1. P.v of A = i – j + 4k,
P.v. of B = 2i + j + 2k
\(\overline {AB}\) = p. v. of B – p. v. of A
= 2i + j + 2k -(i – j + 4k) = i + 2j – 2k.
2. The line L passes through (1, -2, -3) and parallel to \(\overline{AB}\)
3. From (1) of part (ii), we have
xi + yj + zk = (l + λ)i + (-2 + 2λ)j + (-3 – 2λ)k
Put λ = 1
⇒ xi +yj + zk = (1 +1)i + (-2 + 2)j + (-3 – 2 )k
⇒ xi + yj + zk = 2i + 0j – 5k
Therefore a point on line L is (2, 0, -5).