Question

find the vector equation of the line joining the points 2i+j+3k and -4i+3j-k​

Answer 1

$\large\underline{\bf{Solution-}}$

$\rm :\longmapsto\:Let \: \vec{a_1} = 2\hat{i} + \hat{j} + 3\hat{k}$

and

$\rm :\longmapsto\:Let \: \vec{a_2} = - 4\hat{i} + 3\hat{j} - \hat{k}$

Consider,

$\rm :\longmapsto\:\vec{a_2} - \vec{a_1}$

$\:  \: \rm  =  \: \: - 4\hat{i} + 3\hat{j} - \hat{k} - (2\hat{i} + \hat{j} + 3\hat{k})$

$\:  \: \rm  =  \: \: - 4\hat{i} + 3\hat{j} - \hat{k} - 2\hat{i} - \hat{j} - 3\hat{k}$

$\:  \: \rm  =  \: \: - 6\hat{i} + 2\hat{j} -4 \hat{k}$

$\boxed{\bf\implies \:\vec{a_2} - \vec{a_1} = - 6\hat{i} + 2\hat{j} - 4\hat{k}}$

We know,

Equation of line passing through 2 points in vector form is given by

$\rm :\longmapsto\:\vec{r} = \vec{a_1} + \lambda \: (\vec{a_2} - \vec{a_1})$

$\rm :\longmapsto\:\vec{r} = 2\hat{i} + \hat{j} + 3\hat{k} + \lambda \:( - 6\hat{i} + 2\hat{j} - 4\hat{k})$

$\rm :\longmapsto\:\vec{r} = 2\hat{i} + \hat{j} + 3\hat{k} +2 \lambda \:( - 3\hat{i} + \hat{j} - 2\hat{k})$

$\rm :\longmapsto\:\vec{r} = 2\hat{i} + \hat{j} + 3\hat{k} + \mu\:( - 3\hat{i} + \hat{j} - 2\hat{k}) \: \: where \: 2\lambda \: = \mu$

Additional Information :-

Let us consider two lines,

$\rm :\longmapsto\:\vec{r} = \vec{a_1} + \lambda \:\vec{b_1}$

and

$\rm :\longmapsto\:\vec{r} = \vec{a_2} + \mu \:\vec{b_2}$

then,

$\boxed{ \sf \: lines \: are \: \parallel \: if \: \vec{b_1} \times \vec{b_2} = 0}$

$\boxed{ \sf \: lines \: are \: \perp \: if \: \vec{b_1}. \vec{b_2} = 0}$

$\boxed{ \sf \: cos\theta = \dfrac{\vec{b_1}.\vec{b_2}}{ |\vec{b_1}| |\vec{b_2}| }}$

Shortest distance between two skew lines :-

$\boxed{ \sf \: d = \dfrac{ |(\vec{a_2} - \vec{a_1}).(\vec{b_1} \times \vec{b_2})| }{ |\vec{b_1} \times \vec{b_2}| }}$

Distance between two parallel lines :-

$\boxed{ \sf \: d = \dfrac{ |(\vec{a_2} - \vec{a_1}) \times \vec{b}| }{ |\vec{b}| }}$