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Mathematics Question on Three Dimensional Geometry

Equation of the line of the shortest distance between the lines x1=y−1=z1\frac{x}{1} = \frac{y}{-1} = \frac{z}{1} and x−10=y+1−2=z1\frac{x-1}{0} = \frac{y+1}{-2} = \frac{z}{1} is :

A

x1=y−1=z−2\frac{x}{1} = \frac{y}{-1} = \frac{z}{-2}

B

x−11=y+1−1=z−2\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z}{-2}

C

x−11=y+1−1=z−1\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z}{-1}

D

x−2=y1=z2\frac{x}{-2} = \frac{y}{1} = \frac{z}{2}

Answer

x−11=y+1−1=z−2\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z}{-2}

Explanation

Solution

∣i^j^k^\1−11\0−21∣\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\\1 & -1 & 1 \\\0 & -2 & 1\end{vmatrix}
x1=y−1=z1\frac{x}{1}=\frac{y}{-1}=\frac{z}{1}

x−10=y+1−2=z1\frac{x-1}{0}=\frac{y+1}{-2}=\frac{z}{1}
1−λ1=−1−2μ+λ−1=μ−λ−2\frac{1-\lambda}{1}=\frac{-1-2 \mu+\lambda}{-1}=\frac{\mu-\lambda}{-2}
−2=3λ+μ-2=3 \lambda+\mu
−1+λ=−1−2μ+λ-1+\lambda=-1-2 \mu+\lambda
μ=0\mu=0
λ=−23\lambda=-\frac{2}{3}