If the distances of the point (1,2, a) from the line (x-1)/(1)=(y)/(2)=(z-1)/(1) along the…

Question

If the distances of the point $(1,2, a)$ from the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $\mathrm{L}_2: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equ...

Accepted Answer

Let line $L:\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$, point $P(1,2,a)$.

$\begin{gathered} {L}_{1}:\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}={\lambda }_{1}, \ {L}_{2}:\frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}={\lambda }_{2}. \end{gathered}$

Aany point $Q$ on $L$ is $(t+1,2t,t+1)$.

Let ${Q}_{1}({t}_{1}+1,2{t}_{1},{t}_{1}+1)$ be point on $L$ and ${d}_{1}$ be the distance measured from $P(1,2,a)$ along ${L}_{1}$ such that $|P{Q}_{1}|={d}_{1}$.

So, direction ratio of $\overset{	o \leftarrow }{P{Q}_{1}}$ and line ${L}_{1}$ are proportional.

Direction ratio of $\overset{	o \leftarrow }{P{Q}_{1}}$

$\begin{gathered} ={Q}_{1}-P=({t}_{1}+1,2{t}_{1},{t}_{1}+1)-(1,2,a)=({t}_{1},2{t}_{1}-2,{t}_{1}+1-a) \end{gathered}$

Direction ratio of ${L}_{1}=(3,4,b)$.

So,

$\frac{{t}_{1}}{3}=\frac{2{t}_{1}-2}{4}=\frac{{t}_{1}-a+1}{b}$

Equating

$\frac{{t}_{1}}{3}=\frac{2{t}_{1}-2}{4}\, \, \frac{{t}_{1}}{3}=\frac{{t}_{1}-a+1}{b}$

$4{t}_{1}=6{t}_{1}-6\,$ and put ${t}_{1}=3$

${t}_{1}=3\,$ and $\, \frac{3}{3}=\frac{3-a+1}{b}\Rightarrow a+b=4$\ldots \ldots .(1)

Similarly, let ${Q}_{2}({t}_{2}+1,2{t}_{2},{t}_{2}+1)$ be point on $L$ and ${d}_{2}$ be the distance measured from $P(1,2,a)$ along ${L}_{2}$ such that $|P{Q}_{2}|={d}_{2}$.

so, direction ratio of $\overset{	o \leftarrow }{P{Q}_{2}}$ and ${d}_{2}$ are proportional.

direction ratio of $\overset{	o \leftarrow }{PQ}$,

$\begin{gathered} ={Q}_{2}-P=({t}_{2}+1,2{t}_{2},{t}_{2}+1)-(1,2,a) \ =({t}_{2},2{t}_{2}-2,{t}_{2}+1-a) \end{gathered}$

Direction ratio of ${L}_{2}=(1,4,c)$.

$\frac{{t}_{2}}{1}=\frac{2{t}_{2}-2}{4}=\frac{{t}_{2}+1-a}{c}$

Equating

$\frac{{t}_{2}}{1}=\frac{2{t}_{2}-2}{4}\, \, \frac{{t}_{2}}{1}=\frac{{t}_{2}+1-a}{c}$

$4{t}_{2}=2{t}_{2}-2\,$ and put ${t}_{2}=-1$

$2{t}_{2}=-2\,$ and $\, \frac{-1}{1}=\frac{-1+1-a}{c}$

$\Rightarrow$ $-1=\frac{-a}{c}\Rightarrow a=c$ ........(2)

Now, calculating

${d}_{2}^{1}={|P{Q}_{1}|}^{2}={t}_{2}^{1}+{(2{t}_{1}-2)}^{2}+{({t}_{1}+1-a)}^{2},{t}_{1}=3$

$\Rightarrow$ ${d}_{2}^{1}=(3{)}^{2}+(2	imes 3-2{)}^{2}+(3+1-a{)}^{2}$

$=9+16+(4-a{)}^{2}=25+16+{a}^{2}-8a$

$={a}^{2}-8a+41$

Similarly,

${d}_{2}^{2}={|P{Q}_{2}|}^{2}={t}_{2}^{2}+{(2{t}_{2}-2)}^{2}+{({t}_{2}+1-a)}^{2}$

$\Rightarrow$ ${d}_{2}^{2}=(-1{)}^{2}+(2	imes (-1)-2{)}^{2}+(-1+1-a{)}^{2}{$ use ${t}_{2}=-1}$

$\Rightarrow$ ${d}_{2}^{2}=1+16+{a}^{2}=17+{a}^{2}$

It is given that $|P{Q}_{1}|=|P{Q}_{2}|\Rightarrow {d}_{2}^{1}={d}_{2}^{2}$

$\Rightarrow$ ${a}^{2}-8a+41=17+{a}^{2}$

$\Rightarrow -8a+41=17$

$\Rightarrow$ $8a=41-17=24\Rightarrow a=3$

from (1) \& (2)

$b=4-a=4-3=1\, \, c=a=3$

So value of $a+b+c=3+1+3=7$

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