Please please please please solveI have complet answer

Please please please please solveI have complet answer

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Grade:12th pass

3 Answers

Piyush Kumar Behera
436 Points
4 years ago
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}}
=>\sqrt{3^{2}+2^{2}+b^{2}} = 1
=>   {3^{2}+2^{2}+b^{2} = 1
As b will come as a square root of a negative term so it has no real values.
I hope it helps.
Piyush Kumar Behera
12 Points
4 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
3007 Points
4 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.

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