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Position cectors. Using the \(x\) axis as east, the \(y\) axis as north, and the \(z\) axis as up, give the vector representing the following points: (a) \(10 \mathrm{mi}\) northeast and \(2 \mathrm{mi}\) up (b) \(5 \mathrm{yd}\) southeast and \(5 \mathrm{yd}\) down (c) \(1 \mathrm{~cm}\) northwest and \(6 \mathrm{~cm}\) up Find the magnitude of each vector and the expression for the unit vector in that direction.

Short Answer

Expert verified
(a) Vector: \( \left( \frac{10}{\sqrt{2}}, \frac{10}{\sqrt{2}}, 2 \right) \) mi, Mag: \( \sqrt{54} \), Unit: \( \left( \frac{10/\sqrt{2}}{\sqrt{54}}, \frac{10/\sqrt{2}}{\sqrt{54}}, \frac{2}{\sqrt{54}} \right) \). (b) Vector: \( \left( \frac{5}{\sqrt{2}}, -\frac{5}{\sqrt{2}}, -5 \right) \) yd, Mag: \( \sqrt{50} \), Unit: \( \left( \frac{5/\sqrt{2}}{\sqrt{50}}, \frac{-5/\sqrt{2}}{\sqrt{50}}, \frac{-5}{\sqrt{50}} \right) \). (c) Vector: \( \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 6 \right) \) cm, Mag: \( \sqrt{37} \), Unit: \( \left( \frac{-1/\sqrt{2}}{\sqrt{37}}, \frac{1/\sqrt{2}}{\sqrt{37}}, \frac{6}{\sqrt{37}} \right) \).

Step by step solution

01

Decoding Directions

First, we need to understand how the given directions (northeast, southeast, northwest) translate into vector components on the x-y plane. Northeast means equal components in the east and north directions, southeast means equal components in east and south directions, and northwest means equal components in the west and north directions.
02

Calculating Components for (a)

For point (a), '10 mi northeast' means the x-component and y-component are equal. Using trigonometric ratios, each component is \( \frac{10}{\sqrt{2}} \) mi. In vector notation, this becomes \( \left( \frac{10}{\sqrt{2}}, \frac{10}{\sqrt{2}}, 2 \right) \) mi.
03

Calculating Magnitude for (a)

To find the magnitude of the vector from (a), we use the formula for the magnitude of a vector: \[ \sqrt{x^2 + y^2 + z^2} \]. Substituting in the values, we get: \[ \sqrt{\left( \frac{10}{\sqrt{2}} \right)^2 + \left( \frac{10}{\sqrt{2}} \right)^2 + 2^2} = \sqrt{50 + 4} = \sqrt{54} \text{ mi} \].
04

Unit Vector for (a)

The unit vector is found by dividing each component of the vector by its magnitude. For vector (a), the unit vector is: \[ \left( \frac{10/\sqrt{2}}{\sqrt{54}}, \frac{10/\sqrt{2}}{\sqrt{54}}, \frac{2}{\sqrt{54}} \right) \].
05

Calculating Components for (b)

For point (b), '5 yd southeast' means the x-component is positive, and the y-component is negative as southeast is a combination of east and south. Both components are \( \frac{5}{\sqrt{2}} \) yd. Thus the vector is \( \left( \frac{5}{\sqrt{2}}, -\frac{5}{\sqrt{2}}, -5 \right) \) yd.
06

Calculating Magnitude for (b)

The magnitude is: \[ \sqrt{\left( \frac{5}{\sqrt{2}} \right)^2 + \left( -\frac{5}{\sqrt{2}} \right)^2 + (-5)^2} = \sqrt{12.5 + 12.5 + 25} = \sqrt{50} \text{ yd} \].
07

Unit Vector for (b)

For vector (b), the unit vector is: \[ \left( \frac{5/\sqrt{2}}{\sqrt{50}}, \frac{-5/\sqrt{2}}{\sqrt{50}}, \frac{-5}{\sqrt{50}} \right) \].
08

Calculating Components for (c)

For point (c), '1 cm northwest' means both x and y components are negative and equal. The vector components in the x and y directions are \( -\frac{1}{\sqrt{2}} \) cm. Thus, the vector is \( \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 6 \right) \) cm.
09

Calculating Magnitude for (c)

The magnitude is: \[ \sqrt{\left( -\frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2 + 6^2} = \sqrt{0.5 + 0.5 + 36} = \sqrt{37} \text{ cm} \].
10

Unit Vector for (c)

For vector (c), the unit vector is: \[ \left( \frac{-1/\sqrt{2}}{\sqrt{37}}, \frac{1/\sqrt{2}}{\sqrt{37}}, \frac{6}{\sqrt{37}} \right) \].

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Key Concepts

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

Understanding the Magnitude of a Vector
To calculate the magnitude of a vector, you use the Pythagorean theorem but in three dimensions. The formula is straightforward:
  • For a vector \( \mathbf{v} = (x, y, z) \), the magnitude \( |\mathbf{v}| \) is given by \( \sqrt{x^2 + y^2 + z^2} \).

The magnitude tells us the length or size of the vector and is pivotal when you need to understand how far a point is from the origin in a coordinate system.
For example, if a car is moving 10 units northeast and 2 units up, the magnitude of its position vector helps us find its direct distance from its starting point. In our context, the car's vector components are calculated and combined under the square root, yielding the magnitude.

Remember: Magnitudes are always non-negative, and they provide us with a scalar (single value) representation of a vector's size.
Getting Down to Unit Vectors
A unit vector is a vector with a magnitude (or length) of 1. They're particularly useful because they describe direction and negate scalar influences like distance or force size, making calculations involving unit vectors simple and efficient.
  • To convert any vector into a unit vector, divide each of its components by its magnitude.

As an example, suppose we have vector \( \mathbf{v} = (x, y, z) \). You can find the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) by executing the division
\( \mathbf{u} = \left( \frac{x}{|\mathbf{v}|}, \frac{y}{|\mathbf{v}|}, \frac{z}{|\mathbf{v}|} \right) \).
In practice, unit vectors are denoted by adding a "hat" symbol such as \( \mathbf{\hat{v}} \).
Consider vector (a) from our exercise. After calculating its magnitude, we divide each component by \( \sqrt{54} \), finding its unit vector. This shows direction conclusively without regard to magnitude.
Grasping Vector Components
Vector components are crucial for visualizing how a vector can be broken down into its parts along defined axes. Each part or component shows the effect of the vector in one particular direction.
  • For a vector in three-dimensional space \( \mathbf{v} = (x, y, z) \), \( x \), \( y \), and \( z \) correspond to the vector's influence in the east-west, north-south, and up-down directions, respectively.

Breaking down vectors into components is useful as it helps one to calculate or predict the vector's influence, particularly in complex directions like "northeast" or "southeast."
Simple trigonometry can guide us here: directions like northeast suggest equal effects in x and y, both of which can be calculated using \( \frac{magnitude}{\sqrt{2}} \) for the plane.
Understanding components not only aids in drawing a complete picture of the vector's path, but also simplifies many real-world problems, such as resolving forces in physics or adjusting position in navigation.

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Most popular questions from this chapter

Vector algebra. Given two vectors such that \(\mathrm{a}+\mathrm{b}=\) \(11 \hat{\mathrm{x}}-\hat{y}+5 \hat{z}\) and \(a-b=-5 \hat{\mathbf{x}}+11 \hat{y}+9 \hat{z}\) (a) Find a and b. (b) Find the angle included between a and (a + b) using vector methods.

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Scalar and vector products of two vectors. Given two vectors \(\mathbf{a}=3 \hat{\mathbf{x}}+4 \hat{y}-5 \hat{\mathbf{z}}\) and \(\mathbf{b}=-\hat{\mathbf{x}}+2 \hat{\mathbf{y}}+6 \hat{\mathbf{z}}\), calculate by vector methods: (a) The length of each Ans, \(a=\sqrt{50} ; b=\sqrt{41}\). (b) The scalar product a \cdot b Ans. \(-25\). (c) The included angle between them Ans. \(123.5^{\circ}\). (d) The direction cosines for each (e) The vector sum and difference \(a+b\) and \(a-b\)

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