Question

find the equation of a line​

Answer 1

Answer:

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Answer 2

Given that the line passes through the point (1, 2, -1).

Let the equation of the line be, in Cartesian form,

$\longrightarrow\dfrac{x-1}{p}=\dfrac{y-2}{q}=\dfrac{z+1}{r}$

where the vector $\left<p,\ q,\ r\right>$ is a direction ratio of the line.

Given that the line lies on the plane,

$\longrightarrow 2x-3y+4z=-8$

$\longrightarrow\left<2,\ -3,\ 4\right>\cdot\left<x,\ y,\ z\right>=-8$

Here the vector $\left<2,\ -3,\ 4\right>$ is perpendicular to the plane as well as the line, which implies the dot product of this vector with the direction ratio of the line equals zero, i.e.,

$\longrightarrow \left<2,\ -3,\ 4\right>\cdot\left<p,\ q,\ r\right>=0$

$\longrightarrow2p-3q+4r=0$

$\longrightarrow2p-3q=-4r\quad\quad\dots(1)$

Given that the line is perpendicular to,

$\longrightarrow\dfrac{x-1}{3}=\dfrac{y-1}{2}=\dfrac{z-2}{3}$

whose direction ratio is $\left<3,\ 2,\ 3\right>.$ So the dot product of direction ratios of both the lines should be zero, i.e.,

$\longrightarrow \left<3,\ 2,\ 3\right>\cdot\left<p,\ q,\ r\right>=0$

$\longrightarrow3p+2q+3r=0$

$\longrightarrow3p+2q=-3r\quad\quad\dots(2)$

Now,

$\longrightarrow (1)\times3-(2)\times2$

$\longrightarrow(2p-3q)3-(3p+2q)2=(-4r)3-(-3r)2$

$\longrightarrow6p-9q-6p-4q=-12r+6r$

$\longrightarrow-13q=-6r$

$\longrightarrow\dfrac{q}{6}=\dfrac{r}{13}$

Let,

• $q=6k$

• $r=13k$

Then (1) becomes,

$\longrightarrow2p-3(6k)=-4(13k)$

$\longrightarrow2p-18k=-52k$

$\longrightarrow p=-17k$

Hence equation of our line becomes.

$\longrightarrow\dfrac{x-1}{-17k}=\dfrac{y-2}{6k}=\dfrac{z+1}{13k}$

$\longrightarrow\underline{\underline{\dfrac{x-1}{-17}=\dfrac{y-2}{6}=\dfrac{z+1}{13}}}$