(11)

We have, $\frac{x}{1}=\frac{y–1}{2}=\frac{z–2}{3}=t$ (say)

$\overrightarrow{QR}·(\hat{i}+2\hat{j}+3\hat{k})=0$

$\Rightarrow (t–1)+(2t–5)\times 2+(3t–2)\times 3=0 \Rightarrow t=\frac{17}{14}$

$\Rightarrow R\equiv (\frac{17}{14},\frac{48}{14},\frac{79}{14})$

Since, R is the mid point of PQ

$\therefore \frac{\alpha +1}{2}=\frac{17}{14}, \frac{\beta +6}{2}=\frac{48}{14}, \frac{\gamma +4}{2}=\frac{79}{14}$

$\Rightarrow \alpha =\frac{10}{7}, \beta =\frac{6}{7}, \gamma =\frac{51}{7}$

Hence, $2\alpha +\beta +\gamma =2\times \frac{10}{7}+\frac{6}{7}+\frac{51}{7}=\frac{20+6+51}{7}=\frac{77}{7}=11$.

(62)

Let $L : \frac{x–1}{3}=\frac{y}{2}=\frac{z–2}{4}=\lambda$ be the given line.

Any point on L is given by $P(3\lambda +1,2\lambda ,4\lambda +2)$

Direction ratios of L is given by $3\hat{i}+2\hat{j}+4\hat{k}$

Now, AP $\perp$ L so we have

$3(3\lambda +1–6)+2(2\lambda –1)+4(4\lambda +2–5)=0$

$\Rightarrow 9\lambda –15+4\lambda –2+16\lambda –12=0$

$\Rightarrow 29\lambda =29 \Rightarrow \lambda =1$

So, P(4, 2, 6) is the mid point of AB

$\Rightarrow 4=\frac{x+6}{2}, 2=\frac{y+1}{2}, 6=\frac{z+5}{2}$

$\Rightarrow$ B(x, y, z) = (2, 3, 7) is the image of A

Required distance = $\sqrt{4+9+49}=\sqrt{62}$.

(21)

We have, $L_1 : \overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (\hat{i}–\hat{j}+\hat{k})$

$L_2 : \overrightarrow{r}=(4\hat{i}+5\hat{j}+6\hat{k})+\mu (\hat{i}+\hat{j}–\hat{k})$

Here, ${\overrightarrow{b}}_1=\hat{i}–\hat{j}+\hat{k}, {\overrightarrow{b}}_2=\hat{i}+\hat{j}–\hat{k}$

${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=$$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& –1& 1\\ 1& 1& –1\end{array}\right|$$=0\hat{i}+2\hat{j}+2\hat{k}$

Direction ratios of line perpendicular to $L_1$ and $L_2$

$=\left(\right(4+\mu )–(1+\lambda ), (5+\mu )–(2–\lambda ), (6–\mu )–(3+\lambda \left)\right)$

$=3+\mu –\lambda , 3+\mu +\lambda , 3–\mu –\lambda$

$\therefore \frac{3+\mu –\lambda }{0}=\frac{3+\mu +\lambda }{2}=\frac{3–\mu –\lambda }{2}$

On solving, we get

$\lambda =\frac{3}{2}, \mu =\frac{–3}{2}$

Point $P\equiv (\frac{5}{2},\frac{1}{2},\frac{9}{2})$ and $Q\equiv (\frac{5}{2},\frac{7}{2},\frac{15}{2})$

$\therefore$  Mid point of AB = $(\frac{5}{2},2,6)=(\alpha ,\beta ,\gamma )$

$\therefore 2(\alpha +\beta +\gamma )$ = 5 + 4 + 12 = 21.

(108)

We have, $\frac{x–2}{2}=\frac{y}{–2}=\frac{z–7}{16}=\lambda$ (say)

$\Rightarrow x=2\lambda +2, y=–2\lambda , z=16\lambda +7$

Also, $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}=\mu$ (say)

$\Rightarrow x=4\mu –3, y=3\mu –2, z=\mu –2$

Lines are intersecting at point P.

$\therefore 2\lambda +2=4\mu –3 \Rightarrow 2\lambda –4\mu =–5$          ... (i)

    $–2\lambda =3\mu –2 \Rightarrow –2\lambda –3\mu =–2$         ... (ii)

On solving (i) and (ii), we get

$\lambda =–\frac{1}{2}$ and $\mu =1$

$\therefore$  Point P is (1, 1, –1)

Now, $\frac{x+1}{2}=\frac{y–1}{3}=\frac{z–1}{1}=k$ (say)

$\Rightarrow$ x = 2k – 1, y = 3k + 1, z = k + 1

D.r.'s of PQ : 2k – 2, 3k, k + 2

D.r.'s of line $\frac{x+1}{2}=\frac{y–1}{3}=\frac{z–1}{1}$ is 2, 3, 1

As both line are perpendicular to each other.

     2(2k –2) + 3(3k) + 1(k +2) = 0

$\Rightarrow 14k=2 \Rightarrow k=\frac{1}{7}$

Thus, point Q is $(–\frac{5}{7},\frac{10}{7},\frac{8}{7})$

$\therefore PQ=\sqrt{{(1+\frac{5}{7})}^2+{(1–\frac{10}{7})}^2+{(–1–\frac{8}{7})}^2}=\sqrt{\frac{378}{49}}$

Also, $P{Q}^2=l^2=\frac{378}{49} \Rightarrow 14l^2=\frac{378}{49}\times 14=108$.

(65)

We have, $l_1$ : $\frac{x}{1}=\frac{y+2}{1}=\frac{z}{1}=\lambda \in R$

Coordinates of point P are $(\lambda , \lambda –2, \lambda )$

      $l_2$ : $\frac{x+2}{2}=\frac{y}{1}=\frac{z}{1}=\mu \in R$

Coordinates of point Q are $(2\mu –2, \mu , \mu )$

D.r.'s of PQ are $(2\mu –2–\lambda , \mu –\lambda +2, \mu –\lambda )$

Also, D.r.'s of line PQ is (2, 1, 2)

$\therefore \frac{2\mu –2–\lambda }{2}=\frac{\mu –\lambda +2}{1}=\frac{\mu –\lambda }{2}$

$\Rightarrow \lambda =6, \mu =2$

$\therefore$  Coordinates of point P are (6, 4, 6) and coordinates of point Q are (2, 2, 2).

Equations of line PQ is $\frac{x–2}{2}=\frac{y–2}{1}=\frac{z–2}{2}=k\in R$

Now, from condition for perpendicularity,

Since, AB $\perp$ PQ, then,

    2(2k + 1) + (1)k + 2(2k – 10) = 0

$\Rightarrow 9k=18 \Rightarrow k=2$

Therefore, point A is (6, 4, 6)

Now, perpendicular distance from B(1, 2, 12) to line PQ is given by

 $l$$=\sqrt{{(6–1)}^2+{{(4–2)}^2+(6–12)}^2}$$\Rightarrow l^2$ = 25 + 4 + 36 = 65.

(9)

Let, $L_1 : \frac{x–5}{4}=\frac{y–4}{1}=\frac{z–5}{3}=\lambda \in R$

$L_2 : \frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu \in R$

M is a point on $L_1$.

So, $M=(4\lambda +5, \lambda +4, 3\lambda +5)$

and N is a point on $L_2$.

So, $N=(12\mu –8, 5\mu –2, 9\mu –11)$

The direction ratios of MN is

$(12\mu –4\lambda –13, 5\mu –\lambda –6, 9\mu –3\lambda –16)$

$\because MN\perp L_1 \mathrm{and} MN\perp L_2$

$\therefore 4(12\mu –4\lambda –13)+5\mu –\lambda –6+3(9\mu –3\lambda –16)=0$

$\Rightarrow 48\mu –16\lambda –52+5\mu –\lambda -6+27\mu –9\lambda –48=0$

$\Rightarrow 80\mu –26\lambda –106=0 \Rightarrow 40\mu –13\lambda –53=0$          ... (i)

Also,

$12(12\mu –4\lambda –13)+5(5\mu –\lambda –6)+9(9\mu –3\lambda –16)=0$

$144\mu –48\lambda –156+25\mu –5\lambda –30+81\mu –27\lambda –144=0$

$\Rightarrow 250\mu –80\lambda –330=0$$\Rightarrow 25\mu –8\lambda –33=0$          ... (ii)

On solving equation (i) and (ii), we get $\mu =1$ and $\lambda =–1$

M = (1, 3, 2) and N = (4, 3, –2)

$\therefore \overrightarrow{OM} =\hat{i}+3\hat{j}+2\hat{k} \mathrm{and} \overrightarrow{ON} =4\hat{i}+3\hat{j}–2\hat{k}$

    $\overrightarrow{OM}·\overrightarrow{ON}$ = 4 + 9 – 4 = 9.

(16)

We have, $l_1$ : x + 1 = 2y = –12z and $l_2$ : x = y + 2 = 6z – 6

So, $l_1$ : $\frac{x–(–1)}{1}=\frac{y–0}{{\displaystyle \frac{1}{2}}}=\frac{z–0}{{\displaystyle \frac{–1}{12}}}$ and $l_2$ : $\frac{x–0}{1}=\frac{y–(–2)}{{\displaystyle 1}}=\frac{z–1}{{\displaystyle \frac{1}{6}}}$

These lines can be written in vector form as

${\overrightarrow{r}}_1=(–\hat{i}+0\hat{j}+0\hat{k})+\lambda (\hat{i}+\frac{\hat{j}}{2}–\frac{\hat{k}}{12})$ and ${\overrightarrow{r}}_2=(0\hat{i}+2\hat{j}+\hat{k})+\lambda (\hat{i}+\hat{j}+\frac{1}{6}\hat{k})$

For, $\overrightarrow{r_1}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{r_2}=\overrightarrow{a_2}+\lambda \overrightarrow{b_2}$

Shortest distance between $l_1$ and $l_2$ is given by

$d_1=$$\left|\frac{({\overrightarrow{a}}_2–{\overrightarrow{a}}_1)·({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}\right|$$=$$\left|\frac{(\hat{i}+2\hat{j}+\hat{k})·\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1/2& –1/12\\ 1& 1& 1/6\end{array}\right|}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}\right|$

$=$$\left|\frac{(\hat{i}+2\hat{j}+\hat{k})·({\displaystyle \frac{1}{6}}\hat{i}+{\displaystyle \frac{1}{4}}\hat{j}+{\displaystyle \frac{1}{2}}\hat{k})}{\sqrt{{\displaystyle \frac{1}{36}+\frac{1}{16}+\frac{1}{4}}}}\right|$$=$$\left|\frac{{\displaystyle \frac{1}{6}+\frac{1}{2}+\frac{1}{2}}}{\sqrt{{\displaystyle \frac{49}{144}}}}\right|$$=\frac{7}{6}\times \frac{12}{7}=2$

Similarly, $l_3$ : $\frac{x–1}{2}=\frac{y+8}{–7}=\frac{z–4}{5}$ and $l_4$ : $\frac{x–1}{2}=\frac{y–2}{1}=\frac{z–6}{–3}$

These lines can be written in vector form as

${\overrightarrow{r}}_1=(\hat{i}–8\hat{j}+4\hat{k})+\lambda (2\hat{i}–7\hat{j}+5\hat{k})$

${\overrightarrow{r}}_2=(\hat{i}+2\hat{j}+6\hat{k})+\lambda (2\hat{i}+\hat{j}–3\hat{k})$

Shortest distance between $l_3$ and $l_4$ is given as

$d_2=$$\left|\frac{(10\hat{j}+2\hat{k})·\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& –7& 5\\ 2& 1& –3\end{array}\right|}{\left|\right|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& –7& 5\\ 2& 1& –3\end{array}\left|\right|}\right|$

$=$$\left|\frac{(10\hat{j}+2\hat{k})·(16\hat{i}+16\hat{j}+16\hat{k})}{\sqrt{{16}^2+{16}^2+{16}^2}}\right|$$=\frac{160+32}{16\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}$

Now, $\frac{32\sqrt{3}{d}_1}{d_2}=\frac{32\sqrt{3}\times 2}{4\sqrt{3}}=16$.

(196)

We have, $L_1 : \frac{x–1}{3}=\frac{y–2}{2}=\frac{z+1}{–2}=\lambda \in R$

$\therefore M\equiv (3\lambda +1,2\lambda +2,–2\lambda –1)$

$L_2 : \frac{x+2}{–3}=\frac{y–2}{–2}=\frac{z–1}{4}=\mu \in R$

$\therefore N\equiv (–3\mu —2,–2\mu +2,4\mu +1)$

$\alpha +\beta +\gamma =3\lambda +2 \Rightarrow a+b+c=–\mu +1$

Direction ratio's of line AM = $<3\lambda +2,2\lambda ,–2\lambda –4>$

Direction ratio's of line AN = $<–3\mu —1,–2\mu ,4\mu –2>$

$\Rightarrow \frac{3\lambda +2}{–3\mu –1}=\frac{2\lambda }{–2\mu }=\frac{–2\lambda –4}{4\mu –2}$

$\Rightarrow \frac{3\lambda +2}{–3\mu –1}=\frac{\lambda }{–\mu } \mathrm{and} \frac{\lambda }{–\mu }=\frac{–2\lambda –4}{4\mu –2}$

$\Rightarrow –3\lambda \mu –2\mu =–3\mu \lambda –\lambda \mathrm{and} 4\mu \lambda –2\lambda =2\mu \lambda +4\mu$

$\Rightarrow 2\mu =\lambda \mathrm{and} 2\mu \lambda =2\lambda +4\mu$

$\Rightarrow {\lambda }^2=2\lambda +2\lambda \Rightarrow {\lambda }^2=4\lambda$

$\Rightarrow \lambda (\lambda –4)=0 \Rightarrow \lambda =0, \lambda =4$

$\therefore \frac{{(\alpha +\beta +\gamma )}^2}{{(a+b+c)}^2}=\frac{{(3\times 4+2)}^2}{{(–2+1)}^2}=\frac{{\left(14\right)}^2}{1}=196$.

(12)

Line $L_1$ is given by y = x, z = 1 can be expressed as

$L_1 : \frac{x}{1}=\frac{y}{1}=\frac{z–1}{0}=\alpha$

$\Rightarrow x=\alpha ,y=\alpha ,z=1$

Let the coordinate of Q on $L_1$ be $(\alpha , \alpha , 1)$

Line $L_2$ given by y = –x, z = –1 can be expressed as

$L_2 : \frac{x}{1}=\frac{y}{–1}=\frac{z+1}{0}=\beta$ (say)

$\Rightarrow x=\beta ,y=–\beta ,z=–1$

Let the coordinates of R on $L_2$ be $(\beta , –\beta , –1)$

Direction ratios of PQ are $(a–\alpha , a–\alpha , a–1)$.

Now $PQ\perp L_1$

$\therefore 1(a–\alpha )+1(a–\alpha )+0(a–1)=0 \Rightarrow a=\alpha$

Hence Q(a, a, 1)

Direction ratios of PR are $a–\beta , a+\beta , a+1$

Now $PR\perp L_2$

$\therefore 1(a–\beta )+(–1)(a+\beta )+0(a+1)=0 \Rightarrow \beta =0$

Hence R(0, 0, —1)

Now, as $\angle QPR=90°$

$\Rightarrow$ (a – a)(a – 0) + (a – a)(a – 0) + (a – 1)(a + 1) = 0

$\Rightarrow$ (a –1)(a + 1) = 0 $\Rightarrow$ a = 1 or a = –1

    a = 1, rejected as P and Q are different points

$\Rightarrow$ a = –1, then $12a^2=12\times {(–1)}^2=12$.

(22)

Equation of the line through A(4, –6, –2) and B(16, –2, 4) is

$\frac{x–4}{16–4}=\frac{y+6}{–2+6}=\frac{z+2}{4+2}=\lambda , \lambda \in R$

$\Rightarrow \frac{x–4}{12}=\frac{y+6}{4}=\frac{z+2}{6}=\lambda$

$\therefore x=12\lambda +4, y=4\lambda –6,z=6\lambda –2$

Let $a=12\lambda +4, b=4\lambda –6,c=6\lambda –2$.

Now ${(12\lambda +4–4)}^2+{(4\lambda –6+6)}^2+{(6\lambda –2+2)}^2={\left(21\right)}^2$

$144{\lambda }^2+16{\lambda }^2+36{\lambda }^2=441$

$\Rightarrow 196{\lambda }^2=441$

${\lambda }^2=\frac{441}{196} \Rightarrow \lambda =\pm \frac{21}{14}=\pm \frac{3}{2}$

When $\lambda =\frac{3}{2}$

$a=12\times \frac{3}{2}+4=22; b=4\times \frac{3}{2}–6=0; c=6\times \frac{3}{2}–2=7$

When $\lambda =\frac{–3}{2}$

$a=12\times \frac{–3}{2}+4=–14; b=4\times \frac{–3}{2}–6=–12; c=6\times \frac{–3}{2}–2=–11$

Distance between (22, 0, 7) and (4, –12, 3)

$=\sqrt{{18}^2+{12}^2+4^2}=\sqrt{484}=22$.