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:
\[ \frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5} \] and \[ \frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1} \] is:
- \(\frac{187}{\sqrt{563}}\)
- \(\frac{178}{\sqrt{563}}\)
- \(\frac{185}{\sqrt{563}}\)
- \(\frac{179}{\sqrt{563}}\)
The Correct Option is A
Approach Solution - 1
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 \(\frac{187}{\sqrt{563}}\).
Approach Solution -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).
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