Question

find vector equation of line x/1=y-1/2=z-2/3​

Answer 1

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

We know that

The cartesian equation of line is

$\rm :\longmapsto\:\dfrac{x - \alpha }{a_1} = \dfrac{y - \beta }{b_1} = \dfrac{z - \gamma }{c_1} = \lambda \:$

Where,

$\sf \: ( \alpha,\beta,\gamma) \: is \: the \: point \: from \: where \: line \: passes \: having \:$

$\sf \: direction \: ratios \: (a_1,b_1,c_1)$

Its vector form is represented as

$\rm :\longmapsto\: \vec{r} = \vec{a} + \lambda \: \vec{b}$

where

$\rm :\longmapsto\:\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$

and

$\rm :\longmapsto\:\vec{b} = a_1 \hat{i} + b_1 \hat{j} + c_1\hat{k}$

Let's solve the problem now!!

The given equation of line is

$\rm :\longmapsto\:\dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - 2}{3}$

It implies that line passes through the point (0, 1, 2) and having direction ratios (1, 2, 3).

So,

$\rm :\longmapsto\:\vec{a} = 0\hat{i} + 1\hat{j} + 2\hat{k}$

$\rm :\longmapsto\:\vec{a} = \hat{j} + 2\hat{k}$

and

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

So,

Equation of line in vector form is

$\rm :\longmapsto\: \vec{r} = \vec{a} + \lambda \: \vec{b}$

On substituting the values of a and b,

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

Additional Information :-

Let us consider two lines,

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

and

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

Then,

$\sf \: 2 \: lines \: are \parallel \: iff \: \vec{b_1} \times \vec{ b_2} = \vec{0}$

$\sf \: 2 \: lines \: are \perp \: iff \: \vec{b_1}. \vec{b_2} = 0$

Angle between two lines is

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