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Chapter 18: Problem 2

Find the direction cosines of the vector joining the two points \((4,2,2)\) and \((7,6,14)\).

Short Answer

Expert verified
The direction cosines are \(\cos \alpha = \frac{3}{13}\), \(\cos \beta = \frac{4}{13}\), and \(\cos \gamma = \frac{12}{13}\).

Step by step solution

01

Find the components of the vector

Subtract the coordinates of the initial point \(4, 2, 2\) from the coordinates of the terminal point \(7, 6, 14\). The vector components are found by: \[7 - 4, 6 - 2, 14 - 2\]. This gives the vector components \(3, 4, 12\).
02

Calculate the magnitude of the vector

Use the formula for the magnitude of the vector \( \sqrt{x^2 + y^2 + z^2} \). Substituting the vector components \(3, 4, 12\) results in: \ \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \.
03

Find the direction cosines

The direction cosines are \( \cos \alpha, \cos \beta, \cos \gamma \). They are computed as follows: \[\cos \alpha = \frac{x}{\text{magnitude}}, \cos \beta = \frac{y}{\text{magnitude}}, \cos \gamma = \frac{z}{\text{magnitude}}\]. For the vector components \(3, 4, 12\) and magnitude \(13\): \[\cos \alpha = \frac{3}{13}, \cos \beta = \frac{4}{13}, \cos \gamma = \frac{12}{13}\].

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Key Concepts

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

Vector Components
Understanding the components of a vector is essential in the field of vectors and linear algebra. A vector describes both a direction and a magnitude.
To find the components of a vector, you subtract the coordinates of the starting point from the coordinates of the end point.
For example, to find the vector joining the points \((4, 2, 2)\) and \((7, 6, 14)\), we perform the following subtraction:
  • \[7 - 4 = 3\]
  • \[6 - 2 = 4\]
  • \[14 - 2 = 12\]
Thus, we get the vector components \((3, 4, 12)\).
These components indicate the change in the x, y, and z directions, respectively. So, the vector moves 3 units in the x-direction, 4 units in the y-direction, and 12 units in the z-direction.
Vector Magnitude
The magnitude of a vector tells us about the vector's length.
You can think of it as the distance between the initial and the terminal points of the vector.
The formula for the magnitude of a vector \((x, y, z)\) is: \[ \text{{Magnitude}} = \sqrt{x^2 + y^2 + z^2} \] To continue with our example vector \((3, 4, 12)\): \[ \text{{Magnitude}} = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \] As shown, the magnitude (or length) of the vector is 13 units.
This measurement is vital in defining how 'long' our vector is, independent of its direction.
Direction Cosines
Direction cosines provide us with the cosines of the angles that a vector makes with the coordinate axes.
These values are key in understanding the vector's orientation in space.
They are denoted as \(\cos \alpha\), \(\cos \beta\), and \(\cos \gamma\) for the angles with the x, y, and z axes, respectively.
To compute the direction cosines, we divide each component of the vector by the vector's magnitude:
  • \[\cos \alpha = \frac{3}{13}\]
  • \[\cos \beta = \frac{4}{13}\]
  • \[\cos \gamma = \frac{12}{13}\]
These calculations give us the direction cosines:
  • \(\cos \alpha = \frac{3}{13}\)
  • \(\cos \beta = \frac{4}{13}\)
  • \(\cos \gamma = \frac{12}{13}\)
These cosines help in understanding how the vector is oriented relative to each axis.
They are useful in many applications, ranging from physics to computer graphics.

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Most popular questions from this chapter

If \(\mathbf{a}=5 \mathbf{i}+4 \mathbf{j}+2 \mathbf{k}, \mathbf{b}=4 \mathbf{i}-5 \mathbf{j}+3 \mathbf{k}\) and \(\mathbf{c}=2 \boldsymbol{i}-\mathbf{j}-2 \mathbf{k}\), where \(\mathbf{1}, \mathbf{j}, \mathbf{k}\) are the unit vectors, determine: (a) the value of a b and the angle between the vectors a and \(\mathbf{b}\) (b) the magnitude and the direction cosines of the product vector \((\mathbf{a} \times \mathbf{b})\) and also the angle which this product vector makes with the vector \(c\).

Find the direction cosines of the vectors whose direction ratios are \((3,4,5)\) and \((1,2,-3)\). Hence find the angle between the two vectors.
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