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Mathematics Question on Shortest Distance between Two Lines

If the shortest distance between the line joining the points (1,2,3)(1,2,3) and (2,3,4)(2,3,4), and the line x−12=y+1−1=z−20\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0} is α\alpha, then 28α228 \alpha^2 is equal to

Answer

The correct answer is 18.

r⃗=(i^+2j^+3k^)+λ(i^+j^+k^)r⃗=a⃗+λp⃗\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})\vec{r}=\vec{a}+\lambda \vec{p}

r⃗=(+i^−j^+2k^)+μ(2i^−j^)r⃗=b⃗+μq⃗\vec{r}=(+\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}-\hat{j})\vec{r}=\vec{b}+\mu \vec{q}

p⃗×q⃗=∣i^j^k^ 111 2−10∣=i^+2j^−3k^\vec{p}\times\vec{q}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ 1& 1 & 1\\\ 2& -1 & 0 \end{vmatrix}=\hat{i}+2\hat{j}-3\hat{k}

d=∣(b⃗−a⃗).(p⃗×q⃗)∣p⃗×q⃗∣∣d=|\frac{(\vec{b}-\vec{a}).(\vec{p}\times\vec{q})}{|\vec{p}\times\vec{q}|}|

d=∣(−3j^−k^).(i^+2j^−3k^)14∣d=|\frac{(-3\hat{j}-\hat{k}).(\hat{i}+2\hat{j}-3\hat{k})}{\sqrt{14}}|

=∣−6+314∣=314=|\frac{-6+3}{\sqrt{14}}|=\frac{3}{\sqrt{14}}

α=314\alpha =\frac{3}{\sqrt{14}}

Now  28α2≠282×91̸4=18Now\, \, 28\alpha ^{2}=\not{28}^{2}\times\frac{9}{\not{14}}=18