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Question

Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(i) (0, 0, 0) and (1, 0, 2)
(ii) (1, 3, 0) and (0, 3, 0)

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Solution

(i) The equation of the line passing through the points (0, 0, 0) and (1, 0, 2) is

x-01-0=y-00-0=z-02-0=x1=y0=z2

(ii) The equation of the line passing through the points (1, 3, 0) and (0, 3, 0) is

x-10-1=y-33-3=z-00-0=x-1-1=y-30=z0

Since the first line passes through the point (0, 0, 0) and has direction ratios proportional to 1, 0, 2, its vector equation is
r=a1+λb1 ...(1) Here,a1=0i^+0j^+0k^ b1=i^+0j^+2k^

Also, the second line passes through the point (1, 3, 0) and has direction ratios proportional to -1, 0, 0.
Its vector equation is
r=a2+μb2 ...(2) Here,a2=i^+3j^+0k^ b2=- i^+0j^+0k^

Now,

a2- a1=i^+3j^+0k^and b1×b2=i^j^k^102-100=0i^-2j^+0k^ b1×b2=02+-22+02 =0+4+0 =2and a2- a1.b1×b2=i^+3j^+0k^.0i^-2j^+0k^ =-6

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2 =-62 =3 d=3 units

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