1.     Use the definition of scalar product, *a*G ⋅*b*G = *ab* cos θ, and the fact that *a*G ⋅*b*G = *axbx* + *ayby* + *azbz* , calculate the angle between the two vectors given by *a*G= 3.0*i*ˆ+3.0ˆ*j* +3.0*k*ˆ    and    *b*G= 2.0*i*ˆ+1.0ˆ*j* +3.0 .*k*ˆ

We can calculate the angle between the two vectors \(\vec{a}\) and \(\vec{b}\) using the definition of the scalar (dot) product:

\[
\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta
\]

Where:
- \(\vec{a} = 3.0\hat{i} + 3.0\hat{j} + 3.0\hat{k}\)
- \(\vec{b} = 2.0\hat{i} + 1.0\hat{j} + 3.0\hat{k}\)
- \(\theta\) is the angle between the vectors.

### Step 1: Calculate the dot product \(\vec{a} \cdot \vec{b}\)

The dot product of two vectors \(\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}\) and \(\vec{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k}\) is:

\[
\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z
\]

For \(\vec{a} = 3.0\hat{i} + 3.0\hat{j} + 3.0\hat{k}\) and \(\vec{b} = 2.0\hat{i} + 1.0\hat{j} + 3.0\hat{k}\):

\[
\vec{a} \cdot \vec{b} = (3.0)(2.0) + (3.0)(1.0) + (3.0)(3.0)
\]

\[
\vec{a} \cdot \vec{b} = 6.0 + 3.0 + 9.0 = 18.0
\]

### Step 2: Calculate the magnitudes \(|\vec{a}|\) and \(|\vec{b}|\)

The magnitude of a vector \(\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}\) is given by:

\[
|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}
\]

For \(\vec{a} = 3.0\hat{i} + 3.0\hat{j} + 3.0\hat{k}\):

\[
|\vec{a}| = \sqrt{(3.0)^2 + (3.0)^2 + (3.0)^2} = \sqrt{9.0 + 9.0 + 9.0} = \sqrt{27.0} = 3\sqrt{3}
\]

Similarly, for \(\vec{b} = 2.0\hat{i} + 1.0\hat{j} + 3.0\hat{k}\):

\[
|\vec{b}| = \sqrt{(2.0)^2 + (1.0)^2 + (3.0)^2} = \sqrt{4.0 + 1.0 + 9.0} = \sqrt{14.0}
\]

### Step 3: Calculate the angle \(\theta\)

From the dot product formula:

\[
\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta
\]

We can solve for \(\cos \theta\):

\[
\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{18.0}{(3\sqrt{3})(\sqrt{14})}
\]

Simplifying:

\[
\cos \theta = \frac{18.0}{3\sqrt{42}} = \frac{6}{\sqrt{42}}
\]

Now, calculate \(\cos \theta\):

\[
\cos \theta = \frac{6}{\sqrt{42}} \approx \frac{6}{6.4807} \approx 0.9258
\]

Finally, solve for \(\theta\):

\[
\theta = \cos^{-1}(0.9258)
\]

\[
\theta \approx 22.4^\circ
\]

### Conclusion:
The angle between the two vectors \(\vec{a}\) and \(\vec{b}\) is approximately \(22.4^\circ\).

---

Would you like any additional details or have further questions?

Here are 5 questions related to this problem:
1. How does the magnitude of a vector affect the dot product?
2. Can the dot product be negative, and if so, what does that signify about the angle between the vectors?
3. How would the calculation change if one of the vectors was a unit vector?
4. What are the properties of dot products in higher dimensions?
5. How is the dot product used in physics applications such as work or energy?

**Tip:** The dot product of two perpendicular vectors is always zero.

Use the definition of scalar product, aG ⋅ bG = ab cos θ, and the fact that aG ⋅ bG = axbx + ayby + azbz, calculate the angle between the two vectors given by aG = 3.0iˆ + 3.0jˆ + 3.0kˆ and bG = 2.0iˆ + 1.0jˆ + 3.0kˆ.

Learn how to calculate the angle between two vectors a = 3.0i + 3.0j + 3.0k and b = 2.0i + 1.0j + 3.0k using the dot product formula. This solution involves key vector algebra concepts, including calculating dot products, vector magnitudes, and applying the cosine rule. Suitable for students in Grades 10-12.

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