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Given two vectors \(\vec{A}=\) \(4.00 \hat{\imath}+3.00 \hat{\jmath}\) and \(\vec{B}=5.00 \hat{\imath}-\) \(2.00 \hat{\jmath}\) (a) find the magnitude of cach vector; (b) write an expression for the vector difference \(\vec{A}-\vec{B}\) using unit vectors; (c) find the magnitude and direction of the vector difference \(\vec{A}-\vec{B}\). (d) In a vector diagram show \(\vec{A}, \vec{B},\) and \(\vec{A}-\vec{B},\) and also show that your diagram agrees qualitatively with your answer in part (c).

Short Answer

Expert verified
(a) Magnitudes: \(5.00\) and \(5.39\). (b) \(-1.00 \hat{\imath} + 5.00 \hat{\jmath}\). (c) Magnitude: \(5.10\), Direction: \(168.69^\circ\). (d) Diagram matches.

Step by step solution

01

Calculate the Magnitude of Vector A

The magnitude of a vector \(\vec{A} = a \hat{\imath} + b \hat{\jmath}\) is given by the formula \(\|\vec{A}\| = \sqrt{a^2 + b^2}\). For \(\vec{A} = 4.00 \hat{\imath} + 3.00 \hat{\jmath}\), substitute \(a = 4.00\) and \(b = 3.00\):\[\|\vec{A}\| = \sqrt{(4.00)^2 + (3.00)^2} = \sqrt{16.00 + 9.00} = \sqrt{25.00} = 5.00.\]
02

Calculate the Magnitude of Vector B

For \(\vec{B} = 5.00 \hat{\imath} - 2.00 \hat{\jmath}\), use the magnitude formula: \(\|\vec{B}\| = \sqrt{(5.00)^2 + (-2.00)^2} = \sqrt{25.00 + 4.00} = \sqrt{29.00} \approx 5.39.\)
03

Determine the Vector Difference A - B

To find \(\vec{A} - \vec{B}\), subtract each corresponding component of \(\vec{B}\) from \(\vec{A}\):\[\vec{A} - \vec{B} = (4.00 - 5.00) \hat{\imath} + (3.00 - (-2.00)) \hat{\jmath} = -1.00 \hat{\imath} + 5.00 \hat{\jmath}.\]
04

Find the Magnitude of the Vector Difference

Use the magnitude formula on \(\vec{A} - \vec{B} = -1.00 \hat{\imath} + 5.00 \hat{\jmath}\):\[\|\vec{A} - \vec{B}\| = \sqrt{(-1.00)^2 + (5.00)^2} = \sqrt{1.00 + 25.00} = \sqrt{26.00} \approx 5.10.\]
05

Determine the Direction Angle of the Vector Difference

The direction \(\theta\) of a vector \((x \hat{\imath}, y \hat{\jmath})\) is found with \(\theta = \tan^{-1}(\frac{y}{x})\). For \(\vec{A} - \vec{B} = -1.00 \hat{\imath} + 5.00 \hat{\jmath}\), \(\theta = \tan^{-1}\left( \frac{5.00}{-1.00} \right) = \tan^{-1}(-5.00)\). Calculate \(\theta \), keeping in mind the vector lies in the second quadrant, which results in \(\theta \approx 168.69^\circ\).
06

Draw the Vector Diagram

On a graph, sketch \(\vec{A} = 4.00 \hat{\imath} + 3.00 \hat{\jmath}\), \(\vec{B} = 5.00 \hat{\imath} - 2.00 \hat{\jmath}\), and \(\vec{A} - \vec{B} = -1.00 \hat{\imath} + 5.00 \hat{\jmath}\). Check that \(\vec{A} - \vec{B}\) starts where \(\vec{B}\) ends if \(\vec{B}\) is subtracted from \(\vec{A}\). Ensure the angle and vector proportions concord with the calculated magnitude and direction.

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Key Concepts

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

Vector Magnitude
Vector magnitude is a measure that indicates the length or size of the vector. It is always a non-negative number. To find the magnitude of a vector denoted as \(\vec{A} = a \hat{\imath} + b \hat{\jmath}\), you use the Pythagorean theorem. This can be expressed as \(\|\vec{A}\| = \sqrt{a^2 + b^2}\).
For example, if you have a vector \(\vec{A} = 4.00 \hat{\imath} + 3.00 \hat{\jmath}\), the magnitude would be calculated by squaring each component, summing them, and then taking the square root:
  • Calculate \((4.00)^2 = 16.00\)
  • Calculate \((3.00)^2 = 9.00\)
  • Add these results: \(16.00 + 9.00 = 25.00\)
  • Find the square root: \(\sqrt{25.00} = 5.00\)
This process helps you determine how long the vector is when plotted graphically on a two-dimensional plane.
Vector Subtraction
Vector subtraction is the operation of finding the difference between two vectors. It allows you to find the vector pointing from one vector's tip to the other's tip. To perform vector subtraction, you subtract corresponding components.
Consider two vectors, \(\vec{A} = 4.00 \hat{\imath} + 3.00 \hat{\jmath}\) and \(\vec{B} = 5.00 \hat{\imath} - 2.00 \hat{\jmath}\). The difference \(\vec{A} - \vec{B}\) is found by subtracting the components of \(\vec{B}\) from \(\vec{A}\):
  • Subtract the \(\hat{\imath}\) components: \(4.00 - 5.00 = -1.00\)
  • Subtract the \(\hat{\jmath}\) components: \(3.00 - (-2.00) = 5.00\)
Hence, the resulting vector \(\vec{A} - \vec{B}\) is \(-1.00 \hat{\imath} + 5.00 \hat{\jmath}\). This vector effectively describes the direction and distance from the tip of \(\vec{B}\) to the tip of \(\vec{A}\).
Vector Components
Vector components break down the vector along the coordinate axes, making it easier to apply various vector operations like addition, subtraction, or finding the magnitude. Each vector can be viewed as having two parts: the horizontal component along the \(\hat{\imath}\) axis and the vertical component along the \(\hat{\jmath}\) axis.
For example, the vector \(\vec{A} = 4.00 \hat{\imath} + 3.00 \hat{\jmath}\) has:
  • A horizontal component: \(4.00 \hat{\imath}\)
  • A vertical component: \(3.00 \hat{\jmath}\)
Understanding vector components is crucial as they allow you to calculate the vector’s magnitude and perform operations like vector addition or subtraction. Each component acts independently along its respective axis, which simplifies calculations involving vector relationships. By considering components, you can visualize how one vector can project onto others.

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