Question

The line passing through the points $(5,1,a)$ and $(3,b,1)$ crosses the

Answer 1

$\left(a\right)\frac{x-5}{2}=\frac{y-1}{1-b}=\frac{z-a}{a-1}=\mu$(say)

The line crosses the $yz$ plane where $x=0$

$\Rightarrow -5=2\mu \therefore \mu =-\frac{5}{2}$

Again $y=\mu (1-b)+1=\frac{17}{2}$

$\Rightarrow -\frac{5}{2}(1-b)+1=\frac{17}{2}$

$\Rightarrow -\frac{-5}{2}(1-b)=\frac{17}{2}-1=\frac{15}{2}$

$\Rightarrow 1-b=3$

$\therefore b=4$

Again $z=\mu (a-1)+a=-\frac{13}{2}$

$\Rightarrow -\frac{5}{2}(a-1)+a=-\frac{13}{2}$

$\Rightarrow \frac{-5}{2}a-\frac{5}{2}+a=-\frac{13}{2}$

$\Rightarrow \frac{-3}{2}a-\frac{5}{2}=-\frac{13}{2}$

$\Rightarrow \frac{-3}{2}a=-\frac{13}{2}+\frac{5}{2}$

$\Rightarrow \frac{-3}{2}a=-9$

$\Rightarrow a=-9\times \frac{-2}{3}=6$

$\therefore a=6$

$\left(b\right)\frac{x-5}{2}=\frac{y-1}{1-b}=\frac{z-a}{a-1}=\mu$(say)

The line crosses the $zx$ plane where $y=0$

$\Rightarrow \frac{-1}{1-b}=\mu$

Again $b-1=\frac{1}{\mu }$

$\Rightarrow b=\frac{\mu +1}{\mu }$

Again $\frac{x-5}{2}=\mu$

$\Rightarrow x-5=2\mu$

$\Rightarrow 2\mu +5=\frac{17}{3}$

$\Rightarrow 2\mu =\frac{17}{3}-5=\frac{17-15}{3}=\frac{2}{3}$

$\therefore \mu =\frac{1}{3}$

Again $\frac{z-a}{a-1}=\mu$

$\Rightarrow z-a=\mu (a-1)$

$\Rightarrow z=a+\mu (a-1)$

$\Rightarrow a+\frac{1}{3}(a-1)=\frac{7}{3}$

$\Rightarrow a+\frac{a}{3}-\frac{1}{3}=\frac{7}{3}$

$\Rightarrow \frac{4a}{3}=\frac{7+1}{3}$

$\Rightarrow \frac{4a}{3}=\frac{8}{3}$

$\therefore 4a=8$ or $a=2$

$\left(c\right)\frac{x-5}{2}=\frac{y-1}{1-b}=\frac{z-a}{a-1}=\mu$(say)

The line crosses the $xy$ plane where $z=0$

$\Rightarrow \frac{-a}{a-1}=\mu$

$\Rightarrow \mu =\frac{a}{1-a}$

Again $\frac{x-5}{2}=\mu$

$\Rightarrow x=2\mu +5$

$\Rightarrow 2\mu +5=1$

$\Rightarrow 2\mu =1-5=-4$

$\Rightarrow \mu =-2$

Again $\Rightarrow \mu =\frac{a}{1-a}$

$\Rightarrow \frac{a}{1-a}=-2$

$\Rightarrow a=-2+2a$

$\Rightarrow a-2a=-2$

$\Rightarrow a=2$

Again $\frac{y-1}{1-b}=-2$

$\Rightarrow y-1=-2+2b$

$\Rightarrow y=-2+2b+1$

$\Rightarrow -1+2b=1$

$\Rightarrow 2b=2$

$\therefore b=1$

$\left(d\right)$The equation of a line passing through two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Given that the line passed through the points $(5,1,a)$ and $(3,b,1)$ where $x_1=5,y_1=1,z_1=a$ and $x_2=3,y_2=b,z_2=1$

So the equation of the line is

$\frac{x-5}{3-5}=\frac{y-1}{b-1}=\frac{z-a}{1-a}$

$\Rightarrow \frac{x-5}{-2}=\frac{y-1}{b-1}=\frac{z-a}{1-a}=k$ for $k\in R$

So,$x-5=-2k,y-b=k(b-1),z-a=k(1-a)$

$\therefore x=-2k+5,y=k(b-1)+b,z=k(1-a)+a$ ........$\left(1\right)$

Let $(1,1,-4)$ be the coordinates of the point where the line crosses the plane $x+y+z+2=0$

$\Rightarrow x=-2k+5=1,\Rightarrow -2k=1-5=-4\therefore k=2$

$\Rightarrow y=k(b-1)+b=1\Rightarrow 2(b-1)+b=1$

$\Rightarrow 2b-2+b=1$

$\Rightarrow 3b=3\therefore b=1$

$\Rightarrow z=k(1-a)+a=-4$

$\Rightarrow 2(1-a)+a=-4$

$\Rightarrow 2-2a+a=-4$

$\Rightarrow -2a+a=-4-2$

$\Rightarrow -a=-6$

$\therefore a=6$