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Chapter 12: Problem 17

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) $$ \mathbf{a}=\langle 1,-4,1\rangle, \quad \mathbf{b}=\langle 0,2,-2\rangle $$

Short Answer

Expert verified
The angle between the vectors is approximately 151 degrees.

Step by step solution

01

Compute the Dot Product

The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is calculated as \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\). For vectors \(\mathbf{a} = \langle 1, -4, 1 \rangle\) and \(\mathbf{b} = \langle 0, 2, -2 \rangle\), compute:\[\mathbf{a} \cdot \mathbf{b} = (1)(0) + (-4)(2) + (1)(-2) = 0 - 8 - 2 = -10\]
02

Find the Magnitude of Each Vector

The magnitude of a vector \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\) is calculated by the formula \(\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}\).- For \(\mathbf{a} = \langle 1, -4, 1 \rangle\): \[ \|\mathbf{a}\| = \sqrt{1^2 + (-4)^2 + 1^2} = \sqrt{1 + 16 + 1} = \sqrt{18} \]- For \(\mathbf{b} = \langle 0, 2, -2 \rangle\): \[ \|\mathbf{b}\| = \sqrt{0^2 + 2^2 + (-2)^2} = \sqrt{0 + 4 + 4} = \sqrt{8} \]
03

Calculate the Cosine of the Angle

The formula for the cosine of the angle \(\theta\) between two vectors is given by:\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\]Substitute the values from Steps 1 and 2:\[\cos \theta = \frac{-10}{\sqrt{18} \times \sqrt{8}} = \frac{-10}{\sqrt{144}} = \frac{-10}{12} = -\frac{5}{6}\]
04

Find the Exact Angle in Radians

To find the angle \(\theta\), take the inverse cosine (arccos) of the cosine value:\[\theta = \cos^{-1}\left(-\frac{5}{6}\right)\]This gives the exact angle in radians.
05

Convert from Radians to Degrees

To convert the angle from radians to degrees, use the conversion factor: \(180^\circ/\pi\).Calculate:\[\theta_{degrees} = \cos^{-1}\left(-\frac{5}{6}\right) \times \frac{180}{\pi}\]Approximate this value to the nearest degree.

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Key Concepts

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

Dot Product
The dot product, also known as the scalar product, is a fundamental concept in vector algebra. It's a way of multiplying two vectors to obtain a scalar. For vectors \( \mathbf{a} \) and \( \mathbf{b} \), represented as \( \langle a_1, a_2, a_3 \rangle \) and \( \langle b_1, b_2, b_3 \rangle \) respectively, the dot product is calculated using the formula:
  • \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)
In our exercise, substituting the values from vectors \( \mathbf{a} = \langle 1, -4, 1 \rangle \) and \( \mathbf{b} = \langle 0, 2, -2 \rangle \) gives us:
  • \( \mathbf{a} \cdot \mathbf{b} = (1 \times 0) + (-4 \times 2) + (1 \times -2) = -10 \)
This numerical result is integral in determining the angle between vectors, as it shows the relationship in their directional alignment.
Magnitude of a Vector
The magnitude of a vector, often denoted as \( \| \mathbf{v} \| \), represents its length or size. This is calculated using a version of the Pythagorean theorem, which accounts for each component of the vector.
  • For a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), its magnitude is \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \).
In our problem, the magnitude of vectors \( \mathbf{a} \) and \( \mathbf{b} \) are:
  • \( \| \mathbf{a} \| = \sqrt{1^2 + (-4)^2 + 1^2} = \sqrt{18} \)
  • \( \| \mathbf{b} \| = \sqrt{0^2 + 2^2 + (-2)^2} = \sqrt{8} \)
Understanding magnitude is crucial as it provides a scalar context of the vector's extent, necessary for calculating the cosine of the angle between two vectors.
Cosine of the Angle
To find the angle between two vectors, the cosine of the angle \( \theta \) is used. This relationship is a reflection of how closely the vectors align directionally.
  • The formula to find \( \cos \theta \) is: \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|} \)
Inserting previously computed results gives:
  • \( \cos \theta = \frac{-10}{\sqrt{18} \times \sqrt{8}} = \frac{-10}{\sqrt{144}} = \frac{-10}{12} = -\frac{5}{6} \)
These calculations allow us to understand how \( \mathbf{a} \) and \( \mathbf{b} \) are oriented in relation to each other. An important step, as it directly leads us to finding the angle using the inverse cosine function.
Inverse Cosine (Arccos)
The inverse cosine function, also known as arccos, is crucial for determining the actual angle from its cosine value. When we have a cosine value like \( -\frac{5}{6} \), the arccosine function helps us find the angle \( \theta \) for which this cosine value holds true.
  • The formula is: \( \theta = \cos^{-1}\left(-\frac{5}{6}\right) \)
This formula gives us the angle in radians. If degrees are required, conversion is necessary.
  • Convert by multiplying the radian result with \( \frac{180}{\pi} \).
This conversion transforms the value into degrees, a more commonly used unit in many practical scenarios. Understanding both radians and degrees equips you with flexibility in math and physics contexts.

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Most popular questions from this chapter

If \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) are noncoplanar vectors, let \(\mathbf{k}_{1}=\frac{\mathbf{v}_{2} \times \mathbf{v}_{3}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)} \quad \mathbf{k}_{2}=\frac{\mathbf{v}_{3} \times \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}\) \(\mathbf{k}_{3}=\frac{\mathbf{v}_{1} \times \mathbf{v}_{2}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}\) (These vectors occur in the study of crystallography. Vectors of the form \(n_{1} \mathbf{v}_{1}+n_{2} \mathbf{v}_{2}+n_{3} \mathbf{v}_{3},\) where each \(n_{i}\) is an integer, form a lattice for a crystal. Vectors written similarly in terms \(\mathbf{k}_{1}, \mathbf{k}_{2},\) and \(\mathbf{k}_{3}\) form the reciprocal lattice.) (a) Show that \(\mathbf{k}_{i}\) is perpendicular to \(\mathbf{v}_{j}\) if \(i \neq j\) (b) Show that \(\mathbf{k}_{i} \cdot \mathbf{v}_{i}=1\) for \(i=1,2,3\) (c) Show that \(\mathbf{k}_{1} \cdot\left(\mathbf{k}_{2} \times \mathbf{k}_{3}\right)=\frac{1}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}\)

(a) Find parametric equations for the line of intersection of the planes and (b) find the angle between the planes. $$ 3 x-2 y+z=1, \quad 2 x+y-3 z=3 $$

Two tanks are participating in a battle simulation. Tank A is at point \((325,810,561)\) and tank \(B\) is positioned at point \((765,675,599) .\) (a) Find parametric equations for the line of sight between the tanks. (b) If we divide the line of sight into 5 equal segments, the elevations of the terrain at the four intermediate points from tank A to tank B are \(549,566,586,\) and \(589 .\) Can the tanks see each other?

Find an equation of the plane. The plane through the point \(\left(1, \frac{1}{2}, \frac{1}{3}\right)\) and parallel to the plane \(x+y+z=0\)

Find the distance from the point to the given plane. $$ (-6,3,5), \quad x-2 y-4 z=8 $$
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