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
25
Calculating Distance Between Lines and Point from Line
We are given two lines, $L_1$ and $L_2$, and information about the shortest distance between them.
Line Representations
- Line $L_1$: Passes through $P_1 = (1, 2, 3)$ and is parallel to the z-axis. Its direction vector is $\vec{d_1} = \langle 0, 0, 1 \rangle$.
- Line $L_2$: Passes through $P_2 = (\lambda, 5, 6)$ and is parallel to the y-axis. Its direction vector is $\vec{d_2} = \langle 0, 1, 0 \rangle$.
Shortest Distance Between $L_1$ and $L_2$
The vector connecting the points $P_1$ and $P_2$ is $\vec{P_2} - \vec{P_1} = \langle \lambda - 1, 5 - 2, 6 - 3 \rangle = \langle \lambda - 1, 3, 3 \rangle$.
The cross product of the direction vectors is $\vec{d_1} \times \vec{d_2} = \langle 0, 0, 1 \rangle \times \langle 0, 1, 0 \rangle = \langle -1, 0, 0 \rangle$.
The magnitude of the cross product is $|\vec{d_1} \times \vec{d_2}| = \sqrt{(-1)^2 + 0^2 + 0^2} = 1$.
The shortest distance $D$ between two skew lines is given by the formula:
$ D = \frac{|(\vec{P_2} - \vec{P_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} $Substituting the values:
$ D = \frac{|\langle \lambda - 1, 3, 3 \rangle \cdot \langle -1, 0, 0 \rangle|}{1} = |(\lambda - 1)(-1) + 3(0) + 3(0)| = |-( \lambda - 1 )| = |1 - \lambda| $Finding $\lambda_1$ and $\lambda_2$
We are given that the shortest distance $D = 3$. Therefore:
$ |1 - \lambda| = 3 $This leads to two possibilities:
- $1 - \lambda = 3 \implies \lambda = 1 - 3 = -2$
- $1 - \lambda = -3 \implies \lambda = 1 + 3 = 4$
Given that $\lambda_2 < \lambda_1$, we have $\lambda_1 = 4$ and $\lambda_2 = -2$.
Distance of Point $(\lambda_1, \lambda_2, 7)$ from Line $L_1$
The point is $P = (\lambda_1, \lambda_2, 7) = (4, -2, 7)$.
Line $L_1$ passes through $P_1 = (1, 2, 3)$ with direction vector $\vec{d_1} = \langle 0, 0, 1 \rangle$.
The vector from $P_1$ to $P$ is $\vec{P} - \vec{P_1} = \langle 4 - 1, -2 - 2, 7 - 3 \rangle = \langle 3, -4, 4 \rangle$.
The distance from point $P$ to line $L_1$ is given by:
$ \text{Dist} = \frac{|(\vec{P} - \vec{P_1}) \times \vec{d_1}|}{|\vec{d_1}|} $Calculate the cross product:
$ (\vec{P} - \vec{P_1}) \times \vec{d_1} = \langle 3, -4, 4 \rangle \times \langle 0, 0, 1 \rangle = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -4 & 4 \\ 0 & 0 & 1 \end{vmatrix} = \langle -4, -3, 0 \rangle $The magnitude of the cross product is:
$ | \langle -4, -3, 0 \rangle | = \sqrt{(-4)^2 + (-3)^2 + 0^2} = \sqrt{16 + 9} = \sqrt{25} = 5 $The magnitude of the direction vector $\vec{d_1}$ is $|\vec{d_1}| = \sqrt{0^2 + 0^2 + 1^2} = 1$.
The distance is $\text{Dist} = \frac{5}{1} = 5$.
Square of the Distance
The question asks for the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from line $L_1$.
$ \text{Square of Distance} = (\text{Dist})^2 = 5^2 = 25 $Similar Questions
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Important Questions from Coordinate Geometry
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