Question
The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x + 1}{\alpha} = \frac{y - 2}{-1} = \frac{z - 4}{-\alpha}$ and $\frac{x}{\alpha} = \frac{y - 1}{2} = \frac{z - 1}{2\alpha}$ is $\sqrt{2}$, is
The correct answer is
6
Shortest Distance Between Lines Calculation
We are asked to find the sum of all values of $\alpha$ for which the shortest distance between the two given lines is $\sqrt{2}$.
Line 1: $\frac{x + 1}{\alpha} = \frac{y - 2}{-1} = \frac{z - 4}{-\alpha}$ This line passes through point $P_1 = (-1, 2, 4)$ and has direction vector $\vec{d_1} = (\alpha, -1, -\alpha)$.
Line 2: $\frac{x}{\alpha} = \frac{y - 1}{2} = \frac{z - 1}{2\alpha}$ This line passes through point $P_2 = (0, 1, 1)$ and has direction vector $\vec{d_2} = (\alpha, 2, 2\alpha)$.
The formula for the shortest distance $D$ between two skew lines is:
$D = \frac{|(P_2 - P_1) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$Step-by-Step Calculation
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Calculate the vector connecting the points $P_1$ and $P_2$:
$P_2 - P_1 = (0 - (-1), 1 - 2, 1 - 4) = (1, -1, -3)$
- Calculate the cross product of the direction vectors $\vec{d_1} \times \vec{d_2}$: $ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{vmatrix} $ $ = \mathbf{i}(-1(2\alpha) - (-\alpha)(2)) - \mathbf{j}(\alpha(2\alpha) - (-\alpha)(\alpha)) + \mathbf{k}(\alpha(2) - (-1)(\alpha)) $ $ = \mathbf{i}(-2\alpha + 2\alpha) - \mathbf{j}(2\alpha^2 + \alpha^2) + \mathbf{k}(2\alpha + \alpha) $ $ = 0\mathbf{i} - 3\alpha^2\mathbf{j} + 3\alpha\mathbf{k} = (0, -3\alpha^2, 3\alpha) $
- Calculate the magnitude of the cross product: $ |\vec{d_1} \times \vec{d_2}| = \sqrt{0^2 + (-3\alpha^2)^2 + (3\alpha)^2} = \sqrt{9\alpha^4 + 9\alpha^2} $ $ = \sqrt{9\alpha^2(\alpha^2 + 1)} = |3\alpha|\sqrt{\alpha^2 + 1} $ (Note: The line representation implies $\alpha \neq 0$)
- Calculate the scalar triple product $(P_2 - P_1) \cdot (\vec{d_1} \times \vec{d_2})$: $ (1, -1, -3) \cdot (0, -3\alpha^2, 3\alpha) = 1(0) + (-1)(-3\alpha^2) + (-3)(3\alpha) $ $ = 0 + 3\alpha^2 - 9\alpha = 3\alpha^2 - 9\alpha $
- Set up the distance equation and solve for $\alpha$: $ D = \frac{|3\alpha^2 - 9\alpha|}{|3\alpha|\sqrt{\alpha^2 + 1}} = \sqrt{2} $ $ \frac{|3\alpha(\alpha - 3)|}{|3\alpha|\sqrt{\alpha^2 + 1}} = \sqrt{2} $ Since $\alpha \neq 0$, we can simplify $|3\alpha|$: $ \frac{3|\alpha||\alpha - 3|}{3|\alpha|\sqrt{\alpha^2 + 1}} = \sqrt{2} $ $ \frac{|\alpha - 3|}{\sqrt{\alpha^2 + 1}} = \sqrt{2} $ Square both sides: $ \frac{(\alpha - 3)^2}{\alpha^2 + 1} = (\sqrt{2})^2 = 2 $ $ \alpha^2 - 6\alpha + 9 = 2(\alpha^2 + 1) $ $ \alpha^2 - 6\alpha + 9 = 2\alpha^2 + 2 $ $ \alpha^2 + 6\alpha - 7 = 0 $ Factor the quadratic equation: $ (\alpha + 7)(\alpha - 1) = 0 $ The possible values for $\alpha$ are $\alpha = 1$ and $\alpha = -7$.
- Calculate the sum of the values of $\alpha$: Sum = $1 + (-7) = -6$.
The sum of all values of $\alpha$ is $-6$, which corresponds to Option 2.
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