Question

A plane $P$ contains the line $L_1:\frac{y}{b}+\frac{z}{c}=1,x=0$ and is parallel to the line $L_2:\frac{x}{a}-\frac{z}{c}=1,y=0$, then which of the following is/are correct?

• A. If the shortest distance between $L_1$ and $L_2$ is $\frac{1}{4}$, then value of $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$ is $16$
• B. If the shortest distance between $L_1$ and $L_2$ is $\frac{1}{4}$ then value of $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$ is $64$
• C. Distance of image of $A(a,0,0)$ in the plane $P$ from $M(\frac{-5}{3},\frac{8}{3},\frac{11}{3}),$ where $a=b=c=1$ is equal to $2$
• D. Distance of image of $A(a,0,0)$ in the plane $P$ from $M(\frac{-5}{3},\frac{8}{3},\frac{11}{3}),$ where $a=b=c=1$ is equal to $3$

Answer 1

Answer: (B), (D)

The correct options are
B If the shortest distance between $L_1$ and $L_2$ is $\frac{1}{4}$ then value of $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$ is $64$
D Distance of image of $A(a,0,0)$ in the plane $P$ from $M(\frac{-5}{3},\frac{8}{3},\frac{11}{3}),$ where $a=b=c=1$ is equal to $3$

$L_1:\frac{x}{0}=\frac{y-b}{-b}=\frac{z}{c}{L}_2:\frac{x-a}{a}=\frac{y}{0}=\frac{z}{c}$
Equation of plane $P:\left|\begin{array}{ccc}x& y-b& z\\ 0& -b& c\\ a& 0& c\end{array}\right|=0\Rightarrow \frac{x}{a}-\frac{y}{b}-\frac{z}{c}+1=0$
$L_1$ & $L_2$ in vector form
$L_1:\overrightarrow{r}=b\hat{j}+\lambda (-b\hat{j}+c\hat{k})L_2:\overrightarrow{r}=a\hat{i}+\lambda (a\hat{i}+c\hat{k})\therefore \frac{1}{4}=\frac{\left|\begin{array}{ccc}-a& b& 0\\ 0& -b& c\\ a& 0& c\end{array}\right|}{\sqrt{(bc{)}^2+(ac{)}^2+(ab{)}^2}}\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=64$
Image of $A(1,0,0)$ in the plane $x-y-z+1=0$ is $(\frac{-1}{3},\frac{4}{3},\frac{4}{3}).$ Its distance from $M$ is $3$.