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

First, determine the direction vector of the line joining points \((1,2,3)\) and \((2,3,4)\). The direction vector \(\mathbf{b_1}\) is \((2-1, 3-2, 4-3)=(1,1,1)\). Now, consider the given line equation: \( \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0} \). Since \(\frac{z-2}{0}\) implies \(z=2\), this represents a line parallel to the xy-plane with direction vector \(\mathbf{b_2}=(2,-1,0)\). The shortest distance \(\alpha\) between two skew lines is given by:
 

\[\alpha = \frac{|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|}\]

where \(\mathbf{a_1} = (1,2,3)\) and \(\mathbf{a_2} = (1,-1,2)\) is a point on the second line obtained for \(t=0\). The vector (\(\mathbf{a_2} - \mathbf{a_1}\)) equals \((0,-3,-1)\). The cross product \(\mathbf{b_1} \times \mathbf{b_2}\) is \(\left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{array} \right| = \mathbf{i}(1\cdot0 - 1\cdot(-1)) - \mathbf{j}(1\cdot0 - 1\cdot2) + \mathbf{k}(1\cdot(-1) - 1\cdot2)\)
= \( \mathbf{i}(0+1) - \mathbf{j}(0-2) + \mathbf{k}(-1-2) = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\).
The modulus of this cross product is \( \sqrt{1^2+2^2+(-3)^2} = \sqrt{14}\). Calculate the dot product \((\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) = (0,-3,-1)\cdot(1,2,-3) = 0\cdot1 + (-3)\cdot2 + (-1)\cdot(-3) = -6 + 3 = -3\).
Thus, \(\alpha = \frac{|-3|}{\sqrt{14}} = \frac{3}{\sqrt{14}}\).
Then, the value \(28\alpha^2\) is:
 

\[28\left(\frac{3}{\sqrt{14}}\right)^2 = 28 \times \frac{9}{14} = 18\]

Thus, \(28\alpha^2\) is 18, which matches the expected range.