Key Concepts and Formulas

• Standard Form of a Line: A line in 3D space is typically represented in symmetric Cartesian form as $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. Here, $(x_1, y_1, z_1)$ is a point on the line, and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ is its direction vector.
• Vector Form of a Line: Alternatively, a line can be written as $\vec{r} = \vec{a} + t\vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is its direction vector.
• Shortest Distance Between Skew Lines: For two skew lines $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$, the shortest distance (SD) between them is given by the formula: $SD = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{||\vec{b_1} \times \vec{b_2}||}$ The numerator represents the absolute value of the scalar triple product, which can also be computed as a determinant: $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ b_{1x} & b_{1y} & b_{1z} \\ b_{2x} & b_{2y} & b_{2z} \end{vmatrix}$

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Step-by-Step Solution

Step 1: Convert Line Equations to Standard Symmetric Form and Identify Point and Direction Vectors. The first crucial step is to rewrite the given line equations into the standard symmetric form ($\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$) to easily extract a point on each line and its direction vector.

• Line 1: $x - \lambda = 2y - 1 = -2z$ To achieve the standard form, we need the coefficients of $x, y, z$ in the numerators to be $1$. $x - \lambda = 2\left(y - \frac{1}{2}\right) = -2(z - 0)$ Now, divide each part by the appropriate values to make the denominators the direction ratios: $\frac{x - \lambda}{1} = \frac{y - \frac{1}{2}}{\frac{1}{2}} = \frac{z - 0}{-\frac{1}{2}}$ To simplify calculations by avoiding fractions in the direction ratios, we can multiply all denominators by a common factor (e.g., $2$). This scalar multiplication does not change the direction of the vector. $\frac{x - \lambda}{2} = \frac{y - \frac{1}{2}}{1} = \frac{z}{-1}$ From this form, we identify:

  • A point on the first line, $\vec{a_1} = \left(\lambda, \frac{1}{2}, 0\right)$.
  • The direction vector of the first line, $\vec{b_1} = 2\hat{i} + 1\hat{j} - 1\hat{k}$.

• Line 2: $x = y + 2\lambda = z - \lambda$ This equation is already in a form that can be easily converted to standard symmetric form by writing denominators of $1$: $\frac{x - 0}{1} = \frac{y - (-2\lambda)}{1} = \frac{z - \lambda}{1}$ From this form, we identify:

  • A point on the second line, $\vec{a_2} = (0, -2\lambda, \lambda)$.
  • The direction vector of the second line, $\vec{b_2} = 1\hat{i} + 1\hat{j} + 1\hat{k}$.

Step 2: Calculate the Vector $(\vec{a_2} - \vec{a_1})$. This vector connects a point on the first line to a point on the second line. It is a crucial component for the numerator of the shortest distance formula. $\vec{a_2} - \vec{a_1} = (0 - \lambda)\hat{i} + \left(-2\lambda - \frac{1}{2}\right)\hat{j} + (\lambda - 0)\hat{k}$ $\vec{a_2} - \vec{a_1} = -\lambda\hat{i} - \left(2\lambda + \frac{1}{2}\right)\hat{j} + \lambda\hat{k}$

Step 3: Calculate the Cross Product $(\vec{b_1} \times \vec{b_2})$. The cross product of the direction vectors yields a vector that is perpendicular to both lines. This vector's direction defines the line segment of shortest distance, and its magnitude forms the denominator of the shortest distance formula. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end{vmatrix}$ $= \hat{i}((1)(1) - (-1)(1)) - \hat{j}((2)(1) - (-1)(1)) + \hat{k}((2)(1) - (1)(1))$ $= \hat{i}(1 + 1) - \hat{j}(2 + 1) + \hat{k}(2 - 1)$ $\vec{b_1} \times \vec{b_2} = 2\hat{i} - 3\hat{j} + 1\hat{k}$

Step 4: Calculate the Scalar Triple Product (Numerator of SD Formula). Now we compute the dot product of the vector $(\vec{a_2} - \vec{a_1})$ (from Step 2) and the cross product $(\vec{b_1} \times \vec{b_2})$ (from Step 3). $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = \left(-\lambda\hat{i} - \left(2\lambda + \frac{1}{2}\right)\hat{j} + \lambda\hat{k}\right) \cdot (2\hat{i} - 3\hat{j} + 1\hat{k})$ $= (-\lambda)(2) + \left(-2\lambda - \frac{1}{2}\right)(-3) + (\lambda)(1)$ $= -2\lambda + 6\lambda + \frac{3}{2} + \lambda$ $= 5\lambda + \frac{3}{2}$

Step 5: Calculate the Magnitude of the Cross Product (Denominator of SD Formula). Next, we find the magnitude of the vector $\vec{b_1} \times \vec{b_2}$ calculated in Step 3. $||\vec{b_1} \times \vec{b_2}|| = ||2\hat{i} - 3\hat{j} + 1\hat{k}||$ $= \sqrt{(2)^2 + (-3)^2 + (1)^2}$ $= \sqrt{4 + 9 + 1}$ $= \sqrt{14}$

Step 6: Substitute Values into the Shortest Distance Formula and Solve for $\lambda$. We are given that the shortest distance (SD) is $\frac{\sqrt{7}}{2\sqrt{2}}$. We equate this to our calculated expression using the formula: $SD = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{||\vec{b_1} \times \vec{b_2}||}$ $\frac{\sqrt{7}}{2\sqrt{2}} = \frac{\left|5\lambda + \frac{3}{2}\right|}{\sqrt{14}}$ We can simplify $\sqrt{14} = \sqrt{7} \times \sqrt{2}$. $\frac{\sqrt{7}}{2\sqrt{2}} = \frac{\left|5\lambda + \frac{3}{2}\right|}{\sqrt{7}\sqrt{2}}$ To isolate the absolute value term, multiply both sides by $\sqrt{7}\sqrt{2}$: $\frac{\sqrt{7} \cdot \sqrt{7}\sqrt{2}}{2\sqrt{2}} = \left|5\lambda + \frac{3}{2}\right|$ $\frac{7\sqrt{2}}{2\sqrt{2}} = \left|5\lambda + \frac{3}{2}\right|$ $\frac{7}{2} = \left|5\lambda + \frac{3}{2}\right|$

Step 7: Solve the Absolute Value Equation for $\lambda$ and Find $|\lambda|$. The absolute value equation implies two possible cases for the expression inside:

• Case 1: $5\lambda + \frac{3}{2} = \frac{7}{2}$ $5\lambda = \frac{7}{2} - \frac{3}{2}$ $5\lambda = \frac{4}{2}$ $5\lambda = 2$ $\lambda = \frac{2}{5}$
• Case 2: $5\lambda + \frac{3}{2} = -\frac{7}{2}$ $5\lambda = -\frac{7}{2} - \frac{3}{2}$ $5\lambda = -\frac{10}{2}$ $5\lambda = -5$ $\lambda = -1$

The problem statement specifies that $\lambda$ is an integer. Therefore, $\lambda = -1$ is the correct value. Finally, we need to find the value of $|\lambda|$. $|\lambda| = |-1| = 1$

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Common Mistakes & Tips

• Standard Form Conversion: Always ensure the coefficients of $x, y, z$ in the numerator are exactly $1$. If you have $3y-6$, rewrite it as $3(y-2)$ and then divide by $3$ in the denominator to get $\frac{y-2}{1/3}$. Errors here are very common.
• Sign Errors: Be extremely careful with negative signs during vector operations (subtraction, cross product, dot product). A single sign error can propagate through the entire calculation.
• Absolute Value Handling: Remember that $|A|=B$ implies $A=B$ or $A=-B$. Both cases must be considered when solving for the unknown variable.
• Simplifying Radicals: Look for common factors in radicals, like $\sqrt{14} = \sqrt{7}\sqrt{2}$, to simplify expressions and make calculations easier.

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Summary

This problem required finding the value of an integer $\lambda$ given the shortest distance between two skew lines. The solution involved converting the line equations into their standard symmetric form to identify points and direction vectors. These vectors were then used in the shortest distance formula, which requires computing the vector connecting points, the cross product of direction vectors, and their dot product and magnitude. Equating the derived shortest distance expression to the given value led to an absolute value equation for $\lambda$. Solving this equation and applying the integer constraint for $\lambda$ yielded the final answer.

The final answer is $\boxed{1}$.