Problem 35
Question
Find the areas of the parallelograms whose vertices are given in Exercises \(35-38 .\) $$ A(1,0), \quad B(0,1), \quad C(-1,0), \quad D(0,-1) $$
Step-by-Step Solution
Verified Answer
The area of the parallelogram is 4 square units.
1Step 1: Determine Diagonals
To find the area of the parallelogram, we begin by calculating the vectors that represent the diagonals. Compute vector \( \overrightarrow{AC} \) using points A(1,0) and C(-1,0): \( \overrightarrow{AC} = (-1-1, 0-0) = (-2, 0) \). Similarly, calculate vector \( \overrightarrow{BD} \) using points B(0,1) and D(0,-1): \( \overrightarrow{BD} = (0-0, -1-1) = (0, -2) \).
2Step 2: Find Cross Product
The area of the parallelogram can be calculated using the magnitude of the cross product of vectors \( \overrightarrow{AC} \) and \( \overrightarrow{BD} \). For vectors in 2D, treat them as 3D vectors with \(z\)-component zero: \((x_1, y_1, 0)\) and \((x_2, y_2, 0)\). The cross product is \((0, 0, 2 \cdot -2 - 0 \cdot 0) = (0, 0, -4)\).
3Step 3: Calculate Magnitude
Compute the magnitude of the resulting cross product. The magnitude is the length of the vector \((0, 0, -4)\) which is \(\sqrt{0^2 + 0^2 + (-4)^2} = \sqrt{16} = 4\).
4Step 4: State the Area
The magnitude of the cross product gives the area of the parallelogram formed by the points. Hence, the area is 4 square units.
Key Concepts
DiagonalsCross ProductMagnitudeVectors in Geometry
Diagonals
In the context of a parallelogram, a diagonal is a line segment connecting opposite corners. These diagonals are essential for dividing the shape into triangles, helping with calculations such as area determination.
To find the length of a diagonal represented as a vector, we calculate from one vertex to the opposite vertex. Consider diagonal \( \overrightarrow{AC} \) from point A to C. If you have the coordinates of these two points, simply subtract the position values of A from C. In our case, \( \overrightarrow{AC} = (-1-1, 0-0) = (-2, 0) \). Similarly, compute diagonal \( \overrightarrow{BD} \) as \( (0-0, -1-1) = (0, -2) \). By representing diagonals as vectors, we prepare them for further calculations like the cross product.
Diagonals play a pivotal role in vector geometry, helping harness knowledge of vector operations to determine geometrical properties like area.
To find the length of a diagonal represented as a vector, we calculate from one vertex to the opposite vertex. Consider diagonal \( \overrightarrow{AC} \) from point A to C. If you have the coordinates of these two points, simply subtract the position values of A from C. In our case, \( \overrightarrow{AC} = (-1-1, 0-0) = (-2, 0) \). Similarly, compute diagonal \( \overrightarrow{BD} \) as \( (0-0, -1-1) = (0, -2) \). By representing diagonals as vectors, we prepare them for further calculations like the cross product.
Diagonals play a pivotal role in vector geometry, helping harness knowledge of vector operations to determine geometrical properties like area.
Cross Product
The cross product is a mathematical operation performed on two vectors. In geometry, it helps us find a vector perpendicular to the plane containing the original vectors.
When dealing with 2D vectors like \( \overrightarrow{AC} = (-2, 0) \) and \( \overrightarrow{BD} = (0, -2) \), we treat them as 3D vectors with zero in the z-component: \( (-2, 0, 0) \) and \( (0, -2, 0) \). The cross product formula for vectors \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is given by:
\[ (y_1z_2 - z_1y_2, z_1x_2 - x_1z_2, x_1y_2 - y_1x_2) \]
Applying this to our vectors results in \((0, 0, -4)\), simplifying to a single component vector because the other components are zero. The magnitude of this gives us the area of the parallelogram.
Mastering the cross product equips you to handle similar exercises and explore more complex shapes in vector geometry.
When dealing with 2D vectors like \( \overrightarrow{AC} = (-2, 0) \) and \( \overrightarrow{BD} = (0, -2) \), we treat them as 3D vectors with zero in the z-component: \( (-2, 0, 0) \) and \( (0, -2, 0) \). The cross product formula for vectors \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is given by:
\[ (y_1z_2 - z_1y_2, z_1x_2 - x_1z_2, x_1y_2 - y_1x_2) \]
Applying this to our vectors results in \((0, 0, -4)\), simplifying to a single component vector because the other components are zero. The magnitude of this gives us the area of the parallelogram.
Mastering the cross product equips you to handle similar exercises and explore more complex shapes in vector geometry.
Magnitude
Magnitude is the length or size of a vector. Using the magnitude helps convert vector quantities into scalar values which are easier to understand in geometric terms like distance or area.
For a vector \(\mathbf{v} = (x, y, z)\), its magnitude \(|\mathbf{v}|\) is calculated using the formula:
\[ |\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} \]
In our example, the cross product from vectors \( (-2, 0) \) and \( (0, -2) \) resulted in \((0, 0, -4)\). The magnitude becomes \(\sqrt{0^2 + 0^2 + (-4)^2} = 4\).
Understanding magnitude is crucial as it translates abstract vectors into practical applications, providing meaningful numerical values. It helps you visualize the true dimension of vectors, whether they're in 2D or 3D.
For a vector \(\mathbf{v} = (x, y, z)\), its magnitude \(|\mathbf{v}|\) is calculated using the formula:
\[ |\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} \]
In our example, the cross product from vectors \( (-2, 0) \) and \( (0, -2) \) resulted in \((0, 0, -4)\). The magnitude becomes \(\sqrt{0^2 + 0^2 + (-4)^2} = 4\).
Understanding magnitude is crucial as it translates abstract vectors into practical applications, providing meaningful numerical values. It helps you visualize the true dimension of vectors, whether they're in 2D or 3D.
Vectors in Geometry
Vectors are fundamental tools in geometry, allowing the representation and analysis of various physical quantities like displacement, velocity, and force. In geometry, vectors provide a means to represent lines, planes, and even shapes like a parallelogram.
By using vectors, complex geometric problems become more manageable. They offer a straightforward approach to manipulating and calculating properties like area and volume. Vectors \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\) in our exercise illustrate the simplicity they bring.
By using vectors, complex geometric problems become more manageable. They offer a straightforward approach to manipulating and calculating properties like area and volume. Vectors \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\) in our exercise illustrate the simplicity they bring.
- Vectors have direction and magnitude.
- They are easily manipulated using basic operations like addition or multiplication.
- Vectors transform geometric problems into algebraic form.
Other exercises in this chapter
Problem 34
Sketch the surfaces in Exercises \(13-76\) $$ 9 x^{2}+4 y^{2}=36 z^{2} $$
View solution Problem 35
Sketch the surfaces in Exercises \(13-76\) $$ x^{2}+y^{2}-z^{2}=1 $$
View solution Problem 35
In Exercises 33–38, find the distance from the point to the line. $$ (2,1,3) ; \quad x=2+2 t, \quad y=1+6 t, \quad z=3 $$
View solution Problem 35
In Exercises \(35-40,\) find the distance between points \(P_{1}\) and \(P_{2}\) $$ P_{1}(1,1,1), \quad P_{2}(3,3,0) $$
View solution