Question

If shortest distance between the lines \[ \frac{x+1}{\alpha}=\frac{y-2}{-2}=\frac{z-4}{-2\alpha} \quad \text{and} \quad \frac{x}{\alpha}=\frac{y-1}{1}=\frac{z-1}{\alpha} \] is $\sqrt{2}$, then find the sum of all possible values of $\alpha$.

Hint:

For skew lines, always use vector cross product to compute shortest distance.

Answer 1

Correct Answer: -6

Step 1: Use formula for shortest distance between skew lines.
\[ D=\frac{|(\vec r_2-\vec r_1)\cdot(\vec d_1\times\vec d_2)|} {|\vec d_1\times\vec d_2|} \] Step 2: Write direction vectors.
\[ \vec d_1=(\alpha,-2,-2\alpha),\quad \vec d_2=(\alpha,1,\alpha) \] Step 3: Compute cross product magnitude.
\[ |\vec d_1\times\vec d_2| =\sqrt{9\alpha^4+9\alpha^2} \] Step 4: Use given distance $\sqrt2$.
\[ \sqrt2=\frac{| -3\alpha^2+9\alpha |}{\sqrt{9\alpha^4+9\alpha^2}} \] Squaring both sides and simplifying gives: \[ (\alpha+7)(\alpha-1)=0 \] Step 5: Find sum of solutions.
\[ \alpha=-7,\ 1 \Rightarrow \text{sum}=-6 \] Final conclusion.
The required sum is $-6$.