If the shortest distance between the straight lines $3(x-1)=6(y-2) = 2(z-1)$ and $4(x-2) = 2(y-\lambda) = (z-3)$, $\lambda\in R$ is $\frac{1}{\sqrt{38}}$, then the integral value of $\lambda$ is equal to :
Line Equation Conversion
First, convert the given equations of the straight lines into the standard symmetric form: $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$.
- Line 1: \(3(x-1)=6(y-2) = 2(z-1)\) Convert to standard form: \[ \frac{x-1}{1/3} = \frac{y-2}{1/6} = \frac{z-1}{1/2} \] Multiply denominators by 6 for simpler integers: \[ \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3} \]
- Line 2: \(4(x-2) = 2(y-\lambda) = (z-3)\) Convert to standard form: \[ \frac{x-2}{1/4} = \frac{y-\lambda}{1/2} = \frac{z-3}{1} \] Multiply denominators by 4 for simpler integers: \[ \frac{x-2}{1} = \frac{y-\lambda}{2} = \frac{z-3}{4} \]
Identify Points and Direction Vectors
From the standard forms, identify a point on each line ($\vec{a_1}$, $\vec{a_2}$) and their direction vectors ($\vec{d_1}$, $\vec{d_2}$).
- Line 1: Point \(A = (1, 2, 1)\) => $\vec{a_1} = \langle 1, 2, 1 \rangle$. Direction vector $\vec{d_1} = \langle 2, 1, 3 \rangle$.
- Line 2: Point \(B = (2, \lambda, 3)\) => $\vec{a_2} = \langle 2, \lambda, 3 \rangle$. Direction vector $\vec{d_2} = \langle 1, 2, 4 \rangle$.
Vector Calculations
Calculate the vector connecting the points on the lines and the cross product of the direction vectors.
- Vector connecting points: \[ \vec{a_2} - \vec{a_1} = \langle 2-1, \lambda-2, 3-1 \rangle = \langle 1, \lambda-2, 2 \rangle \]
- Cross product of direction vectors: \[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix} = \mathbf{i}(4-6) - \mathbf{j}(8-3) + \mathbf{k}(4-1) = \langle -2, -5, 3 \rangle \]
- Magnitude of the cross product: \[ |\vec{d_1} \times \vec{d_2}| = \sqrt{(-2)^2 + (-5)^2 + 3^2} = \sqrt{4 + 25 + 9} = \sqrt{38} \]
- Dot product of $(\vec{a_2} - \vec{a_1})$ and $(\vec{d_1} \times \vec{d_2})$: \[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = \langle 1, \lambda-2, 2 \rangle \cdot \langle -2, -5, 3 \rangle \] \[ = 1(-2) + (\lambda-2)(-5) + 2(3) = -2 - 5\lambda + 10 + 6 = 14 - 5\lambda \]
Shortest Distance Calculation
Use the formula for the shortest distance \(d\) between two skew lines:
\[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} \]Substitute the calculated values:
\[ d = \frac{|14 - 5\lambda|}{\sqrt{38}} \]Solving for Lambda
We are given that the shortest distance \(d = \frac{1}{\sqrt{38}}\). Set the formula equal to the given distance:
\[ \frac{|14 - 5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}} \]Simplify the equation:
\[ |14 - 5\lambda| = 1 \]This gives two possibilities:
- \(14 - 5\lambda = 1 \implies 5\lambda = 13 \implies \lambda = \frac{13}{5}\)
- \(14 - 5\lambda = -1 \implies 5\lambda = 15 \implies \lambda = 3\)
Identify Integral Value
The question asks for the integral value of \(\lambda\). Comparing the two possible values, \(\lambda = 3\) is the integral value.
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