If →a = 2ˆi - ˆj - ˆk and →b = 7ˆi + 2ˆj + ˆk, then express →b in the form of →b = →b1 + →b2, where →b1 is parallel to →a and →b2 is perpendicular to →a.

Given that →a = 2ˆi - ˆj - ˆk and →b = 7ˆi + 2ˆj + ˆk
Given that →b = →b1 + →b2 where →b1 is parallel to →a and →b2 is perpendicular to →a
Let →b1 = k→a, we have →a . →b = →a . →b1 + →a . →b2 = k→a . →a + 0 = k→a . →a
⇒ k = (→a . →b) / (→a . →a) = (14 - 2 + 1) / (4 + 1 + 1) = 13 / 6
So we get →b1 = (13/6)(2ˆi - ˆj - ˆk) = (13/3)ˆi - (13/6)ˆj - (13/6)ˆk
Now we have →b x →a = →b1 x →a + →b2 x →a = →b2 x →a
Let →b2 = xˆi + yˆj + zˆk, we get
(14ˆi + 8ˆj + 15ˆk) = (-y - z)ˆi + (2x + 2z)ˆj + (-x - 2y)ˆk
By comparing we get -y - z = 14, 2x + 2z = 8 and -x - 2y = 15
By solving , we get x = -2, y = -13/2, z = 27/2
So we get →b2 = -2ˆi - (13/2)ˆj + (27/2)ˆk