Write the vector equation of each of the following lines and hence determine the distance between them : (x - 1)/2 = (y - 2)/3 = (z + 4)/6

Question

Write the vector equation of each of the following lines and hence determine the distance between them : 

\(\cfrac{\text x-1}2=\cfrac{y-2}3 = \cfrac{z + 4}6\) and \(\cfrac{\text x-3}4=\cfrac{y-3}6=\cfrac{z+5}{12}\).

HINT: The given lines are

(x - 1)/2 = (y - 2)/3 = (z + 4)/6, (x - 3)/4 = (y - 3)/6 = (z + 5)/12.

Now, find the distance between the parallel lines L₁ and L₂.

Answer 1

To Find : i) vector equations of given lines

ii) distance d

Formulae :

1. Equation of line :

Equation of line passing through point A (a₁, a₂, a₃) and having direction ratios (b₁, b₂, b₃) is

\(\vec r=\bar a_1+\lambda \bar{b}\)

4. Shortest distance between two parallel lines :

The shortest distance between the parallel lines \(\vec r=\bar a_1+\lambda \bar{b}\) and

\(\vec r=\bar a_2+\lambda \bar {b}\) is given by,

Given Cartesian equations of lines

Line L1 is passing through point (1, 2, -4) and has direction ratios (2, 3, 6)

Therefore, vector equation of line L1 is

Line L2 is passing through point (3, 3, -5) and has direction ratios (4, 6, 12)

Therefore, vector equation of line L2 is

Therefore, the shortest distance between the given lines is