Question

The image of line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2x-y+z+3=0$

• A. $\frac{x-3}{3}=\frac{y+5}{1}=\frac{z-2}{-5}$
• B. $\frac{x-3}{-3}=\frac{y+5}{-1}=\frac{z-2}{5}$
• C. $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$
• D. $\frac{x+3}{-3}=\frac{y-5}{-1}=\frac{z+2}{5}$

Answer 1

The correct option is C $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$

Given,

Image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$

in the plane $2x-y+z+3=0$

firstly.

let us find out the given line is whether parallel or perpendicular.

if the given line is parallel to the given plane them

we get,

$\Rightarrow 3\times x$ coefficient $+1\times y$ co-efficient $+(-5)\times z$ co-efficient

$\Rightarrow 3\times 2+1(-1)+(-5)\times 1=6-1-5$

$=6-6=0\to$ equation $\left(1\right)$

equation $\left(1\right)$ show that given lime is parallel to the given plane

Now, let is represent it through a figure.

Representation of given data diagramatically for the ${\perp }^{lar}$ line, equation will be,

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=k$ [where ${}^{\prime }{k}^{\prime }$ is coefficient]

Now in finding ${}^{\prime }{k}^{\prime }$ value we get

$\frac{x-1}{2}=k;\frac{y-3}{-1}=k;\frac{z-4}{1}=k$

$\Rightarrow x-1=2ky-3=-kz-4=k$

$\Rightarrow x-2k+1;\Rightarrow y=-k+3;\Rightarrow z=k+4$

Let us assume ${}^{\prime }{N}^{\prime }$ as a point of the ${\perp }^{lar}$

as we have plane equation as $2x-y+z+3=0$

Now, let us substitude $x,y,z$ Values thus obtained in the plane equation to find ${}^{\prime }{k}^{\prime }$ we get.

$2(2k+1)-1[-k+3]+1[k+4]+3=0$

$\Rightarrow 4k+2-3+k+k+4+3=0$

$\Rightarrow 6k+6\Rightarrow 0\Rightarrow 6k=-6\Rightarrow k=\frac{-6}{6}=-1$

if $k=-1$ then, let us find out $x,y,z$ values,

$x=2(-1)+1=-1,y=-(-1)+3=4,z=(-1)+4=3$

$\Rightarrow N(x,y,z)=N(-1,4,3)$

As ${}^{\prime }{Q}^{\prime }$ is the image of line ${}^{\prime }{P}^{\prime }$ then ${}^{\prime }{N}^{\prime }$ will be the point bisecting both ${}^{\prime }{Q}^{\prime }$ and ${}^{\prime }{N}^{\prime }$ will be the mid point.

As ${}^{\prime }{N}^{\prime }$ is the midpoint for $P$ and $Q$ we get,

$-1=\frac{x_1+1}{2},4\frac{y_1+3}{2},3=\frac{z_1+4}{2}$

$\Rightarrow x_1=-2-1;y_1=8-3;z_1=6-4$

$\Rightarrow x_1=-3y_1=5z_1=2$

So, now the line equation ${}^{\prime }{Q}^{\prime }$ will be.

$\frac{x-x_1}{3}=\frac{y-y_1}{1}=\frac{z-z_1}{-5}$

$\Rightarrow \frac{x-(-3)}{3}=\frac{y-\left(5\right)}{1}=\frac{z-\left(2\right)}{-5}\Rightarrow \frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$

$\therefore$ the line equation of the image ${}^{\prime }{Q}^{\prime }$ is

$\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$ so, correct option is $\left[C\right]$