Question

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda }$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$, then the sum of all possible values of $\lambda$ is :

• A. 16
• B. 6
• C. 12
• D. 15

Answer 1

Answer: (A)

SHORTEST distance $\frac{|\left(a_2-a_1\right)\cdot \left(b_1\times b_2\right)|}{|b_1\times b_2|}$
$a_1=(1,2,3)$
$a_2=(2,4,5)$
${\overrightarrow{b}}_2=2\hat{i}+3\hat{j}+\lambda \hat{k}$
${\overrightarrow{b}}_2=\hat{i}+4\hat{j}+5\hat{k}$
S.D. $=\frac{|((2-1)\hat{i}+(4-2)\hat{j}+(5-3)\hat{k})\cdot \left({\vec{b}}_1\times {\vec{b}}_2\right)|}{|b_1\times b_2|}$
${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& \lambda \\ 1& 4& 5\end{array}\right|$
$=\hat{i}(15-4\lambda )+\hat{j}(\lambda -10)+\hat{k}(5)$
$=(15-4\lambda )\hat{i}+(\lambda -10)\hat{j}+5\hat{k}$
$\left|{\vec{b}}_1\times {\vec{b}}_2\right|=\sqrt{(15-4\lambda {)}^2+(\lambda -10{)}^2+25}$
Now
S.D. $=\frac{|(\hat{i}+2\hat{j}+2\hat{k})\cdot [(15-4\lambda )\hat{i}+(\lambda -10)\hat{j}+5\hat{k}]|}{\sqrt{(15-4\lambda {)}^2+(\lambda -10{)}^2+25}}$
$\frac{|15-4\lambda +2\lambda -20+10|}{\sqrt{(15-4\lambda {)}^2+(\lambda -10{)}^2+25}}=\frac{1}{\sqrt{3}}$
square both side
$3(5-2\lambda {)}^2=225+16{\lambda }^2-120\lambda +{\lambda }^2+100-20\lambda +25$
$12{\lambda }^2+75-60\lambda =17{\lambda }^2-140\lambda +350$
$5{\lambda }^2-80\lambda +275=0$
${\lambda }^2-16\lambda +55=0$
$(\lambda -5)(\lambda -11)=0$
$\Rightarrow \lambda =5,11$