Math Problem Statement

Jika \(\vec{A} = A_1 \hat{i} + A_2 \hat{j}\) dan \(\vec{B} = B_1 \hat{i} + B_2 \hat{j}\) adalah dua vektor pada bidang dengan sistem koordinat tegak lurus XY dan jika didefinisikan \(\vec{A} \cdot \vec{B} = A_1 B_1 + A_2 B_2\) maka \(\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta\), di mana \(\theta\) adalah sudut yang dibentuk oleh \(\vec{A}\) dan \(\vec{B}\), 0 ≤ \(\theta\) ≤ \(\pi\), buktikanlah!

Solution

The problem involves two vectors, A\vec{A} and B\vec{B}, in a 2D Cartesian coordinate system (XY). The vectors are expressed as:

A=A1i^+A2j^,B=B1i^+B2j^\vec{A} = A_1 \hat{i} + A_2 \hat{j}, \quad \vec{B} = B_1 \hat{i} + B_2 \hat{j}

The dot product of these vectors is defined as:

AB=A1B1+A2B2\vec{A} \cdot \vec{B} = A_1 B_1 + A_2 B_2

We are also given the general definition of the dot product in terms of magnitudes and the cosine of the angle θ\theta between them:

AB=ABcosθ\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta

where θ\theta is the angle between vectors A\vec{A} and B\vec{B}, and 0θπ0 \leq \theta \leq \pi.

We are asked to prove the relationship between these two expressions for the dot product.

Steps for the Proof:

  1. Calculate the Magnitudes of A\vec{A} and B\vec{B}: The magnitude (or length) of a vector A\vec{A} is given by: A=A12+A22|\vec{A}| = \sqrt{A_1^2 + A_2^2} Similarly, for vector B\vec{B}: B=B12+B22|\vec{B}| = \sqrt{B_1^2 + B_2^2}

  2. Dot Product in Cartesian Form: The dot product AB\vec{A} \cdot \vec{B} in terms of components is: AB=A1B1+A2B2\vec{A} \cdot \vec{B} = A_1 B_1 + A_2 B_2

  3. Dot Product in Geometric Form: The dot product can also be written as: AB=ABcosθ\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta where θ\theta is the angle between A\vec{A} and B\vec{B}.

  4. Equating the Two Forms of the Dot Product: Set the two expressions for the dot product equal to each other: A1B1+A2B2=ABcosθA_1 B_1 + A_2 B_2 = |\vec{A}| |\vec{B}| \cos \theta Thus, you can solve for cosθ\cos \theta: cosθ=A1B1+A2B2AB\cos \theta = \frac{A_1 B_1 + A_2 B_2}{|\vec{A}| |\vec{B}|} where: A=A12+A22,B=B12+B22|\vec{A}| = \sqrt{A_1^2 + A_2^2}, \quad |\vec{B}| = \sqrt{B_1^2 + B_2^2}

This proves the relationship between the two forms of the dot product.


Would you like further details on this proof? If so, I can break it down more or explain specific parts in greater detail.

Related Questions:

  1. How do you compute the angle θ\theta if the components of both vectors are known?
  2. What is the geometric interpretation of the dot product in 2D?
  3. Can the dot product ever be negative? If so, under what conditions?
  4. What happens to the dot product when the two vectors are orthogonal?
  5. How can this method be extended to 3D vectors?

Tip:

Remember that the dot product provides a way to find the angle between two vectors and determine if they are orthogonal (dot product = 0).

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Dot Product
Trigonometry

Formulas

Dot product in Cartesian form: \(\vec{A} \cdot \vec{B} = A_1 B_1 + A_2 B_2\)
Magnitude of a vector: \(|\vec{A}| = \sqrt{A_1^2 + A_2^2}\)
Dot product in geometric form: \(\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta\)

Theorems

Dot product properties
Relationship between dot product and angle between vectors

Suitable Grade Level

Grades 11-12