The shortest distance between the lines r=((1)/(3)i+2j+(8)/(3)k)+ (2i-5j+6k) and r=(-(2)/(…

Question

The shortest distance between the lines

$\vec{r}=(\frac{1}{3}\hat{i}+2\hat{j}+\frac{8}{3}\hat{k})+\lambda (2\hat{i}-5\hat{j}+6\hat{k})$

and $\vec{r}=(-\frac{2}{3}\hat{i}-\frac{1}{3}\hat{k})+\mu (\hat{j}-\hat{k}),\lambda ,\mu \in R$, is:

Accepted Answer

For two skew lines,

$\vec{r}={\vec{a}}_{1}+\lambda {\vec{b}}_{1}\, \, \vec{r}={\vec{a}}_{2}+\mu {\vec{b}}_{2},$

the shortest distance is

$d=\frac{|({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}	imes {\vec{b}}_{2})|}{|{\vec{b}}_{1}	imes {\vec{b}}_{2}|}.$

Now compare with the given lines.

First line:

$\vec{r}=(\frac{1}{3}\hat{i}+2\hat{j}+\frac{8}{3}\hat{k})+\lambda (2\hat{i}-5\hat{j}+6\hat{k})$

So,

${\vec{a}}_{1}=(\frac{1}{3},2,\frac{8}{3}),\, {\vec{b}}_{1}=(2,-5,6)$

Second line:

$\vec{r}=(-\frac{2}{3}\hat{i}-\frac{1}{3}\hat{k})+\mu (\hat{j}-\hat{k})$

So,

${\vec{a}}_{2}=(-\frac{2}{3},0,-\frac{1}{3}),\, {\vec{b}}_{2}=(0,1,-1)$

Now find

${\vec{a}}_{2}-{\vec{a}}_{1}=(-\frac{2}{3}-\frac{1}{3},\, 0-2,\, -\frac{1}{3}-\frac{8}{3})=(-1,-2,-3)$

Next, calculate ${\vec{b}}_{1}	imes {\vec{b}}_{2}$:

${\vec{b}}_{1}	imes {\vec{b}}_{2}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 2 \quad  -5 \quad  6 \ 0 \quad  1 \quad  -1 \end{vmatrix}$

$=\hat{i}[(-5)(-1)-6(1)]-\hat{j}[2(-1)-6(0)]+\hat{k}[2(1)-(-5)(0)]$

$=\hat{i}(5-6)-\hat{j}(-2)+\hat{k}(2)=-\hat{i}+2\hat{j}+2\hat{k}$

So,

${\vec{b}}_{1}	imes {\vec{b}}_{2}=(-1,2,2)$

Its magnitude is

$|{\vec{b}}_{1}	imes {\vec{b}}_{2}|=\sqrt{(-1{)}^{2}+{2}^{2}+{2}^{2}}=\sqrt{1+4+4}=3$

Now find the scalar triple product:

$({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}	imes {\vec{b}}_{2})=(-1,-2,-3)\cdot (-1,2,2)$

$=(-1)(-1)+(-2)(2)+(-3)(2)=1-4-6=-9$

Therefore,

$d=\frac{|-9|}{3}=3$

Hence, the shortest distance is

$3$

So, the correct option is Option B.

Solutions · 1