Question

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

Answer 1

Correct Answer: 18

The shortest distance between two skew lines is:

\( d = \frac{|(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{p} \times \mathbf{q})|}{|\mathbf{p} \times \mathbf{q}|}. \)

Here:

\( \mathbf{p} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}, \quad \mathbf{q} = 2\mathbf{i} - \mathbf{j}, \quad \mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}. \)

Calculate \(\mathbf{p} \times \mathbf{q}\), simplify \(d\), and compute:

\( \alpha = \frac{3}{\sqrt{14}}, \quad 28\alpha^2 = 18. \)

Answer 2

The correct answer is 18.

\(\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})\vec{r}=\vec{a}+\lambda \vec{p}\)

\(\vec{r}=(+\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}-\hat{j})\vec{r}=\vec{b}+\mu  \vec{q}\)

\(\vec{p}\times\vec{q}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\   1& 1 & 1\\   2& -1 & 0 \end{vmatrix}=\hat{i}+2\hat{j}-3\hat{k}\)

\(d=|\frac{(\vec{b}-\vec{a}).(\vec{p}\times\vec{q})}{|\vec{p}\times\vec{q}|}|\)

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

\(=|\frac{-6+3}{\sqrt{14}}|=\frac{3}{\sqrt{14}}\)

\(\alpha =\frac{3}{\sqrt{14}}\)

\(Now\, \, 28\alpha ^{2}=\not{28}^{2}\times\frac{9}{\not{14}}=18\)