Let O be the origin, and M and N be the points on the lines (x-5)/(4)=(y-4)/(1)=(z-5)/(3) …

Question

Let O be the origin, and M and $N$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $MN$ is the shortest distance between the given lines. Then $\overset{	o \leftarrow }{OM}\cdot \overset{	o \leftarrow }{ON}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 9

Accepted Answer

${L}_{1}:\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}=\lambda \, drs(4,1,3)={b}_{1}$

$\begin{gathered} M(4\lambda +5,\lambda +4,3\lambda +5) \ {L}_{2}:\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu \ N(12\mu -8,5\mu -2,9\mu -11) \ \overset{	o \leftarrow }{MN}=(4\lambda -12\mu +13,\lambda -5\mu +6,3\lambda -9\mu +16)\, \end{gathered}$

Now

${\overset{	o }{b}}_{1}	imes {\overset{	o }{b}}_{2}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 4 \quad  1 \quad  3 \ 12 \quad  5 \quad  9 \end{vmatrix}=-6\hat{i}+8\hat{k}\,$

Equation (1) and (2)

$	herefore \frac{4\lambda -12\mu +13}{-6}=\frac{\lambda -5\mu +6}{0}=\frac{3\lambda -9\mu +16}{8}$

I and II

$\lambda -5\mu +6=0\,$

I and III

$\lambda -3\mu +4=0\,$

Solve (3) and (4) we get

$\begin{gathered} \lambda = -1,\mu =1 \ 	herefore \, M(1,3,2) \ N(4,3,-2) \ 	herefore \, \overset{	o \leftarrow }{OM}\cdot \overset{	o \leftarrow }{ON}=4+9-4=9 \end{gathered}$

Solutions · 1