Math Problem Statement

Виріши задачу на фото

Знайти площу паралелограма, побудованого на векторах a = 2j + k та b = i + 2k.

Solution

To solve the problem in the image, we are tasked with finding the area of the parallelogram formed by the vectors a=2j+k\vec{a} = 2\vec{j} + \vec{k} and b=i+2k\vec{b} = \vec{i} + 2\vec{k}.

Solution:

The area of a parallelogram formed by two vectors a\vec{a} and b\vec{b} is given by the magnitude of their cross product:

Area=a×b\text{Area} = |\vec{a} \times \vec{b}|

Step 1: Define the vectors

Given the vectors: a=2j+k\vec{a} = 2\vec{j} + \vec{k} b=i+2k\vec{b} = \vec{i} + 2\vec{k}

In component form, these vectors can be written as: a=(0,2,1)\vec{a} = (0, 2, 1) b=(1,0,2)\vec{b} = (1, 0, 2)

Step 2: Compute the cross product a×b\vec{a} \times \vec{b}

The cross product of two vectors a=(a1,a2,a3)\vec{a} = (a_1, a_2, a_3) and b=(b1,b2,b3)\vec{b} = (b_1, b_2, b_3) is given by: a×b=i^j^k^a1a2a3b1b2b3\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}

Substitute the components of a\vec{a} and b\vec{b}: a×b=i^j^k^021102\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{vmatrix}

Now, expand the determinant: a×b=i^2102j^0112+k^0210\vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 2 & 1 \\ 0 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 2 \\ 1 & 0 \end{vmatrix}

a×b=i^(2×20×1)j^(0×21×1)+k^(0×01×2)\vec{a} \times \vec{b} = \hat{i} (2 \times 2 - 0 \times 1) - \hat{j} (0 \times 2 - 1 \times 1) + \hat{k} (0 \times 0 - 1 \times 2)

a×b=i^(4)j^(1)+k^(2)\vec{a} \times \vec{b} = \hat{i} (4) - \hat{j} (-1) + \hat{k} (-2)

Thus, the cross product is: a×b=4i^+j^2k^\vec{a} \times \vec{b} = 4\hat{i} + \hat{j} - 2\hat{k} or in component form: a×b=(4,1,2)\vec{a} \times \vec{b} = (4, 1, -2)

Step 3: Compute the magnitude of the cross product

The magnitude of the vector a×b=(4,1,2)\vec{a} \times \vec{b} = (4, 1, -2) is given by: a×b=42+12+(2)2|\vec{a} \times \vec{b}| = \sqrt{4^2 + 1^2 + (-2)^2}

a×b=16+1+4=21|\vec{a} \times \vec{b}| = \sqrt{16 + 1 + 4} = \sqrt{21}

Final Answer:

The area of the parallelogram is 21\sqrt{21}.


Would you like further details or have any questions?

Here are 5 related questions you can explore:

  1. How is the cross product of two vectors geometrically interpreted?
  2. How do you calculate the volume of a parallelepiped formed by three vectors?
  3. What is the difference between the dot product and the cross product of vectors?
  4. Can the area of a parallelogram be negative? Why or why not?
  5. How does the scalar triple product relate to the cross product?

Tip: Always verify your vector calculations, especially signs, when computing cross products!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vector Algebra
Cross Product
Geometry of Vectors

Formulas

Area of Parallelogram = |a × b|
Cross Product: a × b = |a||b|sin(θ)

Theorems

Properties of the Cross Product
Magnitude of a Vector

Suitable Grade Level

Grade 10-12