}{\sqrt{6}\cdot 5\sqrt{2}}=\frac{16}{10\sqrt{3}}=\frac{8}{5\sqrt{3}}$.

(i) $\theta ={\cos }^{-1}\left(\frac{19}{21}\right)$ (ii) $\theta ={\cos }^{-1}\left(\frac{8}{5\sqrt{3}}\right)$

(i) $(2,5,-3)$ and $(-1,8,4)$: $\cos \theta =\frac{-2+40-12}{\sqrt{38}\sqrt{81}}=\frac{26}{9\sqrt{38}}$.

(ii) $(2,2,1)$ and $(4,1,8)$: $\cos \theta =\frac{8+2+8}{3\times 9}=\frac{2}{3}$.

(i) $\theta ={\cos }^{-1}\left(\frac{26}{9\sqrt{38}}\right)$ (ii) $\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$

Standard forms: $\frac{x-1}{-3}=\frac{y-2}{2p/7}=\frac{z-3}{2}$ and $\frac{x-1}{-3p/7}=\frac{y-5}{1}=\frac{z-6}{-5}$.

Perpendicular: $(-3)\left(-\frac{3p}{7}\right)+\frac{2p}{7}(1)+2(-5)=0\Rightarrow \frac{11p}{7}=10$.

$p=\frac{70}{11}$

$7(1)+(-5)(2)+1(3)=7-10+3=0$.

Perpendicular.

${\vec{a}}_1=(1,2,1)$, ${\vec{b}}_1=(1,-1,1)$; ${\vec{a}}_2=(2,-1,-1)$, ${\vec{b}}_2=(2,1,2)$.

${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ 2& 1& 2\end{array}\right|=-3\hat{i}+0\hat{j}+3\hat{k}$, magnitude $3\sqrt{2}$.

${\vec{a}}_2-{\vec{a}}_1=\hat{i}-3\hat{j}-2\hat{k}$; $({\vec{b}}_1\times {\vec{b}}_2)\cdot ({\vec{a}}_2-{\vec{a}}_1)=-3+0-6=-9$.

$d=\frac{|-9|}{3\sqrt{2}}$

$\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$ units

${\vec{a}}_1=(-1,-1,-1)$, ${\vec{b}}_1=(7,-6,1)$; ${\vec{a}}_2=(3,5,7)$, ${\vec{b}}_2=(1,-2,1)$.

${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 7& -6& 1\\ 1& -2& 1\end{array}\right|=-4\hat{i}-6\hat{j}-8\hat{k}$, magnitude $\sqrt{116}=2\sqrt{29}$.

${\vec{a}}_2-{\vec{a}}_1=(4,6,8)$; dot $=-16-36-64=-116$.

$d=\frac{116}{2\sqrt{29}}=\frac{58}{\sqrt{29}}$

$2\sqrt{29}$ units

${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 2\\ 2& 3& 1\end{array}\right|=-9\hat{i}+3\hat{j}+9\hat{k}$, magnitude $\sqrt{171}=3\sqrt{19}$.

${\vec{a}}_2-{\vec{a}}_1=3\hat{i}+3\hat{j}+3\hat{k}$; dot $=-27+9+27=9$.

$d=\frac{9}{3\sqrt{19}}$

$\frac{3}{\sqrt{19}}$ units

Rewrite: $\vec{r}=(\hat{i}-2\hat{j}+3\hat{k})+t(-\hat{i}+\hat{j}-2\hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2\hat{j}-2\hat{k})$.

${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 1& -2\\ 1& 2& -2\end{array}\right|=2\hat{i}-4\hat{j}-3\hat{k}$, magnitude $\sqrt{29}$.

${\vec{a}}_2-{\vec{a}}_1=\hat{j}-4\hat{k}$; dot $=-4+12=8$.

$\frac{8}{\sqrt{29}}$ units

$a(b-c)+b(c-a)+c(a-b)=ab-ac+bc-ab+ac-bc=0$, so $\cos \theta =0$.

90°

Direction ratios of the x-axis: 1, 0, 0; point (0, 0, 0).

$\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$ (vector form $\vec{r}=\lambda \hat{i}$)

$(-3)(3k)+(2k)(1)+2(-5)=0\Rightarrow -7k-10=0$.

$k=-\frac{10}{7}$

${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 2\\ 3& -2& -2\end{array}\right|=8\hat{i}+8\hat{j}+4\hat{k}$, magnitude 12.

${\vec{a}}_2-{\vec{a}}_1=-10\hat{i}-2\hat{j}-3\hat{k}$; dot $=-80-16-12=-108$.

$d=\frac{108}{12}$

9 units

The direction is along ${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -16& 7\\ 3& 8& -5\end{array}\right|=24\hat{i}+36\hat{j}+72\hat{k}=12(2\hat{i}+3\hat{j}+6\hat{k})$.

$\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$