Question

The sum of all values of $\alpha$, for which the shortest distance between the lines $\dfrac{x+1}{\alpha}=\dfrac{y-2}{-1}=\dfrac{z-4}{-\alpha}$ and $\dfrac{x}{\alpha}=\dfrac{y-1}{2}=\dfrac{z-1}{2\alpha}$ is $\sqrt{2}$, is

• A. 6
• B. $-6$
• C. $-8$
• D. 8

Hint:

For skew lines, always use vector cross product method to find the shortest distance.

Answer 1

The Correct Option is B

Step 1: Identify direction vectors.
\[ \vec{d_1}=\langle \alpha,-1,-\alpha\rangle,\quad \vec{d_2}=\langle \alpha,2,2\alpha\rangle \] Step 2: Find vector joining points.
\[ \vec{r}=\langle 1,-1,3\rangle \] Step 3: Apply shortest distance formula.
\[ D=\frac{|\vec{r}\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|} \] Step 4: Equate to $\sqrt{2}$.
Solving gives \[ \alpha=-2,\ -4 \] Step 5: Sum of all values.
\[ -2+(-4)=-6 \]