Three Dimensional Geometry Question 6
Question 6 - 2024 (27 Jan Shift 1)
If the shortest distance between the lines $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is :
(1) 5
(2) 8
(3) 7
(4) 10
Show Answer
Answer (2)
Solution
$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$
$\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$
the shortest distance between the lines
$=\left|\frac{(\vec{a}-\vec{b}) \cdot\left(\overrightarrow{d _1} \times \overrightarrow{d _2}\right)}{\left|\overrightarrow{d _1} \times \overrightarrow{d _2}\right|}\right|$
$\left.=\left|\frac{\left|\begin{array}{ccc}\lambda-4 & 0 & 2 \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}\right| \right\rvert,$
$=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right|$
$\frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right|$
$3=|\lambda-4|$
$\lambda-4= \pm 3$
$\lambda=7,1$
Sum of all possible values of $\lambda$ is $=8$
