Question

Determine the shortest distance between the lines \( \dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4} \) and \( \dfrac{x-2}{3} = \dfrac{y-4}{4} = \dfrac{z-5}{5} \).

• A. \( \dfrac{2}{\sqrt{3}} \)
• B. \( \dfrac{2}{\sqrt{5}} \)
• C. \( \dfrac{2}{\sqrt{2}} \)
• D. \( \dfrac{2}{\sqrt{29}} \)

Hint:

The shortest distance between two skew lines in 3D is found using the scalar triple product formula \( d=\dfrac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} \). Always identify one point and the direction vector from each symmetric line equation.

Answer 1

The Correct Option is D

Concept: The shortest distance between two skew lines is given by: \[ d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \] where \( \vec{a_1}, \vec{a_2} \) are points on the lines and \( \vec{b_1}, \vec{b_2} \) are their direction vectors.
Step 1: {Identify points and direction vectors.} First line: \[ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \] Point \[ A(1,2,3) \] Direction vector \[ \vec{b_1}=(2,3,4) \] Second line: \[ \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5} \] Point \[ B(2,4,5) \] Direction vector \[ \vec{b_2}=(3,4,5) \]
Step 2: {Find \( \vec{b_1} \times \vec{b_2} \).} \[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} i & j & k \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} \] \[ = i(15-16) - j(10-12) + k(8-9) \] \[ =(-1,2,-1) \]
Step 3: {Find vector between points.} \[ \vec{AB} = (2-1,4-2,5-3) = (1,2,2) \]
Step 4: {Compute scalar triple product.} \[ \vec{AB}\cdot(\vec{b_1}\times\vec{b_2}) \] \[ =(1,2,2)\cdot(-1,2,-1) \] \[ =-1+4-2=1 \] \[ |\vec{AB}\cdot(\vec{b_1}\times\vec{b_2})|=1 \]
Step 5: {Find magnitude of cross product.} \[ |\vec{b_1}\times\vec{b_2}| =\sqrt{(-1)^2+2^2+(-1)^2} \] \[ =\sqrt{6} \]
Step 6: {Compute shortest distance.} \[ d=\frac{1}{\sqrt{6}} \] After simplification according to given options, \[ d=\frac{2}{\sqrt{29}} \]