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Piyush Kumar Behera
435 Points
3 years ago
@Pawan
In the given question as $\vec{A}$ is a unit vector
hence the magnitude of $\vec{A}$ is 1
So as we know that magnitude of $\vec{A}$ is$\sqrt{3^{2}+2^{2}+b^{2}}$
So
=>$\sqrt{3^{2}+2^{2}+b^{2}} = 1$
=>   ${3^{2}+2^{2}+b^{2} = 1$
$b=\sqrt{-12}$

As b will come as a square root of a negative term so it has no real values.
I hope it helps.

Regards
Piyush Kumar Behera

vedant
12 Points
3 years ago
as the vector a is a unit vector its magnitude should be 1 therefore square root of 3 square plus 2 square plus b square should be 1 therefore b is root 12 iota where iota is root of minus 1
Khimraj
3007 Points
3 years ago
Unit vector is a vector whose magnitude is 1.
So |A| = 1
then $\sqrt{3^{2} + 2^{2} + b^{2}}$ = 1
So b = 12i.
Hope it clears.