Chapter 14: Problem 2
Sketch the position vector of a and find \(\|\mathrm{a}\|\). $$ a=\langle-4,-7\rangle $$
Short Answer
Expert verified
The magnitude of \( a \) is \( \sqrt{65} \).
Step by step solution
01
Understanding the Vector
The vector \( \mathbf{a} = \langle -4, -7 \rangle \) is in two-dimensional space. This means it has two components: \( x = -4 \) and \( y = -7 \). The vector provides a direction and magnitude from the origin \((0, 0)\) to the point \((-4, -7)\).
02
Sketching the Vector
To sketch the vector, plot the point \((-4, -7)\) on a coordinate plane. Start from the origin \((0, 0)\) and draw an arrow to the point \((-4, -7)\). The direction of the arrow represents the direction of the vector \( \mathbf{a} \).
03
Using the Magnitude Formula
The magnitude (or length) of a vector \( \mathbf{a} = \langle x, y \rangle \) is calculated using the formula: \[ \| \mathbf{a} \| = \sqrt{x^2 + y^2} \] Substitute the values of \( x \) and \( y \): \[ \| \mathbf{a} \| = \sqrt{(-4)^2 + (-7)^2} \]
04
Calculate the Magnitude
First, square each component: \(-4\) squared is \(16\) and \(-7\) squared is \(49\). Then add these two results: \(16 + 49 = 65\). Use this sum in the magnitude formula: \[ \| \mathbf{a} \| = \sqrt{65} \]
05
Simplifying the Result
\( \sqrt{65} \) is the magnitude of the vector \( \mathbf{a} \). It does not simplify nicely to a whole number, so leave it in this square root form or use a calculator to approximate it as \( \approx 8.06 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Position Vector
A position vector represents the position of a point in space relative to the origin. In our case, the vector \( \mathbf{a} = \langle -4, -7 \rangle \) is a position vector in two-dimensional space. This vector has two components: an \(x\)-component of \(-4\) and a \(y\)-component of \(-7\). These components tell us how far and in which direction to move from the origin \((0, 0)\) to reach the point \((-4, -7)\).
When sketching a position vector in a coordinate plane, like \( \mathbf{a} \):
When sketching a position vector in a coordinate plane, like \( \mathbf{a} \):
- Start at the origin \((0, 0)\).
- Move horizontally by \(-4\) units (since \(x = -4\)).
- Then, move vertically by \(-7\) units (since \(y = -7\)).
- Draw an arrow pointing from the origin to the location \((-4, -7)\) to represent the vector.
Magnitude of a Vector
The magnitude of a vector refers to its length or size. It's a measure of how far the vector extends from the origin to its pointed end. To find the magnitude of a vector \( \mathbf{a} = \langle x, y \rangle \), apply the Pythagorean theorem:\[\| \mathbf{a} \| = \sqrt{x^2 + y^2}\]
In our example, the vector \( \mathbf{a} \) has components \(x = -4\) and \(y = -7\). Substituting these into the formula gives:\[\| \mathbf{a} \| = \sqrt{(-4)^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65}\]
This \( \sqrt{65} \) is the magnitude of vector \( \mathbf{a} \), and it represents how long the vector is. Magnitudes are always positive or zero; they are like distances. For practical use, you may approximate \( \sqrt{65} \approx 8.06 \) if needed.
In our example, the vector \( \mathbf{a} \) has components \(x = -4\) and \(y = -7\). Substituting these into the formula gives:\[\| \mathbf{a} \| = \sqrt{(-4)^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65}\]
This \( \sqrt{65} \) is the magnitude of vector \( \mathbf{a} \), and it represents how long the vector is. Magnitudes are always positive or zero; they are like distances. For practical use, you may approximate \( \sqrt{65} \approx 8.06 \) if needed.
Two-Dimensional Space
Two-dimensional space can be visualized like a flat sheet or a plane. It has two directions: horizontal (or \(x\)-axis) and vertical (or \(y\)-axis). Vectors in this space, like \( \mathbf{a} = \langle -4, -7 \rangle \), have two components which are aligned along these axes.
Understanding two-dimensional space is crucial because:
Understanding two-dimensional space is crucial because:
- Most basic geometry problems are solved in this plane.
- It's foundational for visualizing graphs, shapes, and vectors.
- Understanding it makes transitioning to three dimensions or more easier.