Math Problem Statement

Untuk tiga vektor berikut, berapakah nilai dari 3\vec{C} \cdot (2\vec{A} \times \vec{B})?

Solution

The problem is asking to calculate the value of 3C⃗⋅(A⃗×B⃗)3\vec{C} \cdot (\vec{A} \times \vec{B}), where the vectors are:

A⃗=2i^+3j^−4k^,B⃗=−3i^+4j^−2k^,C⃗=7i^−8j^.\vec{A} = 2\hat{i} + 3\hat{j} - 4\hat{k}, \quad \vec{B} = -3\hat{i} + 4\hat{j} - 2\hat{k}, \quad \vec{C} = 7\hat{i} - 8\hat{j}.

Step 1: Find A⃗×B⃗\vec{A} \times \vec{B} (the cross product)

The cross product of two vectors A⃗=a1i^+a2j^+a3k^\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} and B⃗=b1i^+b2j^+b3k^\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} is given by the determinant:

\hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \hat{i} \left( a_2b_3 - a_3b_2 \right) - \hat{j} \left( a_1b_3 - a_3b_1 \right) + \hat{k} \left( a_1b_2 - a_2b_1 \right)$$ Substituting the components of $$\vec{A}$$ and $$\vec{B}$$: $$\vec{A} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix}, \quad \vec{B} = \begin{pmatrix} -3 \\ 4 \\ -2 \end{pmatrix}$$ So, $$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -4 \\ -3 & 4 & -2 \end{vmatrix}$$ Expanding the determinant: $$\vec{A} \times \vec{B} = \hat{i} \left( 3 \times (-2) - (-4) \times 4 \right) - \hat{j} \left( 2 \times (-2) - (-4) \times (-3) \right) + \hat{k} \left( 2 \times 4 - 3 \times (-3) \right)$$ Calculating the components: - $$\hat{i}$$ component: $$3 \times (-2) - (-4) \times 4 = -6 + 16 = 10$$ - $$\hat{j}$$ component: $$2 \times (-2) - (-4) \times (-3) = -4 - 12 = -16$$ - $$\hat{k}$$ component: $$2 \times 4 - 3 \times (-3) = 8 + 9 = 17$$ Thus, the cross product is: $$\vec{A} \times \vec{B} = 10\hat{i} - 16\hat{j} + 17\hat{k}$$ ### Step 2: Find $$3\vec{C}$$ Next, we scale vector $$\vec{C}$$ by 3: $$3\vec{C} = 3(7\hat{i} - 8\hat{j}) = 21\hat{i} - 24\hat{j}$$ ### Step 3: Find the dot product $$3\vec{C} \cdot (\vec{A} \times \vec{B})$$ Now, calculate the dot product between $$3\vec{C}$$ and $$\vec{A} \times \vec{B}$$: $$3\vec{C} \cdot (\vec{A} \times \vec{B}) = (21\hat{i} - 24\hat{j}) \cdot (10\hat{i} - 16\hat{j} + 17\hat{k})$$ This dot product is: $$= 21 \times 10 + (-24) \times (-16) + (0 \times 17)$$ $$= 210 + 384 + 0 = 594$$ ### Final Answer: The value of $$3\vec{C} \cdot (\vec{A} \times \vec{B})$$ is **594**. Would you like further details or clarifications? --- Here are some related questions: 1. How is the cross product of two vectors different from the dot product? 2. What are the geometric interpretations of the dot and cross products? 3. How would the calculation change if one of the vectors was in 2D instead of 3D? 4. Can the result of a cross product ever be zero? If so, under what conditions? 5. How is the magnitude of the cross product related to the area of a parallelogram? **Tip**: Always keep track of signs when dealing with cross products, as they directly affect the direction of the resulting vector.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Cross Product
Dot Product
Scalar Multiplication

Formulas

Cross product: \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}
Dot product: \vec{C} \cdot \vec{D} = C_x D_x + C_y D_y + C_z D_z
Scalar multiplication: k \cdot \vec{C} = (kC_x)\hat{i} + (kC_y)\hat{j} + (kC_z)\hat{k}

Theorems

Properties of Vector Operations

Suitable Grade Level

Undergraduate Level or Advanced High School

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