Determine the shortest distance between the lines l1 and l2 whose vector equations are \(\vec{r}=2 \hat{\imath}+\hat{\jmath}+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \quad \ldots(1)\) \(\vec{r}=3 \hat{\imath}+\hat{\jmath}-\hat{k}+\mu(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}) \quad \ldots(2)\)
To determine the shortest distance between two lines given their vector equations, we utilize a specific formula that applies to skew lines (lines that are not parallel and do not intersect). The process involves identifying key vectors from the line equations and then performing vector operations.
Shortest Distance Formula
The vector equations of the two lines are typically given in the form \(\vec{r} = \vec{a_1} + \lambda \vec{b_1}\) and \(\vec{r} = \vec{a_2} + \mu \vec{b_2}\). Here, \(\vec{a_1}\) and \(\vec{a_2}\) are the position vectors of points on the respective lines, and \(\vec{b_1}\) and \(\vec{b_2}\) are vectors parallel to the lines. The shortest distance \(D\) between these two skew lines is calculated using the formula:
\[ D = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| \]
Line Equations Identification
Let's first identify the vectors \(\vec{a_1}, \vec{b_1}, \vec{a_2},\) and \(\vec{b_2}\) from the given vector equations:
- For Line 1: \(\vec{r}=2 \hat{\imath}+\hat{\jmath}+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k})\)
From this equation, we can identify:
- The position vector of a point on Line 1: \(\vec{a_1} = 2 \hat{\imath}+\hat{\jmath}\)
- The direction vector parallel to Line 1: \(\vec{b_1} = 2 \hat{\imath}-\hat{\jmath}+\hat{k}\)
- For Line 2: \(\vec{r}=3 \hat{\imath}+\hat{\jmath}-\hat{k}+\mu(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k})\)
From this equation, we can identify:
- The position vector of a point on Line 2: \(\vec{a_2} = 3 \hat{\imath}+\hat{\jmath}-\hat{k}\)
- The direction vector parallel to Line 2: \(\vec{b_2} = 3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}\)
Vector Operations for Distance Calculation
Now, we will perform the necessary vector operations step-by-step as required by the shortest distance formula.
Difference Vector
First, calculate the vector difference between the position vectors \(\vec{a_2}\) and \(\vec{a_1}\):
\[ \vec{a_2} - \vec{a_1} = (3 \hat{\imath}+\hat{\jmath}-\hat{k}) - (2 \hat{\imath}+\hat{\jmath}) \]
\[ \vec{a_2} - \vec{a_1} = (3-2)\hat{\imath} + (1-1)\hat{\jmath} + (-1-0)\hat{k} \]
\[ \vec{a_2} - \vec{a_1} = \hat{\imath} - \hat{k} \]
Cross Product of Direction Vectors
Next, compute the cross product of the direction vectors \(\vec{b_1}\) and \(\vec{b_2}\). This vector \(\vec{b_1} \times \vec{b_2}\) is perpendicular to both lines.
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} \]
Expanding the determinant to find the components:
\[ \vec{b_1} \times \vec{b_2} = \hat{\imath}((-1)(2) - (1)(-5)) - \hat{\jmath}((2)(2) - (1)(3)) + \hat{k}((2)(-5) - (-1)(3)) \]
\[ \vec{b_1} \times \vec{b_2} = \hat{\imath}(-2 + 5) - \hat{\jmath}(4 - 3) + \hat{k}(-10 + 3) \]
\[ \vec{b_1} \times \vec{b_2} = 3\hat{\imath} - \hat{\jmath} - 7\hat{k} \]
Magnitude of the Cross Product
Now, calculate the magnitude (length) of the cross product vector \(\vec{b_1} \times \vec{b_2}\):
\[ |\vec{b_1} \times \vec{b_2}| = |3\hat{\imath} - \hat{\jmath} - 7\hat{k}| \]
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(3)^2 + (-1)^2 + (-7)^2} \]
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{9 + 1 + 49} \]
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{59} \]
Dot Product (Scalar Triple Product Numerator)
Next, find the dot product of the difference vector \((\vec{a_2} - \vec{a_1})\) with the cross product vector \((\vec{b_1} \times \vec{b_2})\):
\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (\hat{\imath} - \hat{k}) \cdot (3\hat{\imath} - \hat{\jmath} - 7\hat{k}) \]
\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (1)(3) + (0)(-1) + (-1)(-7) \]
\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 3 + 0 + 7 \]
\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 10 \]
Final Shortest Distance Calculation
Substitute all the calculated values into the shortest distance formula:
\[ D = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| \]
\[ D = \left| \frac{10}{\sqrt{59}} \right| \]
\[ D = \frac{10}{\sqrt{59}} \]
Thus, the shortest distance between the given lines is \(\frac{10}{\sqrt{59}}\) units.
| Component | Calculated Value |
|---|---|
| \(\vec{a_1}\) | \(2 \hat{\imath}+\hat{\jmath}\) |
| \(\vec{b_1}\) | \(2 \hat{\imath}-\hat{\jmath}+\hat{k}\) |
| \(\vec{a_2}\) | \(3 \hat{\imath}+\hat{\jmath}-\hat{k}\) |
| \(\vec{b_2}\) | \(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}\) |
| \(\vec{a_2} - \vec{a_1}\) | \(\hat{\imath} - \hat{k}\) |
| \(\vec{b_1} \times \vec{b_2}\) | \(3\hat{\imath} - \hat{\jmath} - 7\hat{k}\) |
| Magnitude \(|\vec{b_1} \times \vec{b_2}|\) | \(\sqrt{59}\) |
| Dot Product \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\) | \(10\) |
| Shortest Distance \(D\) | \(\frac{10}{\sqrt{59}}\) |
Important Questions from Distance between points
Find the distance between the points P (2, -1, 3) and Q (-5, 2, 1)?
The sum of distances from origin to (0, 5, 5) and (5, 8, 6) is:
The distance between the points (2, 3) and (4, 1) is.
The vertices of a triangle are \(A(2, 0, 0)\), \(B(0, 6, 0)\) and \(C(0, 0, 4)\). If \(AD\), \(BE\) and \(CF\) are the medians of the triangle, then what is \(AD^2 + BE^2 + CF^2\) equal to?
The distance of a point from the x-axis is called that point.
