Grade 11Mechanics
0.4i^+0.8j^+bk^ is unit vector. ,what is the value of b?
0.4i^+0.8j^+bk^ is unit vector. ,what is the value of b?
5 Answers
Vikas TU
Hiii Alisa
for a unit vector , its magnitude should be equal to 1.
so here we will find the magnitude of vector and put that equal to 1.
so square root (0.42+0.82+b2) = 1
the value of b would be 0.44
Hope this helps.
Jiya
Unit vector means its magnitude is equal to 1
so magnitude:
(0.4^2 +0.8^2 + b^2) = 1
(0.16 + 0.64 + b^2) = 1
(0.80 + b^2) = 1
B^2 = 1- 0.8
b^2 = 0.2
b= 0.2 ^ 1/2
Harsh
Unit vector is a vector whose magnitude is equal to one
Hence
(0.4)²+(0.8)²+(b)²=1
1.6+6.4+b²=1
b²=1-0.8=0.2
Khimraj
Unit vector means magnitude of vector is unity .
1=root {(0.4)^2+(0.8)^2+(b)^2}
take square both side
1=0.16 +0.64 +b^2
b^2 =0.2
b=+_0.45
1=root {(0.4)^2+(0.8)^2+(b)^2}
take square both side
1=0.16 +0.64 +b^2
b^2 =0.2
b=+_0.45
aswanth nayak
Dear student,
Since it is a unit vector, it's magnitude = 1.
(0.4)²+(0.8)²+(b)²=1
(0.4)²+(0.8)²+(b)²=1
1.6+6.4+b²=1
b²=1-0.8=0.2
Hope this helps you
Regards
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