Answer:
[tex]\large\boxed{\large\boxed{70.6\º}}[/tex]
Explanation:
1. Calculate the scalar product using the coordinates
[tex]A\cdot B=(3.0\hat i-4.0\hat j+0\hat k)\cdot (-2.0\hat i+0\hat j+3.0\hat k)\\\\A\cdot B=(3)(-2)+(-4)(0)+(0)(3)=-6[/tex]
2. Write the scalar product using the cosine of the angle
[tex]A\cdot B=|A|\cdot |B|\cdot cos\theta[/tex]
[tex]|A|=\sqrt{(3.0)^2+(-4.0)^2}=\sqrt{9.0+16.}=5.0[/tex]
[tex]|B|=\sqrt{(-2.0)^2+(3.0)^2}=\sqrt{4.0+9.0}=\sqrt{13.0}[/tex]
3. Equal the two scalar products and solve for cos(θ)
[tex]-6.0=(5.0)(\sqrt{13.0})cos\theta\\\\cos\theta=0.3328\\\\\theta = arccos(0.3328)=70.6\º[/tex]
