Question

The cartesian equation of a line is $\frac{x-5}{3} = \frac{y+4}{7}=\frac{z-6}{2}$. Write its vector form.

Answer 1

Answer:

The line can be represented bu the cartesian equation

$l:(1, \frac{7}{3}, \frac{2}{3})\lambda+(0,-\frac{47}{3}, -\frac{8}{3})$

Step-by-step explanation:

Let's write y in funtion of x. Note that

$\frac{x-5}{3} = \frac{y+4}{7} \Rightarrow y = \frac{7x-35}{3}-4 = \frac{7}{3}x-\frac{47}{3}$

Now, writing z in function of x, we get

$\frac{x-5}{3} = \frac{z-6}{2} \Rightarrow z = \frac{2x-10}{3}+6 = \frac{2}{3}x-\frac{8}{3}$

Letting $x = \lambda$, the line can be represented bu the cartesian equation

$l: (\lambda, \frac{7}{3}\lambda-\frac{47}{3}, \frac{2}{3}\lambda-\frac{8}{3}) = (1, \frac{7}{3}, \frac{2}{3})\lambda+(0,-\frac{47}{3}, -\frac{8}{3})$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

The cartesian equation of a line is

$\displaystyle \sf{ \frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} \: }$

TO DETERMINE

The vector form of the line

CALCULATION

The given equation of the line is

$\displaystyle \sf{ \frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} \: } = \lambda \: \: (say)$

Which gives

$\sf{ x = 5 + 3 \lambda\: }$

$\sf{ y = - 4+ 7 \lambda\: }$

$\sf{ z = 6 + 2 \lambda\: }$

So the vector equation of the line is

$\sf{ \vec{r} = x \: \hat{i} + y \: \hat{j} \: + z \: \hat{k}\: }$

$\implies \: \sf{ \vec{r} = (5 + 3 \lambda) \: \hat{i} + ( - 4 + 7 \lambda) \: \hat{j} \: + (6 + 2 \lambda) \: \hat{k}\: }$

$\implies \: \sf{ \vec{r} = (5 \: \hat{i} - 4 \: \hat{j} \: + 6 \: \hat{k}\: ) + \: ( 3 \lambda\: \hat{i} + 7 \lambda \: \hat{j} \: + 2 \lambda) \: \hat{k}\: }$

$\therefore\sf{ \vec{r} = (5 \: \hat{i} - 4 \: \hat{j} \: + 6 \: \hat{k}\: ) + \:\lambda ( 3 \hat{i} + 7 \: \hat{j} \: + 2 \: \hat{k}\: )}$

RESULT

The required vector equation of the given line is

$\boxed{\sf{ \: \: \vec{r} = (5 \: \hat{i} - 4 \: \hat{j} \: + 6 \: \hat{k}\: ) + \:\lambda ( 3 \hat{i} + 7 \: \hat{j} \: + 2 \: \hat{k}\: ) \: \: }}$

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