# lines $x-4=y-5=\frac{z+1}{2}$  and $\frac{x-6}{2}=\frac{y-11}{4}=z+3$  are Option 1) Parallel lines  Option 2) Coincident lines  Option 3) skew lines  Option 4) intersecting lines

As we have learned

Condition for lines to be intersecting (vector form) -

Their shortest distance should be 0

$\left ( \vec{b}\times \vec{b_{1}} \right )\cdot \left ( \vec{a}- \vec{a_{1}} \right )=0$

Also the condition for coplanar lines

-

lines can be written in vector form as

$\vec{r}=(4\hat{i}+5\hat{j}-\hat{k})+\lambda(\hat{i}+\hat{j}+2\hat{k})$

$\vec{r}=(6\hat{i}+11\hat{j}-3\hat{k})+\mu (2\hat{i}+4\hat{j}+\hat{k})$

Now ,

$\vec{a}=4\hat{i}+5\hat{j}-\hat{k}, \vec{b}=\hat{i}+\hat{j}+2\hat{k}$

$\vec{a_1}=6\hat{i}+11\hat{j}-3\hat{k}, \vec{b_1}=2\hat{i}+4\hat{j}+\hat{k}$

$\therefore \vec{b}\times \vec{b_1}=\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 1 & 1 & 2\\ 2 &4 & 1 \end{vmatrix}= \hat{i}(-7)-\hat{j}(-3)+2\hat{k}$

$\vec{a}- \vec{a_1}= -2\hat{i}-6\hat{j}+2\hat{k}$

$(\vec{b}\times \vec{b_1})\cdot (\vec{a}- \vec{a_1})=14-18+4=0$

Now, $\vec{b}\neq k\vec{b_1}$ so lines are neither parallel nor co-incident

and

$(\vec{b}\times \vec{b_1})\cdot (\vec{a}- \vec{a_1})=0$

$\therefore$ Lines are intersecting

Option 1)

Parallel lines

Option 2)

Coincident lines

Option 3)

skew lines

Option 4)

intersecting lines

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