Question

If \( d_1 \) is the shortest distance between the lines  \(\frac{x+1}{2} = \frac{y-1}{-12} = \frac{z-1}{-7} \quad \text{and} \quad \frac{x-1}{7} = \frac{y+8}{2} = \frac{z-4}{5},\) and \( d_2 \) is the shortest distance between the lines \(\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3} \quad \text{and} \quad \frac{x+2}{1} = \frac{y+2}{2} = \frac{z-1}{6},\) then the value of \( \frac{32\sqrt{3}d_1}{d_2} \) is:

Answer 1

Correct Answer: 16

To find d₁, the shortest distance between the lines L₁ and L₂:

L₁ : \( \frac{x + 1}{2} = \frac{y - 1}{-12} = \frac{z}{1} \), L₂ : \( \frac{x - 1}{-7} = \frac{y + 8}{2} = \frac{z - 4}{5} \)

Using the formula for the distance between two skew lines \( d = \frac{|(\vec{a_2} - \vec{a_1}) \times (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \), we calculate:

\( d_1 = 2 \)

Similarly, to find d₂ for lines L₃ and L₄:

L₃ : \( \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{-3} \), L₄ : \( \frac{x + 2}{1} = \frac{y + 2}{1} = \frac{z - 1}{6} \)

we get: \( d_2 = \frac{12}{\sqrt{3}} \)

Finally,

\( \frac{32 \sqrt{3} d_1}{d_2} = \frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}} = 16 \)

Answer 2

The problem asks for the value of the expression \( \frac{32\sqrt{3}d_1}{d_2} \), where \(d_1\) and \(d_2\) are the shortest distances between two specified pairs of skew lines.

Concept Used:

The shortest distance between two skew lines given by the vector equations \( \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \) is calculated using the formula:

\[ d = \frac{\left| (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \right|}{\left| \vec{b}_1 \times \vec{b}_2 \right|} \]

We will first convert the Cartesian equations of the lines into standard form to identify the point vectors (\(\vec{a}\)) and direction vectors (\(\vec{b}\)) for each line, and then apply this formula to find \(d_1\) and \(d_2\).

Step-by-Step Solution:

Part 1: Calculation of the shortest distance \(d_1\)

Step 1: Convert the equations for the first pair of lines to standard form.

Line 1: \( x + 1 = 2y = -12z \)

To put this in the form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), we divide by the LCM of the coefficients' denominators (which is 12):

\[ \frac{x - (-1)}{12} = \frac{2y}{12} = \frac{-12z}{12} \implies \frac{x - (-1)}{12} = \frac{y - 0}{6} = \frac{z - 0}{-1} \]

Line 2: \( x = y + 2 = 6z - 6 \)

We rewrite this as:

\[ \frac{x - 0}{6} = \frac{y - (-2)}{6} = \frac{6(z - 1)}{6} \implies \frac{x - 0}{6} = \frac{y - (-2)}{6} = \frac{z - 1}{1} \]

Step 2: Identify the point and direction vectors for the first pair of lines.

\( \vec{a}_1 = -\hat{i} + 0\hat{j} + 0\hat{k} \), \( \vec{b}_1 = 12\hat{i} + 6\hat{j} - \hat{k} \)

\( \vec{a}_2 = 0\hat{i} - 2\hat{j} + \hat{k} \), \( \vec{b}_2 = 6\hat{i} + 6\hat{j} + \hat{k} \)

Step 3: Compute the necessary vector products.

\[ \vec{a}_2 - \vec{a}_1 = (0 - (-1))\hat{i} + (-2 - 0)\hat{j} + (1 - 0)\hat{k} = \hat{i} - 2\hat{j} + \hat{k} \] \[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 12 & 6 & -1 \\ 6 & 6 & 1 \end{vmatrix} = \hat{i}(6 - (-6)) - \hat{j}(12 - (-6)) + \hat{k}(72 - 36) = 12\hat{i} - 18\hat{j} + 36\hat{k} \]

Step 4: Calculate \(d_1\).

Numerator: \( \left| (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \right| = \left| (1)(12) + (-2)(-18) + (1)(36) \right| = \left| 12 + 36 + 36 \right| = 84 \)

Denominator: \( \left| \vec{b}_1 \times \vec{b}_2 \right| = \sqrt{12^2 + (-18)^2 + 36^2} = \sqrt{144 + 324 + 1296} = \sqrt{1764} = 42 \)

\[ d_1 = \frac{84}{42} = 2 \]

Part 2: Calculation of the shortest distance \(d_2\)

Step 5: Identify the point and direction vectors for the second pair of lines.

Line 3: \( \frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5} \)

Line 4: \( \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3} \)

\( \vec{a}_3 = \hat{i} - 8\hat{j} + 4\hat{k} \), \( \vec{b}_3 = 2\hat{i} - 7\hat{j} + 5\hat{k} \)

\( \vec{a}_4 = \hat{i} + 2\hat{j} + 6\hat{k} \), \( \vec{b}_4 = 2\hat{i} + \hat{j} - 3\hat{k} \)

Step 6: Compute the necessary vector products for the second pair.

\[ \vec{a}_4 - \vec{a}_3 = (1 - 1)\hat{i} + (2 - (-8))\hat{j} + (6 - 4)\hat{k} = 0\hat{i} + 10\hat{j} + 2\hat{k} \] \[ \vec{b}_3 \times \vec{b}_4 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21 - 5) - \hat{j}(-6 - 10) + \hat{k}(2 - (-14)) = 16\hat{i} + 16\hat{j} + 16\hat{k} \]

Step 7: Calculate \(d_2\).

Numerator: \( \left| (\vec{a}_4 - \vec{a}_3) \cdot (\vec{b}_3 \times \vec{b}_4) \right| = \left| (0)(16) + (10)(16) + (2)(16) \right| = \left| 160 + 32 \right| = 192 \)

Denominator: \( \left| \vec{b}_3 \times \vec{b}_4 \right| = \sqrt{16^2 + 16^2 + 16^2} = \sqrt{3 \times 16^2} = 16\sqrt{3} \)

\[ d_2 = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3} \]

Final Computation & Result:

Step 8: Substitute the values of \(d_1\) and \(d_2\) into the final expression.

We need to find the value of \( \frac{32\sqrt{3}d_1}{d_2} \).

We found \( d_1 = 2 \) and \( d_2 = 4\sqrt{3} \).

\[ \frac{32\sqrt{3} \times 2}{4\sqrt{3}} = \frac{64\sqrt{3}}{4\sqrt{3}} \]

The \( \sqrt{3} \) terms cancel out.

\[ \frac{64}{4} = 16 \]

The value of the expression is 16.