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 $\frac{\left|\left(a_{2}-a_{1}\right) \cdot\left(b_{1} \times b_{2}\right)\right|}{\left|b_{1} \times b_{2}\right|}$
$a_{1}=(1,2,3)$
$a_{2}=(2,4,5)$
$\overrightarrow{ b }_{2}=2 \hat{ i }+3 \hat{ j }+\lambda \hat{ k }$
$\overrightarrow{ b }_{2}=\hat{ i }+4 \hat{ j }+5 \hat{ k }$
S.D. $=\frac{\left|((2-1) \hat{i}+(4-2) \hat{j}+(5-3) \hat{k}) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)\right|}{\left|b_{1} \times b_{2}\right|}$
$\overrightarrow{ b }_{1} \times \overrightarrow{ b }_{2}=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 3 & \lambda \\ 1 & 4 & 5\end{vmatrix}$
$=\hat{ i }(15-4 \lambda)+\hat{ j }(\lambda-10)+\hat{ k }(5)$
$=(15-4 \lambda) \hat{ i }+(\lambda-10) \hat{ j }+5 \hat{ k }$
$\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{(15-4 \lambda)^{2}+(\lambda-10)^{2}+25}$
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 $
$\Rightarrow \lambda=5,11$