If →a = ^i + ^j + ^k and →b = ^j − ^k, find a vector →c such that →a × →c = →b and →a ⋅ →c = 3.

Let →c = x→i + y→j + z→k
→a × →c = | | | →i →j →k | | | 1 1 1 | | | x y z | | | = →i(z − y) − →j(z − x) + →k(y − x)
Now, →a × →c = →b
Given →b = →j − →k
Therefore, z − y = 0, x − z = 1, y − x = −1
y = z and x − y = 1 — (1)
Again, →a ⋅ →c = 3
x + y + z = 3
x + y + y = 3
x + 2y = 3 —(2)
Solving 1 and 2,
x + 2y = 3
x − y = 1

3y = 2
y = 2/3 = z
Therefore,
x = 1 + y = 5/3
So, →c = 5/3→i + 2/3→j + 2/3→k