Given: |a→| = 1, |b→| = sqrt(3). ∠(a→, b→) = (2*pi)/3. c→ = (a→ + b→) × (3b→ - a→). Find |c→| step by step

Ответ:

Given: \(|\vec{a}| = 1\), \(|\vec{b}| = \sqrt{3}\), \(\angle(\vec{a}, \vec{b}) = \frac{2\pi}{3}\), and \(\vec{c} = (\vec{a} + \vec{b}) \times (3\vec{b} - \vec{a})\).

Let's find \(|\vec{c}|\) step by step:

1. **Vector Addition:**

\[ \vec{a} + \vec{b} = \begin{bmatrix} a_1 + b_1 \\ a_2 + b_2 \end{bmatrix} \]

2. **Scalar Multiplication:**

\[ 3\vec{b} - \vec{a} = 3 \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} - \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} = \begin{bmatrix} 3b_1 - a_1 \\ 3b_2 - a_2 \end{bmatrix} \]

3. **Vector Cross Product:**

\[ \vec{c} = (\vec{a} + \vec{b}) \times (3\vec{b} - \vec{a}) = \begin{vmatrix} \hat{i} & \hat{j} \\ a_1 + b_1 & a_2 + b_2 \\ 3b_1 - a_1 & 3b_2 - a_2 \end{vmatrix} \]

Expand the determinant:

\[ \vec{c} = \hat{i}((a_2 + b_2)(3b_2 - a_2)) - \hat{j}((a_1 + b_1)(3b_1 - a_1)) \]

4. **Calculate \(|\vec{c}|\):**

\[ |\vec{c}| = \sqrt{(\text{coefficient of }\hat{i})^2 + (\text{coefficient of }\hat{j})^2} \]

Now, perform the calculations step by step according to the given values of \(\vec{a}\) and \(\vec{b}\).