Question

given vector ( u = langle 1, sqrt{3}
angle ) and ( v = langle 1, -1
angle ), how are the vectors related?
the vectors form an acute angle of approximately ( 75^circ ).
the vectors form an acute angle of approximately ( 81^circ ).
the vectors form an obtuse angle of approximately ( 99^circ ).
the vectors form an obtuse angle of approximately ( 105^circ ).

Explanation:

Step1: Calculate the dot - product

The dot - product formula is \(\mathbf{u}\cdot\mathbf{v}=u_1v_1 + u_2v_2\).
For \(\mathbf{u}=\langle1,\sqrt{3}
angle\) and \(\mathbf{v}=\langle1, - 1
angle\), we have \(\mathbf{u}\cdot\mathbf{v}=(1\times1)+(\sqrt{3}\times(-1))=1-\sqrt{3}\).

Step2: Calculate the magnitudes

The magnitude formula is \(\vert\mathbf{u}\vert=\sqrt{u_1^{2}+u_2^{2}}\).
For \(\mathbf{u}\), \(\vert\mathbf{u}\vert=\sqrt{1^{2}+(\sqrt{3})^{2}}=\sqrt{1 + 3}=2\).
For \(\mathbf{v}\), \(\vert\mathbf{v}\vert=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}\).

Step3: Use the dot - product formula for the angle

The formula \(\mathbf{u}\cdot\mathbf{v}=\vert\mathbf{u}\vert\vert\mathbf{v}\vert\cos\theta\), so \(\cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{\vert\mathbf{u}\vert\vert\mathbf{v}\vert}\).
Substitute the values: \(\cos\theta=\frac{1-\sqrt{3}}{2\sqrt{2}}\approx\frac{1 - 1.732}{2\times1.414}=\frac{- 0.732}{2.828}\approx - 0.259\).
Then \(\theta=\cos^{-1}(-0.259)\approx105^{\circ}\).

Answer:

The vectors form an obtuse angle of approximately \(105^{\circ}\).