The angle between a and b is 0.999°
Given,
a = 8i - 3j
b = -4i + 6j
We have to find the angle between the resultant of a, b and the x axis.
We know that,
Vec A . Vec B = |A||B|cosθ
Let the angle between Vec a and Vec b be θ, then
Vec a = 8i - 3j
Vec b = -4i + 6j
The modulus is given as,
|| Vec a || = [tex]\sqrt{8^{2}+(-3)^{2} }[/tex] = [tex]\sqrt{64 - 9}[/tex] = [tex]\sqrt{55}[/tex]
|| Vec b || = [tex]\sqrt{(-4)^{2}+6^{2} }[/tex] = [tex]\sqrt{16+36}[/tex] = [tex]\sqrt{52}[/tex]
For getting angle, we have to find the scalar product also.
So, the scalar product is,
Vec a . Vec b = (8) (-4) + (-3)(6)
= -32 - 18
= -50
By using Vec A . Vec B = || A || . || B || . cosθ we have,
-50 = [tex]\sqrt{55}[/tex] × [tex]\sqrt{52}[/tex] × cosθ
cosθ = [tex]\frac{-50}{\sqrt{55} \sqrt{52} }[/tex]
cosθ = 2.510659539
θ = 0.9990400892
θ ≈ 0.999
The angle between a and b is 0.999°
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