Let the lines L_1:r=i+2j+3k+ (2i+3j+4k), R and L_2:r=(4i+j)+ (5i+2j+k), R, intersect at th…

Question

Let the lines ${L}_{1}:\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (2\hat{i}+3\hat{j}+4\hat{k}),\lambda \in R$ and ${L}_{2}:\vec{r}=(4\hat{i}+\hat{j})+\mu (5\hat{i}+2\hat{j}+\hat{k}),\mu \in R$, intersect at the point R . Let P and Q be the points lying on lines ${L}_{1}$ and ${L}_{2}$, respectively, such that $|\overset{	o \leftarrow }{PR}|=\sqrt{29}$ and $|\overset{	o \leftarrow }{PQ}|=\sqrt{\frac{47}{3}}$. If the point P lies in the first octant, then $27(QR{)}^{2}$ is equal to

Accepted Answer

$\begin{gathered} {L}_{1} :\vec{r}=\hat{i}+\hat{j}+3\hat{k}+\lambda (2\hat{i}+3\hat{j}+4\hat{k}) \ {L}_{2} :\vec{r}=(4\hat{i}+\hat{j})+\mu (5\hat{i}+2\hat{j}+\hat{k}) \end{gathered}$

at the intersection point R , position vectors are equal.

$(\hat{i}+2\hat{j}+3\hat{k})+\lambda (2\hat{i}+3\hat{j}+4\hat{k})=(4\hat{i}+\hat{j})+\mu (5\hat{i}+2\hat{j}+\hat{k})$

equating components of $\hat{i},\hat{j}$ and $\hat{k}$.

$\begin{gathered} 1+2\lambda =4+5\mu \implies 2\lambda -5\mu =3 \ 2+3\lambda =1+2\mu \implies 3\lambda -2\mu =-1 \ 3+4\lambda =\mu \implies \mu -4\lambda =3......(3) \ 	imes 3	imes 2 \ -15\mu +4\mu =9+2\implies -11\mu =11\implies \mu =-1 \end{gathered}$

substitute in equation (1)

$2\lambda -5(-1)=3\implies \lambda =\frac{3-5}{2}=-1$

Now, checking equation (3) $(-1)-4(-1)=3=$ R.H.S consistent.

Now finding point R using $\lambda =-1$ in ${L}_{1}$

$\begin{gathered} \overset{	o \leftarrow }{OR}=(\hat{i}+2\hat{j}+3\hat{k})-(2\hat{i}+3\hat{j}+4\hat{k}) \ \overset{	o \leftarrow }{OR}=-\hat{i}-\hat{i}-\hat{k} \end{gathered}$

Point P lies on ${L}_{1}$ so its position vector is $\, \overset{	o \leftarrow }{OP}=(1+2\lambda )\hat{i}+(2+3\lambda )\hat{j}+(3+4\lambda )\hat{k}$ It is given that

$\begin{gathered} |\overset{	o \leftarrow }{PR}|=\sqrt{29}\Rightarrow |\overset{	o \leftarrow }{PR}{|}^{2}=29 \ \overset{	o \leftarrow }{PR}=\overset{	o \leftarrow }{OR}-\overset{	o \leftarrow }{OP} \ \overset{	o \leftarrow }{PR}=(-2-2\lambda )\hat{i}+(-3-3\lambda )\hat{j}+(-4-4\lambda )\hat{k} \ |\overset{	o \leftarrow }{PR}{|}^{2}=4(1+\lambda {)}^{2}+9(1+\lambda {)}^{2}+16(1+\lambda {)}^{2}=29 \ 29(1+\lambda {)}^{2}=29\Rightarrow (1+\lambda {)}^{2}=1\Rightarrow \lambda =0,-2 \end{gathered}$

point $P$ lies in first octant, means position vector of $P$ must be positive, so $\lambda =0$

$\overset{	o \leftarrow }{OP}=\hat{i}+2\hat{j}+3\hat{k}$

similarly Q lies on ${L}_{2}$, so position vector of Q is

$\begin{gathered} \overset{	o \leftarrow }{OQ}=(4+5\mu )\hat{i}+(1+2\mu )\hat{j}+\mu \hat{k} \ |\overset{	o \leftarrow }{PQ}|=\sqrt{\frac{47}{3}}\Rightarrow |\overset{	o \leftarrow }{PQ}{|}^{2}=\frac{47}{3}\, \ \overset{	o \leftarrow }{PQ}=\overset{	o \leftarrow }{OQ}-\overset{	o \leftarrow }{OP} \ \overset{	o \leftarrow }{PQ}=(3+5\mu )\hat{i}+(-1+2\mu )\hat{j}+(\mu -3)\hat{k} \ (3+5\mu {)}^{2}+(-1+2\mu {)}^{2}+(\mu -3{)}^{2}=\frac{47}{3} \ 30{\mu }^{2}+20\mu +19=\frac{47}{3} \ 90{\mu }^{2}+60\mu +57=47 \ 90{\mu }^{2}+60\mu +10=0 \ 9{\mu }^{2}+6\mu +1=0 \ (3\mu +1{)}^{2}=0\Rightarrow \mu =-\frac{1}{3} \end{gathered}$

finding position vector of Q using $\mu =-1/3$

$\begin{gathered} \overset{	o \leftarrow }{OQ}=(4-\frac{5}{3})\hat{i}+(1-\frac{2}{3})\hat{j}-\frac{1}{3}\hat{k} \ \overset{	o \leftarrow }{OQ}=\frac{7}{3}\hat{i}+\frac{1}{3}\hat{j}-\frac{1}{3}\hat{k} \end{gathered}$

$\begin{gathered} 27|QR{|}^{2} \ \, \overset{	o \leftarrow }{QR}=\overset{	o \leftarrow }{OR}-\overset{	o \leftarrow }{OQ} \ \overset{	o \leftarrow }{QR}=-\frac{10}{3}\hat{i}-\frac{4}{3}\hat{j}-\frac{2}{3}\hat{k} \ |\overset{	o \leftarrow }{QR}{|}^{2}=\frac{9}{100}+\frac{9}{16}+\frac{9}{4}=\frac{9}{120} \ 27|\overset{	o \leftarrow }{QR}{|}^{2}=\frac{9}{120}	imes 27=360 \end{gathered}$

Solutions · 1