At what angle the two vectors a+b and a-b act so that their resultant is under root 3a^2+b^2
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Nikku0609:
I did not understand how it become a^2 - b^2
if you multiply (a+b)(a-b)
a (a-b)+b (a-b)
which is a^2-b^2
Bro it’s 3a^2 + b^2 = 2 ( a^2+b^2)+ 2(a^2-b^2)
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Answer:
Ф=π/3
Explanation:
R²=a²+b²+2abcosФ
3A²+B²=(A+B)²+(A-B)²+2(A+B)(A-B)cosФ
3A²+B²=A²+B²+2AB+A²+B²-2AB+2(A²-B²)cosФ
3A²-2A²+B²-2B²=2(A²-B²)cosФ
(A²-B²)=2(A²-B²)cosФ
1=2cosФ
cosФ=1/2
Ф=60degrees
=π/3
#SPJ2
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