Question

the magnitude of vector u is 34, and the magnitude of vector v is 20. the dot product u · v = -10. what is the approximate angle between the vectors? 79° 89° 91° 101°

Explanation:

Step1: Recall the dot - product formula

The dot - product formula is \(\mathbf{u}\cdot\mathbf{v}=\vert\mathbf{u}\vert\vert\mathbf{v}\vert\cos\theta\), where \(\theta\) is the angle between the vectors \(\mathbf{u}\) and \(\mathbf{v}\), \(\vert\mathbf{u}\vert\) is the magnitude of vector \(\mathbf{u}\), and \(\vert\mathbf{v}\vert\) is the magnitude of vector \(\mathbf{v}\).
We are given that \(\vert\mathbf{u}\vert = 34\), \(\vert\mathbf{v}\vert=20\), and \(\mathbf{u}\cdot\mathbf{v}=- 10\).
Substitute these values into the formula: \(-10=(34)(20)\cos\theta\).

Step2: Solve for \(\cos\theta\)

First, simplify the right - hand side: \((34)(20)\cos\theta = 680\cos\theta\).
So, the equation becomes \(\cos\theta=\frac{-10}{680}=-\frac{1}{68}\approx - 0.0147\).

Step3: Find \(\theta\)

We know that \(\theta=\cos^{-1}(\cos\theta)\).
Using a calculator, \(\theta=\cos^{-1}(-0.0147)\approx91^{\circ}\).

Answer:

C. \(91^{\circ}\)