Use a dot product to find the angle between each pair of
vectors (to the nearest tenth of a degree).
13.
14.

Ask by Wagner Robbins. in the United States

Jan 23,2025

Question

Use a dot product to find the angle between each pair of
vectors (to the nearest tenth of a degree).
13.
14.

Ask by Wagner Robbins. in the United States

Jan 23,2025

UpStudy AI · Accepted Answer

To find the angle between two vectors using the **dot product**, you can use the following formula:

$$
\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \, |\mathbf{b}|}
$$

Where:
- $$\mathbf{a} \cdot \mathbf{b}$$ is the dot product of vectors $$\mathbf{a}$$ and $$\mathbf{b}$$.
- $$|\mathbf{a}|$$ and $$|\mathbf{b}|$$ are the magnitudes (lengths) of the vectors.
- $$\theta$$ is the angle between the vectors.

The angle $$\theta$$ can then be found by taking the inverse cosine ($$\cos^{-1}$$) of the result.

Let's apply this method to the given problems.

---

### Problem 13
**Given Vectors:**
$$
\mathbf{u} = \langle 1, 4 \rangle, \quad \mathbf{v} = \langle -3, -2 \rangle
$$

**Step 1: Compute the Dot Product ($$\mathbf{u} \cdot \mathbf{v}$$)**
$$
\mathbf{u} \cdot \mathbf{v} = (1)(-3) + (4)(-2) = -3 - 8 = -11
$$

**Step 2: Compute the Magnitudes ($$|\mathbf{u}|$$ and $$|\mathbf{v}|$$)**
$$
|\mathbf{u}| = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.1231
$$
$$
|\mathbf{v}| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.6056
$$

**Step 3: Compute $$\cos \theta$$**
$$
\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| \, |\mathbf{v}|} = \frac{-11}{4.1231 \times 3.6056} \approx \frac{-11}{14.837} \approx -0.7415
$$

**Step 4: Compute the Angle ($$\theta$$)**
$$
\theta = \cos^{-1}(-0.7415) \approx 136.3^\circ
$$

**Answer:** The angle between $$\mathbf{u}$$ and $$\mathbf{v}$$ is approximately **136.3 degrees**.

---

### Problem 14
**Given Vectors:**
$$
\mathbf{r} = \langle 7, 6 \rangle, \quad \mathbf{s} = \langle 2, 9 \rangle
$$

**Step 1: Compute the Dot Product ($$\mathbf{r} \cdot \mathbf{s}$$)**
$$
\mathbf{r} \cdot \mathbf{s} = (7)(2) + (6)(9) = 14 + 54 = 68
$$

**Step 2: Compute the Magnitudes ($$|\mathbf{r}|$$ and $$|\mathbf{s}|$$)**
$$
|\mathbf{r}| = \sqrt{7^2 + 6^2} = \sqrt{49 + 36} = \sqrt{85} \approx 9.2195
$$
$$
|\mathbf{s}| = \sqrt{2^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85} \approx 9.2195
$$

**Step 3: Compute $$\cos \theta$$**
$$
\cos \theta = \frac{\mathbf{r} \cdot \mathbf{s}}{|\mathbf{r}| \, |\mathbf{s}|} = \frac{68}{9.2195 \times 9.2195} = \frac{68}{85} \approx 0.8
$$

**Step 4: Compute the Angle ($$\theta$$)**
$$
\theta = \cos^{-1}(0.8) \approx 36.9^\circ
$$

**Answer:** The angle between $$\mathbf{r}$$ and $$\mathbf{s}$$ is approximately **36.9 degrees**.

---
**Problem 13:** - Vectors: $$ \mathbf{u} = \langle 1, 4 \rangle $$, $$ \mathbf{v} = \langle -3, -2 \rangle $$ - Angle: Approximately **136.3 degrees** **Problem 14:** - Vectors: $$ \mathbf{r} = \langle 7, 6 \rangle $$, $$ \mathbf{s} = \langle 2, 9 \rangle $$ - Angle: Approximately **36.9 degrees**

Use a dot product to find the angle between each pair of
vectors (to the nearest tenth of a degree).
13.
14.

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Problem 13

Problem 14

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Use a dot product to find the angle between each pair of vectors (to the nearest tenth of a degree). 13. 14.

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Use a dot product to find the angle between each pair of
vectors (to the nearest tenth of a degree).
13.
14.