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fredd [130]
2 years ago
14

Question 15 (5 points) Find the angle between u = <7, -2> and v= (-1,2>.

Mathematics
1 answer:
Tema [17]2 years ago
8 0

Answer:

Approximately 2.3127 radians, which is approximately 132.51^{\circ}.

Step-by-step explanation:

Dot product between the two vectors:

\begin{aligned}& u\cdot v \\ =\; & 7 \times (-1) + (-2) \times 2 \\ =\; & (-11) \end{aligned}.

Magnitude of the two vectors:

\begin{aligned} \| u \| &= \sqrt{{7}^{2} + {(-2)}^{2}} \\ &= \sqrt{53} \end{aligned}.

\begin{aligned} \| v \| &= \sqrt{{(-1)}^{2} + {2}^{2}} \\ &= \sqrt{5} \end{aligned}.

Let \theta denote the angle between these two vectors. By the property of dot products:

\begin{aligned} \cos(\theta) &= \frac{u \cdot v}{\|u\| \, \| v \|} \\ &= \frac{(-11)}{(\sqrt{53})\, (\sqrt{5})} \\ &= \frac{(-11)}{\sqrt{265}}\end{aligned}.

Apply the inverse cosine function {\rm arccos} to find the value of this angle:

\begin{aligned} \theta &= \arccos\left(\frac{u \cdot v}{\| u \| \, \| v \|}\right) \\ &= \arccos\left(\frac{(-11)}{\sqrt{265}}\right) \\ & \approx \text{$2.3127$ radians} \\ &= 2.3127 \times \frac{180^{\circ}}{\pi} \\ &\approx 132.51^{\circ}\end{aligned}.

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