Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Please please please please solveI have complet answer

Please please please please solveI have complet answer

Question Image
Grade:12th pass

3 Answers

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.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free