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Chapter 2: Problem 209

Find the area of the parallelogram with adjacent sides \(\mathbf{u}=\langle 3,2,0\rangle\) and \(\mathbf{v}=\langle 0,2,1\rangle.\)

Short Answer

Expert verified
The area of the parallelogram is 7.

Step by step solution

01

Understand the Geometry

The area of a parallelogram formed by two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by the magnitude of their cross product \( \mathbf{u} \times \mathbf{v} \).
02

Calculate the Cross Product

The cross product \( \mathbf{u} \times \mathbf{v} \) is calculated using the formula for vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), given by:\[ \mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \rangle \].For \( \mathbf{u} = \langle 3, 2, 0 \rangle \) and \( \mathbf{v} = \langle 0, 2, 1 \rangle \):\[ \mathbf{u} \times \mathbf{v} = \langle 2 \times 1 - 0 \times 2, 0 \times 0 - 3 \times 1, 3 \times 2 - 2 \times 0 \rangle = \langle 2, -3, 6 \rangle \].
03

Calculate the Magnitude of the Cross Product

The magnitude of the cross product \( \mathbf{u} \times \mathbf{v} = \langle 2, -3, 6 \rangle \) is calculated using:\[ |\mathbf{u} \times \mathbf{v}| = \sqrt{2^2 + (-3)^2 + 6^2} \].Compute:\[ |\mathbf{u} \times \mathbf{v}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \].Hence, the magnitude is 7.
04

Determine the Area

The area of the parallelogram is equal to the magnitude of \( \mathbf{u} \times \mathbf{v} \), which is 7.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

parallelogram area
To find the area of a parallelogram formed by two vectors, you utilize the cross product. The length of the cross product gives the magnitude essential for area calculations. This concept is crucial in vector calculus as it allows determining the "actual size" of shapes created by vectors. Imagine vectors as arrows originating from a common point; these vectors form parallelograms in 3D space. The area is not just about the boundaries, but about the comprehensive space covered. This is best appreciated with a real-world example:
  • The vectors \( \mathbf{u} = \langle 3, 2, 0 \rangle \) and \( \mathbf{v} = \langle 0, 2, 1 \rangle \) define two sides of a parallelogram.
  • Utilizing their cross product and its magnitude gives insight into the size of the parallelogram.
  • The geometrics of parallelograms involve understanding this spatial structure.
Even without actual images, we calculate areas using math operations, bringing complex structures to simpler understanding.
cross product
The cross product is a vital operation in vector calculus when working with three-dimensional vectors. It produces a new vector orthogonal (perpendicular) to the two input vectors. The cross product of vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) is calculated as:\[ \mathbf{u} \times \mathbf{v} = \langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \rangle \]This operation highlights several properties:
  • Direction: The resultant vector's direction is dictated by the right-hand rule. This means if you point your index finger in the direction of the first vector and your middle finger in the direction of the second, your thumb will point in the direction of the cross product.
  • Magnitude: The length of this new vector corresponds to the parallelogram area formed by the original vectors.
Understanding the cross product's direction and magnitude is key to solving problems in engineering and physics where forces and motions interact.
magnitude of vectors
The magnitude of a vector is its length or size, often requiring calculation when determining the spatial effects or dimensions related to the vector. For a given vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), its magnitude is defined as:\[ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]Applying this to the cross product \( \mathbf{u} \times \mathbf{v} = \langle 2, -3, 6 \rangle \), we compute:
  • First: Square each component and sum them up: \( 2^2 + (-3)^2 + 6^2 = 4 + 9 + 36 \)
  • Second: Take the square root of this result: \( \sqrt{49} \)
  • Finally: The magnitude is \( 7 \), illustrating the area of the parallelogram formed by the two original vectors.
Calculating magnitudes helps quantify abstract concepts in vector calculus, enabling precise interpretations and applications in our tangible world.

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Most popular questions from this chapter

Consider the planes of equations \(x+y+z=1\) and \(x+z=0\) a. Show that the planes intersect. b. Find symmetric equations of the line passing through point \(P(1,4,6)\) that is parallel to the line of intersection of the planes.

For the following exercises, points \(P, Q,\) and \(R\) are given. a. Find the general equation of the plane passing through \(P, Q,\) and \(R .\) b. Write the vector equation \(\mathbf{n} \cdot \overrightarrow{P S}=0\) of the plane at a, where \(S(x, y, z)\) is an arbitrary point of the plane. c. Find parametric equations of the line passing through the origin that is perpendicular to the plane passing through \(P, Q,\) and \(R\) \(P(-2,1,4), Q(3,1,3), and R(-2,1,0)\)

For the following exercises, the equation of a plane is given. a. Find normal vector \(\mathbf{n}\) to the plane. Express \(\mathbf{n}\) using standard unit vectors. b. Find the intersections of the plane with the axes of coordinates. c. Sketch the plane. \(3 x+4 y-12=0\)

\([T]\) The force vector \(F\) acting on a proton with anelectric charge of \(1.6 \times 10^{-19} \mathrm{C}\) moving in a magnetic field \(B\) where the velocity vector \(V\) is given by\(\mathbf{F}=1.6 \times 10^{-19}(\mathbf{v} \times \mathbf{B})\) (here, v is expressed in meters per second, \(B\) in \(T\), and \(F\) in \(N\)). If the magnitude of force \(F\) acting on a proton is\(5.9 \times 10^{-17} \mathrm{~N}\) and the proton is moving at the speed of 300 m/sec in magnetic field \(B\) of magnitude \(2.4 T\), find the angle between velocity vector \(V\) of the proton and magnetic field \(B\). Express the answer in degrees rounded to the nearest integer.

For the following exercises, lines \(L_{1}\) and \(L_{2}\) are given. Determine whether the lines are equal, parallel but not equal, skew, or intersecting. \(L_{1} : 3 x=y+1=2 z\) and \(L_{2} : x=6+2 t, y=17+6 t, z=9+3 t, \quad t \in \mathbb{R}\)
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