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The angle between the vectors vec A = ha...

The angle between the vectors `vec A = hati+hatj and vec B =hati+hatj+cveck" is "30^(@)` What is the value of c ?

A

0

B

`pm 1`

C

`pm sqrt((2)/(3))`

D

`pm (1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`vec A cdot vec B = AB cos theta therefore cos theta =(vec A cdot vec B)/(|A| |B|)" …(1)"`
But `vec A cdot vec B =(hati + hatj) cdot (hati+hatj+ chatk)=1+1=2`
`|A|=sqrt (1^(2) +1^(2))= sqrt(2)`
`|B| = sqrt (1^(2)+1^(2)+c^(2)) = sqrt (2+c^(2))`
`therefore` From (1)
`cos theta = cos 30^(@) =(sqrt(3))/(2) =(2)/((sqrt(2) xx sqrt(2+ c^(2)))`
Squaring both sides, we get
`(3)/(4)=(4)/(2 cdot (2+ c^(2)))`
`therefore 3 (4+ 2c^(2)) =16" "therefore 6c^(2) =16 -12 =4`
`therefore c^(2) =(4)/(6) =(2)/(3)" "therefore c = pm sqrt((2)/(3))`
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