Key Concepts and Formulas

• Skew Lines: In three-dimensional space, two lines are defined as skew lines if they are neither parallel nor intersecting. They lie in different planes. The shortest distance between them is the length of the unique line segment that is perpendicular to both lines.
• Vector Form of a Line: A line passing through a point with position vector $\vec{a}$ and parallel to a direction vector $\vec{b}$ can be represented as $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter.
• Shortest Distance Formula: For two skew lines given in vector form: Line 1 ($L_1$): $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ Line 2 ($L_2$): $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ The shortest distance $D$ between them is given by the formula: $D = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ This formula essentially calculates the scalar projection of the vector connecting a point on $L_1$ to a point on $L_2$ onto the direction vector of the common perpendicular (which is $\vec{b_1} \times \vec{b_2}$).

Step-by-Step Solution

Step 1: Convert Cartesian Equations to Standard Vector Form The first crucial step is to convert the given Cartesian equations of the lines into their standard vector form, $\vec{r} = \vec{a} + \lambda \vec{b}$. This allows us to identify a point on each line ($\vec{a}$) and its direction vector ($\vec{b}$) correctly.

• For Line 1 ($L_1$): The given equation is $\frac{x+7}{-6}=\frac{y-6}{7}=z$. To match the standard Cartesian form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, we rewrite it as: $\frac{x - (-7)}{-6} = \frac{y - 6}{7} = \frac{z - 0}{1}$ From this, we identify:

  • A point on $L_1$, $(x_1, y_1, z_1)$: $(-7, 6, 0)$. So, its position vector is $\vec{a_1} = -7\hat{i} + 6\hat{j} + 0\hat{k}$.
  • The direction vector of $L_1$, $(a, b, c)$: $(-6, 7, 1)$. So, $\vec{b_1} = -6\hat{i} + 7\hat{j} + 1\hat{k}$.

• For Line 2 ($L_2$): The given equation is $\frac{7-x}{2}=y-2=z-6$. It is vital to ensure the terms in the numerator are in the form $(x-x_1)$, $(y-y_1)$, and $(z-z_1)$. The term $\frac{7-x}{2}$ must be rewritten as $\frac{-(x-7)}{2}$, which is equivalent to $\frac{x-7}{-2}$. So, the standard Cartesian equation for $L_2$ becomes: $\frac{x - 7}{-2} = \frac{y - 2}{1} = \frac{z - 6}{1}$ From this, we identify:

  • A point on $L_2$, $(x_2, y_2, z_2)$: $(7, 2, 6)$. So, its position vector is $\vec{a_2} = 7\hat{i} + 2\hat{j} + 6\hat{k}$.
  • The direction vector of $L_2$, $(a, b, c)$: $(-2, 1, 1)$. So, $\vec{b_2} = -2\hat{i} + 1\hat{j} + 1\hat{k}$.

Step 2: Calculate the Vector Connecting Points on the Lines ($\vec{a_2} - \vec{a_1}$) This vector connects the identified point on $L_1$ to the identified point on $L_2$. It is a crucial component for the numerator of the shortest distance formula. $\vec{a_2} - \vec{a_1} = (7\hat{i} + 2\hat{j} + 6\hat{k}) - (-7\hat{i} + 6\hat{j} + 0\hat{k})$ $\vec{a_2} - \vec{a_1} = (7 - (-7))\hat{i} + (2 - 6)\hat{j} + (6 - 0)\hat{k}$ $\vec{a_2} - \vec{a_1} = 14\hat{i} - 4\hat{j} + 6\hat{k}$

Step 3: Calculate the Cross Product of Direction Vectors ($\vec{b_1} \times \vec{b_2}$) The cross product of the direction vectors, $\vec{b_1} \times \vec{b_2}$, gives a vector that is perpendicular to both $\vec{b_1}$ and $\vec{b_2}$. This vector defines the direction of the common perpendicular between the two skew lines. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -6 & 7 & 1 \\ -2 & 1 & 1 \end{vmatrix}$ $= \hat{i}((7)(1) - (1)(1)) - \hat{j}((-6)(1) - (1)(-2)) + \hat{k}((-6)(1) - (7)(-2))$ $= \hat{i}(7 - 1) - \hat{j}(-6 - (-2)) + \hat{k}(-6 - (-14))$ $= 6\hat{i} - (-4)\hat{j} + 8\hat{k}$ $\vec{b_1} \times \vec{b_2} = 6\hat{i} + 4\hat{j} + 8\hat{k}$

Step 4: Calculate the Magnitude of the Cross Product ($|\vec{b_1} \times \vec{b_2}|$ ) The magnitude of the cross product forms the denominator of the shortest distance formula. It normalizes the direction vector of the common perpendicular. $|\vec{b_1} \times \vec{b_2}| = \sqrt{(6)^2 + (4)^2 + (8)^2}$ $= \sqrt{36 + 16 + 64}$ $= \sqrt{116}$ We can simplify $\sqrt{116}$ by factoring out perfect squares: $\sqrt{4 \times 29} = 2\sqrt{29}$. So, $|\vec{b_1} \times \vec{b_2}| = 2\sqrt{29}$.

Step 5: Calculate the Scalar Triple Product ($(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$) This dot product forms the numerator of the formula. It represents the scalar projection of the vector $(\vec{a_2} - \vec{a_1})$ onto the direction of the common perpendicular, which is precisely the shortest distance (before normalization). $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (14\hat{i} - 4\hat{j} + 6\hat{k}) \cdot (6\hat{i} + 4\hat{j} + 8\hat{k})$ $= (14)(6) + (-4)(4) + (6)(8)$ $= 84 - 16 + 48$ $= 68 + 48$ $= 116$

Step 6: Apply the Shortest Distance Formula Now, substitute all the calculated values into the shortest distance formula: $D = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ $D = \left| \frac{116}{2\sqrt{29}} \right|$ Since 116 and $2\sqrt{29}$ are positive, the absolute value can be removed: $D = \frac{116}{2\sqrt{29}}$ $D = \frac{58}{\sqrt{29}}$ To rationalize the denominator, multiply the numerator and denominator by $\sqrt{29}$: $D = \frac{58}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}}$ $D = \frac{58\sqrt{29}}{29}$ $D = 2\sqrt{29}$

Common Mistakes & Tips

• Sign Errors in Direction Vectors: A very common mistake is not correctly converting terms like $(7-x)$ to $(x-7)$ and adjusting the denominator's sign. Always ensure the numerators are in the form $(x-x_1)$, $(y-y_1)$, and $(z-z_1)$.
• Calculation Errors: Be meticulous with vector cross products, dot products, and magnitude calculations. A single arithmetic error can lead to an incorrect final answer.
• Forgetting Absolute Value: Distance must always be non-negative. Remember to take the absolute value of the scalar triple product divided by the magnitude of the cross product.

Summary

To find the shortest distance between two skew lines, we first convert their Cartesian equations into standard vector form to identify a point and a direction vector for each line. Then, we calculate the vector connecting the two points, the cross product of the direction vectors, and the magnitude of this cross product. Finally, we apply the shortest distance formula by taking the absolute value of the scalar triple product (dot product of the connecting vector with the cross product of direction vectors) divided by the magnitude of the cross product. Careful attention to algebraic signs and vector operations is essential for accuracy. Following these steps, the shortest distance between the given lines is $2\sqrt{29}$.

The final answer is \boxed{\text{2 \sqrt{29}}}, which corresponds to option (A).