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For each pair of vectors, find \(\mathbf{U}+\mathbf{V}, \mathbf{U}-\mathbf{V}\), and \(3 \mathbf{U}+2 \mathbf{V}\). $$\mathbf{U}=2 \mathbf{i}+5 \mathbf{j}, \mathbf{V}=5 \mathbf{i}+2 \mathbf{j}$$

Short Answer

Expert verified
\( \mathbf{U} + \mathbf{V} = 7\mathbf{i} + 7\mathbf{j} \), \( \mathbf{U} - \mathbf{V} = -3\mathbf{i} + 3\mathbf{j} \), \( 3\mathbf{U} + 2\mathbf{V} = 16\mathbf{i} + 19\mathbf{j} \).

Step by step solution

01

Add the Vectors

To find \( \mathbf{U} + \mathbf{V} \), we simply add the corresponding components of vectors \( \mathbf{U} \) and \( \mathbf{V} \). \( \mathbf{U} = 2\mathbf{i} + 5\mathbf{j} \), and \( \mathbf{V} = 5\mathbf{i} + 2\mathbf{j} \).\[ \mathbf{U} + \mathbf{V} = (2+5)\mathbf{i} + (5+2)\mathbf{j} = 7\mathbf{i} + 7\mathbf{j} \]
02

Subtract the Vectors

For \( \mathbf{U} - \mathbf{V} \), subtract the corresponding components of vector \( \mathbf{V} \) from \( \mathbf{U} \).\[ \mathbf{U} - \mathbf{V} = (2-5)\mathbf{i} + (5-2)\mathbf{j} = -3\mathbf{i} + 3\mathbf{j} \]
03

Multiply and Add the Vectors

To find \( 3\mathbf{U} + 2\mathbf{V} \), first multiply each vector by the scalar, then add the results.\[ 3\mathbf{U} = 3(2\mathbf{i} + 5\mathbf{j}) = 6\mathbf{i} + 15\mathbf{j} \]\[ 2\mathbf{V} = 2(5\mathbf{i} + 2\mathbf{j}) = 10\mathbf{i} + 4\mathbf{j} \]\[ 3\mathbf{U} + 2\mathbf{V} = (6 + 10)\mathbf{i} + (15 + 4)\mathbf{j} = 16\mathbf{i} + 19\mathbf{j} \]

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Key Concepts

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

Vector Subtraction
Vector subtraction is much like vector addition but involves taking the components of one vector away from another. If you have two vectors, say \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \) and \( \mathbf{V} = c\mathbf{i} + d\mathbf{j} \), the subtraction \( \mathbf{U} - \mathbf{V} \) is calculated by subtracting the corresponding components:
\[ \mathbf{U} - \mathbf{V} = (a-c)\mathbf{i} + (b-d)\mathbf{j} \]

  • First, subtract the \( \mathbf{i} \) components of \( \mathbf{V} \) from \( \mathbf{U} \).
  • Next, subtract the \( \mathbf{j} \) components.
Using the exercise's example vectors, \( \mathbf{U} = 2\mathbf{i} + 5\mathbf{j} \) and \( \mathbf{V} = 5\mathbf{i} + 2\mathbf{j} \), we compute:
  • \( \mathbf{U} - \mathbf{V} = (2-5)\mathbf{i} + (5-2)\mathbf{j} = -3\mathbf{i} + 3\mathbf{j} \).
This simple process helps visually understand the direction and magnitude that result when one vector is removed from another.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar (a real number). This process scales the vector's magnitude without changing its direction, unless the scalar is negative—which will reverse the vector's direction.
For a vector \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \) and a scalar \( k \), the resulting vector is given by:
\[ k\mathbf{U} = k(a\mathbf{i} + b\mathbf{j}) = (ka)\mathbf{i} + (kb)\mathbf{j} \]
  • Multiply each component of the vector by the scalar.
  • The length of the vector changes depending on the magnitude of the scalar.
In our exercise, we scaled each vector before adding them:
  • \( 3\mathbf{U} = 3(2\mathbf{i} + 5\mathbf{j}) = 6\mathbf{i} + 15\mathbf{j} \)
  • \( 2\mathbf{V} = 2(5\mathbf{i} + 2\mathbf{j}) = 10\mathbf{i} + 4\mathbf{j} \)
This manipulation is crucial in vector operations, especially when working with forces or other physical quantities.
Component Form of Vectors
Vectors in component form are expressed as a sum of their horizontal (\( \mathbf{i} \)) and vertical (\( \mathbf{j} \)) components. This form is extremely handy in calculations because it allows us to work with vectors mathematically, just as we do with other number pairs.
Consider vector \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \). This vector has:
  • A horizontal component \( a \)
  • A vertical component \( b \)
This separation simplifies addition, subtraction, and scalar multiplication:
  • Just operate on corresponding components independently.
In the exercise, \( \mathbf{U} = 2\mathbf{i} + 5\mathbf{j} \) and \( \mathbf{V} = 5\mathbf{i} + 2\mathbf{j} \), each vector is fully characterized by its components.
The component form breaks down complex vector operations into simple arithmetic, enabling easy computation and visualization in physics, engineering, and computer graphics.

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Most popular questions from this chapter

In triangle \(A B C, A=40^{\circ}, b=19 \mathrm{ft}\), and \(a=18 \mathrm{ft}\). Use the law of sines to find \(\sin B\) and then give two possible values for \(B\).

These questions are available for instructors to help assess if you have successfully met the learning objectives for this section. Find \(c\) for triangle \(A B C\) if \(a=6.8\) meters, \(b=8.4\) meters, and \(C=48^{\circ}\). a. \(5.6 \mathrm{~m}\) b. \(40 \mathrm{~m}\) c. \(8.6 \mathrm{~m}\) d. \(6.4 \mathrm{~m}\)

For Problems 37 through 42, use your knowledge of bearing, heading, and true course to sketch a diagram that will help you solve each problem. True Course and Speed A plane is flying with an airspeed of 244 miles per hour with heading \(272.7^{\circ}\). The wind currents are running at a constant \(45.7\) miles per hour in the direction \(262.6^{\circ}\). Find the ground speed and true course of the plane.

Pulling a Crate A large crate is pulled across the ice with two ropes. A force of 47 pounds is applied to the first rope in the direction \(80^{\circ}\), and a force of 55 pounds is applied to the second rope in the direction \(105^{\circ}\). What are the magnitude and direction of the resultant force acting on the crate?

Find the work performed when a force \(\mathbf{F}=15 \mathbf{i}-9 \mathbf{j}\) is applied to an object whose resulting motion is represented by displacement vector \(\mathbf{d}=80 \mathrm{i}+12 \mathrm{j}\). Assume the force is measured in pounds and the displacement in feet. a. \(1,415 \mathrm{ft}-\mathrm{lb}\) b. \(552 \mathrm{ft}-\mathrm{lb}\) c. \(1,092 \mathrm{ft}-\mathrm{lb}\) d. \(1,308 \mathrm{ft}-\mathrm{lb}\)

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