Question

given vectors ( u = (-2, 3) ) and ( v = (-6, 9) ), what is the measure of the angle between ( u ) and ( v )?
( 0^{circ} )
( 86^{circ} )
( 91^{circ} )
( 180^{circ} )

Explanation:

Step1: Calculate the dot - product

The dot - product formula is \(\mathbf{u}\cdot\mathbf{v}=u_1v_1 + u_2v_2\).
For \(\mathbf{u}=(-2,3)\) and \(\mathbf{v}=(-6,9)\), \(\mathbf{u}\cdot\mathbf{v}=(-2)\times(-6)+3\times9 = 12 + 27=39\).

Step2: Calculate the magnitudes

The magnitude formula is \(|\mathbf{a}|=\sqrt{a_1^{2}+a_2^{2}}\).
\(|\mathbf{u}|=\sqrt{(-2)^{2}+3^{2}}=\sqrt{4 + 9}=\sqrt{13}\), \(|\mathbf{v}|=\sqrt{(-6)^{2}+9^{2}}=\sqrt{36+81}=\sqrt{117}=3\sqrt{13}\).

Step3: Use the dot - product formula for the angle \(\theta\)

The formula \(\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta\), so \(\cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\).
Substitute the values: \(\cos\theta=\frac{39}{\sqrt{13}\times3\sqrt{13}}=\frac{39}{3\times13}=1\).
Since \(\cos\theta = 1\) and \(0^{\circ}\leq\theta\leq180^{\circ}\), then \(\theta = 0^{\circ}\).

Another way:
We can check if the vectors are scalar multiples. Let \(\mathbf{v}=k\mathbf{u}\), then \((-6,9)=k(-2,3)\).
From the first component: \(-6=-2k\Rightarrow k = 3\).
From the second component: \(9 = 3k\Rightarrow k=3\). So \(\mathbf{v}=3\mathbf{u}\), which means the vectors are parallel (in the same direction), and the angle between them is \(0^{\circ}\).

Answer:

A. \(0^{\circ}\)