Question

The shortest distance between the line:
\[ \frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5} \] and \[ \frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1} \] is:

• A. \(\frac{187}{\sqrt{563}}\)
• B. \(\frac{178}{\sqrt{563}}\)
• C. \(\frac{185}{\sqrt{563}}\)
• D. \(\frac{179}{\sqrt{563}}\)

Answer 1

The Correct Option is A

To find the shortest distance between two skew lines, we use the formula for the distance between two skew lines:

Let's consider the lines: \(L₁: \frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5}\) 
\(L₂: \frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1}\)

The direction vectors for these lines are:

• For \(L₁\): \(\mathbf{d₁} = (4, -11, 5)\)
• For \(L₂\): \(\mathbf{d₂} = (3, -6, 1)\)

The formula for the distance \(d\) between two skew lines is: \(d = \frac{|(\mathbf{a₂} - \mathbf{a₁}) \cdot (\mathbf{d₁} \times \mathbf{d₂})|}{|\mathbf{d₁} \times \mathbf{d₂}|}\)

Substitute the initial points on each line: \(\mathbf{a₁} = (3, -7, 1)\) and \(\mathbf{a₂} = (5, 9, -2)\).

First, find the vector between points on the lines:

\(\mathbf{a₂} - \mathbf{a₁} = (5 - 3, 9 + 7, -2 - 1) = (2, 16, -3)\)

Next, calculate the cross product of \(\mathbf{d₁}\) and \(\mathbf{d₂}\):

\(\mathbf{d₁} \times \mathbf{d₂} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{vmatrix}\)

\(= \mathbf{i}((-11)(1) - (5)(-6)) - \mathbf{j}((4)(1) - (5)(3)) + \mathbf{k}((4)(-6) - (-11)(3))\)

\(= \mathbf{i}(-11 + 30) - \mathbf{j}(4 - 15) + \mathbf{k}(-24 + 33)\)

\(= (19, 11, 9)\)

Now, calculate the dot product \((\mathbf{a₂} - \mathbf{a₁}) \cdot (\mathbf{d₁} \times \mathbf{d₂})\):

\((2, 16, -3) \cdot (19, 11, 9) = (2)(19) + (16)(11) + (-3)(9)\)

\(= 38 + 176 - 27 = 187\)

Next, find the magnitude of the cross product \(|\mathbf{d₁} \times \mathbf{d₂}|\):

\(=\sqrt{19^2 + 11^2 + 9^2}\)

\(=\sqrt{361 + 121 + 81}\)

\(=\sqrt{563}\)

Finally, calculate the shortest distance \(d\):

\(d = \frac{|187|}{\sqrt{563}}\) = \(\frac{187}{\sqrt{563}}\)

Therefore, the shortest distance between the given lines is \(\frac{187}{\sqrt{563}}\).

Answer 2

Step 1: Represent the lines in vector form The first line can be written as:

\(\vec{r_1} = \vec{a_1} + \lambda \vec{p}, \quad \text{where } \vec{a_1} = 3\hat{i} - 7\hat{j} + \hat{k}, \quad \vec{p} = 4\hat{i} - 11\hat{j} + 5\hat{k}.\)

The second line can be written as:

\(\vec{r_2} = \vec{a_2} + \mu \vec{q}, \quad \text{where } \vec{a_2} = 5\hat{i} + 9\hat{j} - 2\hat{k}, \quad \vec{q} = 3\hat{i} - 6\hat{j} + \hat{k}.\)

Step 2: Find the direction vector perpendicular to both lines The direction vector perpendicular to both lines is:

\(\vec{n} = \vec{p} \times \vec{q}.\)

Using the determinant method for the cross product:

\(\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{vmatrix}.\)

Expanding the determinant:

\(\vec{n} = \hat{i}((-11)(1) - (-6)(5)) - \hat{j}((4)(1) - (3)(5)) + \hat{k}((4)(-6) - (3)(-11)).\)

\(\vec{n} = \hat{i}(-11 + 30) - \hat{j}(4 - 15) + \hat{k}(-24 + 33).\)

\(\vec{n} = 19\hat{i} + 11\hat{j} + 9\hat{k}.\)

Step 3: Find \(\vec{AB}\) The vector \(\vec{AB}\) is:

\(\vec{AB} = \vec{a_2} - \vec{a_1} = (5 - 3)\hat{i} + (9 - (-7))\hat{j} + (-2 - 1)\hat{k}.\)

\(\vec{AB} = 2\hat{i} + 16\hat{j} - 3\hat{k}.\)

Step 4: Shortest distance formula The shortest distance between two skew lines is:

\(\text{S.D.} = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{n}|}.\)

Dot product \(\vec{AB} \cdot \vec{n}\):

\(\vec{AB} \cdot \vec{n} = (2)(19) + (16)(11) + (-3)(9).\)

\(\vec{AB} \cdot \vec{n} = 38 + 176 - 27 = 187.\)

Magnitude of \(\vec{n}\):

\(|\vec{n}| = \sqrt{19^2 + 11^2 + 9^2}.\)

\(|\vec{n}| = \sqrt{361 + 121 + 81} = \sqrt{563}.\)

Shortest distance:

\(\text{S.D.} = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{n}|} = \frac{187}{\sqrt{563}}.\)

Final Answer: Option (1).