Question

The short distance between the $y-$axis and the line $2x+3y+5z+1=3x+4y+6z+2=0$ is $\frac{2}{\sqrt{k}}$, then the value of ${}^{\prime }{k}^{\prime }$ is

• A. $2$
• B. $3$
• C. $5$
• D. $6$

Answer 1

Answer: (C)

$5$

Let $P_1:2x+3y+5z+1=0\Rightarrow$Direction ratios $(2,3,5)$

and $P_2:3x+4y+6z+2=0\Rightarrow$Direction ratios $(3,4,6)$

Direction ratios of the line $=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 5\\ 3& 4& 6\end{array}\right|$

$=(18-20)\hat{i}-(12-15)\hat{j}+(8-9)\hat{k}$

$=-2\hat{i}-3\hat{j}-\hat{k}$

$\therefore$direction ratios of the line is $(-2,-3,-1)$

If the plane lies on $y-$axis then $y=0$ then equation of the line is

$2x+5z=-1$ .....$\left(1\right)$

$3x+6z=-2$ .....$\left(2\right)$

On multiplying eqn$\left(1\right)$ by $3$ and eqn$\left(2\right)$ by $2$ we have

$6x+15z=-3$

$6x+12z=-4$

$\Rightarrow 6x+15z-6x-12z=-3+4$

$\Rightarrow 3z=1$ or $z=\frac{1}{3}$

Put $z=\frac{1}{3}$ in $\left(1\right)$ we get

$2x+5\times \frac{1}{3}=-1$

$\Rightarrow 2x+\frac{5}{3}=-1$

$\Rightarrow 2x=-1-\frac{5}{3}$

$\Rightarrow 2x=\frac{-3-5}{3}$

$\Rightarrow 2x=\frac{-8}{3}$

$\Rightarrow x=\frac{-4}{3}$

$\therefore$ the point is $(\frac{-4}{3},0,\frac{1}{3})$ and

direction ratio is $(-2,-3,-1)$

Equation of the line is $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$

Now the line $L_1$ is $\frac{x-\frac{-4}{3}}{-2}=\frac{y-0}{-3}=\frac{z-\frac{1}{3}}{-1}$

In $y-$axis the point is $(0,y,0)$

$\therefore L_2=\frac{x}{0}=\frac{y}{1}=\frac{z}{0}$

The shortest distance$=\frac{\left|\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|\right|}{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}$

where $x_1=-\frac{4}{3},y_1=0,z_1=\frac{1}{3},a_1=-2,b_1=-3,c_1=-1$ and $x_2=0,y_2=1,z_2=0,a_2=0,b_2=1,c_2=0$

The shortest distance$=\frac{\left|\left|\begin{array}{ccc}\frac{4}{3}& 1& -\frac{1}{3}\\ -2& -3& -1\\ 0& 1& 0\end{array}\right|\right|}{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& -3& -1\\ 0& 1& 0\end{array}\right|}$

$=\frac{\frac{4}{3}(0+1)-1(0-0)-\frac{1}{3}(-2+0)}{|(0+1)\hat{i}-(0-0)\hat{j}+(-2-0)\hat{k}|}$

$=\frac{\frac{4}{3}+\frac{2}{3}}{|\hat{i}-2\hat{k}|}$

$=\frac{\frac{6}{3}}{\sqrt{1+4}}$

$=\frac{2}{\sqrt{5}}$

Given Shortest distance $=\frac{2}{\sqrt{k}}$

$\therefore \frac{2}{\sqrt{k}}=\frac{2}{\sqrt{5}}$

$\therefore k=5$