Chapter 13: Problem 35
Resolve the following vectors into components: (a) The vector in 2 -space of length 2 pointing up and to the right at an angle of \(\pi / 4\) with the \(x\) -axis. (b) The vector in 3 -space of length 1 lying in the \(x z\) plane pointing upward at an angle of \(\pi / 6\) with the positive \(x\) -axis.
Short Answer
Step by step solution
Understanding the Problem
Resolve Vector in 2-Space (a)
Resolve Vector in 3-Space (b)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometry
- The cosine function can help find how much of the vector goes along the x-axis.
- The sine function helps figure out the portion of the vector along the y-axis.
2-Space Vectors
- The x-component of this vector, which represents horizontal magnitude, is determined by multiplying the vector's length by \(\cos(\theta)\).
- The y-component or vertical magnitude is calculated using \(\sin(\theta)\).
- \(v_x = 2 \cos(\pi/4) = \sqrt{2}\)
- \(v_y = 2 \sin(\pi/4) = \sqrt{2}\)
3-Space Vectors
- The x-component is derived using \(\cos(\theta)\), representing horizontal influence along the x-axis.
- The vector in the xz-plane implies the y-component is zero, since it does not move up or down vertically.
- The z-component is determined by \(\sin(\theta)\), showing the vector's depth magnitude.
- \(v_x = 1 \cos(\pi/6) = \frac{\sqrt{3}}{2}\)
- \(v_y = 0\)
- \(v_z = 1 \sin(\pi/6) = \frac{1}{2}\)
Angles with Axes
In the context of 3-space vectors, understanding angles becomes even more significant. Consider a vector in the xz-plane; the angle with the x-axis governs how it spreads along x and z axes, while y-components are dismissed due to the plane of existence.
- An angle of \(\pi/4\) assigns equal components in both x and y directions when in 2-space.
- An angle of \(\pi/6\) ensures a different split between the x and z components in 3-space.