Line $ L_1 $ passes through the point (1, 2, 3) and is parallel to z-axis. Line $ L_2 $ passes through the point $ (\lambda, 5, 6) $ and is parallel to y-axis. Let for $ \lambda = \lambda_1, \lambda_2, \lambda_2<\lambda_1 $, the shortest distance between the two lines be 3. Then the square of the distance of the point $ (\lambda_1, \lambda_2, 7) $ from the line $ L_1 $ is
Show Hint
- 40
- 32
- 25
- 37
The Correct Option is C
Approach Solution - 1
The equation of line \( L_1 \) is: \( \frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1} \)
The equation of line \( L_2 \) is: \( \frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0} \)
The shortest distance (SD) between the lines is given by: \( SD = \frac{ \begin{vmatrix} \lambda - 1 & 5 - 2 & 6 - 3 0 & 0 & 1 \\ 0 & 1 & 0\end{vmatrix} }{ \sqrt{0^2 + 1^2 + 0^2} } \)
\( SD = \begin{vmatrix} \lambda - 1 & 3 & 3 0 & 0 & 1 \\ 0 & 1 & 0 \end{vmatrix} \)
\( SD = |\lambda - 1| \)
Given SD = 3, we have: \( |\lambda - 1| = 3 \) \( \lambda - 1 = \pm 3 \) \( \lambda = 4, -2 \)
Since \( \lambda_2<\lambda_1 \), we have: \( \lambda_1 = 4 \) \( \lambda_2 = -2 \) The point is \( P(4, -2, 7) \).
Let Q be the foot of the perpendicular from P to \( L_1 \). Q is of the form \( (1, 2, 3+t) \).
The direction vector of PQ is \( (1-4, 2-(-2), 3+t-7) = (-3, 4, t-4) \).
The direction vector of \( L_1 \) is \( (0, 0, 1) \).
Since PQ is perpendicular to \( L_1 \), their dot product is 0. \( (-3, 4, t-4) \cdot (0, 0, 1) = 0 \) \( t-4 = 0 \) \( t = 4 \) So, Q is \( (1, 2, 7) \).
Now, \( PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 \) \( PQ^2 = 3^2 + (-4)^2 + 0^2 \) \( PQ^2 = 9 + 16 = 25 \)
Approach Solution -2
To solve this problem, we need to find the square of the distance of the point \((\lambda_1, \lambda_2, 7)\) from the line \(L_1\), given that lines \(L_1\) and \(L_2\) have a shortest distance of 3 for \(\lambda = \lambda_1, \lambda_2\) where \(\lambda_2 < \lambda_1\).
- Line \(L_1\) passes through the point \((1, 2, 3)\) and is parallel to the z-axis. Hence, the vector equation of line \(L_1\) is: \(\mathbf{r_1} = (1, 2, 3) + t(0, 0, 1)\), where \(t\) is a parameter.
- Line \(L_2\) passes through \((\lambda, 5, 6)\) and is parallel to the y-axis. Thus, the vector equation of line \(L_2\) is: \(\mathbf{r_2} = (\lambda, 5, 6) + s(0, 1, 0)\), where \(s\) is a parameter.
- The direction vectors of lines \(L_1\) and \(L_2\) are \(\mathbf{d_1} = (0, 0, 1)\) and \(\mathbf{d_2} = (0, 1, 0)\), respectively.
- The vector joining points on \(L_1\) and \(L_2\) is \(\mathbf{r_1} - \mathbf{r_2} = (1-\lambda, -3, -3)\).
- The shortest distance \(D\) between two skew lines is given by: \(D = \frac{|(\mathbf{b-a}) \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|}\). Here, \(\mathbf{b-a}\) represents the vector joining any points on the two lines.
- The cross product \(\mathbf{d_1} \times \mathbf{d_2} = (0, 0, 1) \times (0, 1, 0) = (-1, 0, 0)\).
- The magnitude of this cross product is: \(|\mathbf{d_1} \times \mathbf{d_2}| = 1\).
- The dot product \((\mathbf{r_1} - \mathbf{r_2}) \cdot (\mathbf{d_1} \times \mathbf{d_2})\) is: \((1-\lambda, -3, -3) \cdot (-1, 0, 0) = -1 + \lambda\).
- Equating the expression for distance to 3, we get: \(|-1 + \lambda| = 3\) leads to \(\lambda = 4\) or \(\lambda = -2\).
- Thus, \(\lambda_1 = 4\) and \(\lambda_2 = -2\) (since \(\lambda_2 < \lambda_1\)). So, the point is \((4, -2, 7)\).
- The squared distance from the point \((4, -2, 7)\) to line \(L_1\) is the distance from \((4, -2, 7)\) to \((1, 2, 3)\) perpendicular to \(\mathbf{d_1} = (0, 0, 1)\).
- The perpendicular vector from \((4, -2, 7)\) to line \(a\) is \((4-1, -2-2, 0) = (3, -4, 0)\), where the z-component is 0 because it's in the direction perpendicular to the z-axis.
- The squared distance is \((3)^2 + (-4)^2 = 9 + 16 = 25\).
Therefore, the square of the distance of the point \((\lambda_1, \lambda_2, 7)\) from \(L_1\) is 25. The correct answer is 25.
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