Let a line L_1 pass through the origin and be perpendicular to the lines L_2: r=(3+t)i+(2t…

Question

Let a line ${L}_{1}$ pass through the origin and be perpendicular to the lines

${L}_{2}:\overset{	o }{r}=(3+t)\hat{i}+(2t-1)\hat{j}+(2t+4)\hat{k}$ and

${L}_{3}:\overset{	o }{r}=(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k},t,s\in R$.

If $(a,b,c),a\in Z$, is the point on ${L}_{3}$ at a distance of $\sqrt{17}$ from the point of intersection of ${L}_{1}$ and ${L}_{2}$, then $(a+b+c{)}^{2}$ is equal to $____$ .

Answer: 4

Accepted Answer

A pair of lines in 3D have the vector equations :

$\begin{gathered} {L}_{2} \quad  {L}_{3} \quad  (t,s) \ \overset{	o }{r}=(3+t)\hat{i}+(2t-1)\hat{j}+(2t+4)\hat{k} \quad  \overset{	o }{r}=(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k} \quad  \in R \end{gathered}$

A third line ${L}_{1}$ passes through the origin and is \perp  to both the lines ${L}_{2}$ and ${L}_{3}$ defined just above

The point of intersection of ${L}_{1}$ and ${L}_{2}$ is at a distance $\sqrt{17}$ units from another point with coordinates $(a,b,c)$ and lying on ${L}_{3}$ where $a\in Z$

$\begin{gathered} {L}_{2}:\frac{x-3}{1}=\frac{y+1}{2}=\frac{z-4}{2}=t \ {L}_{5}:\frac{x-3}{2}=\frac{y-3}{2}=\frac{z-2}{1}=s \ \perp {L}_{2}{L}_{3} \end{gathered}$

$\begin{gathered} \Rightarrow |\begin{array}{ccc}i \quad  j \quad  k \ 1 \quad  2 \quad  2 \ 2 \quad  2 \quad  1\end{array}|=\hat{i}(-2)-\hat{j}(-3)+\hat{k}(-2) \ =-2\hat{i}+3\hat{j}-2\hat{k} \ =-(2\hat{i}-3\hat{j}+2\hat{k}) \end{gathered}$

Now, ${L}_{1}:\frac{x}{2}=\frac{y}{-3}=\frac{z}{2}=l$

intersection of ${L}_{1}$ \& ${L}_{2}$

$\Rightarrow t+3=2l,2t-1=-3l,2t+4=2l$

$\Rightarrow$ $t+3=2t+4$

$\begin{gathered} t=-1 \ \Rightarrow P\equiv (2,-3,2) \end{gathered}$

A point on ${L}_{3}$ is $Q(2s+3,2s+3,s+2)$

$\begin{gathered} \Rightarrow P{Q}^{2}=(2s+1{)}^{2}+(2s+6{)}^{2}+{s}^{2}=17 \ \Rightarrow 9{s}^{2}+28s+20=0 \ \Rightarrow (9s+10)(s+2)=0\Rightarrow s=-2 \ 	herefore Q\equiv (-1,-1,0)\equiv (a,b,c)\Rightarrow (a+b+c{)}^{2}=4 \end{gathered}$

Solutions · 1