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Question

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

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Solution

Let c=xi+yj+zk
a×c=∣ ∣ ∣ijk111xyz∣ ∣ ∣
=i(zy)j(zx)+k(yx)
Now, a×c=b
Given b=jk
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
__________
2y=2
y=23=z
Therefore, x=1+y=53
So, c=53i+23j+23k

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