Respuesta :
Answer:
a. The angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(\sqrt{3}, 1\right)[/tex] and [tex]\mathbf{v}=\left(0, 5\right)[/tex] is [tex]\phi=\frac{\pi}{3}[/tex] or [tex]60\º[/tex].
b. The angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(0, 4, 4\right)[/tex] and [tex]\mathbf{v}=\left(3, -3, 0\right)[/tex] is [tex]\phi=\frac{2 \pi}{3}[/tex] or [tex]120\º[/tex].
Step-by-step explanation:
a. To calculate the angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(\sqrt{3}, 1\right)[/tex] and [tex]\mathbf{v}=\left(0, 5\right)[/tex] you must:
Step 1: Calculate the dot product.
The dot product is given as [tex]\displaystyle{\large{{\left({u}_{{x}},{u}_{{y}}\right)}\cdot{\left({v}_{{x}},{v}_{{y}}\right)}={u}_{{x}}\cdot{v}_{{x}}+{u}_{{y}}\cdot{v}_{{y}}}}[/tex].
So,
[tex]\left(\sqrt{3}, 1\right)\cdot\left(0, 5\right)=\left(\sqrt{3}\right)\cdot\left(0\right)+\left(1\right)\cdot\left(5\right)=5[/tex]
Step 2: Find the lengths of the vectors.
[tex]\left|\mathbf{u}\right|=\sqrt{\left(u_x\right)^2+\left(u_y\right)^2}=\sqrt{\left(\sqrt{3}\right)^2+\left(1\right)^2}=2[/tex]
[tex]\left|\mathbf{v}\right|=\sqrt{\left(v_x\right)^2+\left(v_y\right)^2}=\sqrt{\left(0\right)^2+\left(5\right)^2}=5[/tex]
Step 3: The angle is given by [tex]\cos\left(\phi\right)=\frac{\mathbf{u} \cdot \mathbf{v}}{\left|\mathbf{u}\right| \cdot \left|\mathbf{v}\right|}[/tex]
[tex]\frac{5}{2 \cdot 5}=\frac{1}{2}[/tex]
[tex]\phi=cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}=60^0[/tex]
b. To calculate the angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(0, 4, 4\right)[/tex] and [tex]\mathbf{v}=\left(3, -3, 0\right)[/tex] you must:
Step 1: Calculate the dot product.
[tex]\left(0, 4, 4\right)\cdot\left(3, -3, 0\right)=\left(0\right)\cdot\left(3\right)+\left(4\right)\cdot\left(-3\right)+\left(4\right)\cdot\left(0\right)=-12[/tex]
Step 2: Find the lengths of the vectors.
[tex]\left|\mathbf{u}\right|=\sqrt{\left(u_x\right)^2+\left(u_y\right)^2+\left(u_z\right)^2}=\sqrt{\left(0\right)^2+\left(4\right)^2+\left(4\right)^2}=4 \sqrt{2}[/tex]
[tex]\left|\mathbf{v}\right|=\sqrt{\left(v_x\right)^2+\left(v_y\right)^2+\left(v_z\right)^2}=\sqrt{\left(3\right)^2+\left(-3\right)^2+\left(0\right)^2}=3 \sqrt{2}[/tex]
Step 3: The angle is given by [tex]\cos\left(\phi\right)=\frac{\mathbf{u} \cdot \mathbf{v}}{\left|\mathbf{u}\right| \cdot \left|\mathbf{v}\right|}[/tex]
[tex]\cos\left(\phi\right)=\frac{\mathbf{u} \cdot \mathbf{v}}{\left|\mathbf{u}\right| \cdot \left|\mathbf{v}\right|}=\frac{-12}{4 \sqrt{2} \cdot 3 \sqrt{2}}=- \frac{1}{2}[/tex]
[tex]\phi=cos^{-1}\left(- \frac{1}{2}\right)=\frac{2 \pi}{3}=120^0[/tex]
