Question

a butterfly is moving in a straight line in the space.
let this path be denoted by a line l whose equation is x-2/2=2-y/3=z-3/4 say.
the position vector is?

Answer 1

SOLUTION

GIVEN

A butterfly is moving in a straight line in the space.

Let this path be denoted by a line L whose equation is

$\displaystyle { \frac{x - 2}{2} = \frac{2 - y}{3} = \frac{z - 3}{4} }$

TO DETERMINE

The position vector

EVALUATION

Here the given equation is

$\displaystyle { \frac{x - 2}{2} = \frac{2 - y}{3} = \frac{z - 3}{4} }$

We rewrite the above equation as below

$\displaystyle { \frac{x - 2}{2} = \frac{y - 2}{ - 3} = \frac{z - 3}{4} = t \: (say) }$

Where t is a real parameter

$\therefore \: \: x = 2t + 2$

$\therefore \: \: y = - 3t + 2$

$\therefore \: \: z = 4t + 3$

Hence the required position vector

$\vec{r} = x \hat{ \imath} + y \hat{ \jmath} + z \hat{k}$

$\therefore \: \: \vec{r} = (2t + 2) \hat{ \imath} + ( - 3t + 2) \hat{ \jmath} + (4t + 3)\hat{k}$

$\therefore \: \: \vec{r} =( 2 \hat{ \imath} + 2 \hat{ \jmath} + 3\hat{k}) + t( 2 \hat{ \imath} - 3 \hat{ \jmath} + 4\hat{k})$

Where t is a real parameter

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