Calculus III Advanced (Course) (11.3) (Homework)

Section 11.3 Homework

From Calculus 10e by Larson and Edwards, p. 773. Exercises 4, 10, 14, 42, 57, 62.

Exercise 11.3.4 Finding Dot Products

Find
(a) \( \textbf{u} \cdot \textbf{v}\)
(b) \( \textbf{u} \cdot \textbf{u}\)
(c) \( \| \textbf{u} \|^{2} \)
(d) \( (\textbf{u} \cdot \textbf{v})\textbf{v}\)
(e) \( \textbf{u} \cdot 2(\textbf{v})\)
where \(\textbf{u}=\langle -4,8 \rangle, \: \textbf{v}=\langle 7,5 \rangle\)
Solution
(a) \( \textbf{u} \cdot \textbf{v} = (-4)(7)+(8)(5) = -28 + 40 = 12\)
(b) \( \textbf{u} \cdot \textbf{u} = (-4)(-4)+(8)(8) = 16 + 64 = 80\)
(c) \( \| \textbf{u} \|^{2} = \textbf{u} \cdot \textbf{v} = \langle -4,8 \rangle \cdot \langle -4,8 \rangle= (-4)(-4)+(8)(8) = 16 + 64 = 80 \)
(d) \( (\textbf{u} \cdot \textbf{v})\textbf{v} = 12 \langle 7,5 \rangle = \langle 84,60 \rangle\)
(e) \( \textbf{u} \cdot 2(\textbf{v})= 2(\textbf{u} \cdot \textbf{v}) = 2(12) = 24\)

Exercise 11.3.10 Finding the Angle Between Two Vectors

Find the angle \(\theta\) between the vectors (a) in radians and (b) in degrees, where \( \textbf{u} = \langle 3,1 \rangle\), and \( \textbf{v} = \langle 2,-1 \rangle\).
Solution

$$ \cos \theta = \frac{5}{\sqrt{10} \sqrt{5}} = \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

(a) \( \frac{\pi}{4} \) radians
(b) 45° degrees

Exercise 11.3.14 Finding the Angle Between Two Vectors

Find the angle \(\theta\) between the vectors (a) in radians and (b) in degrees, where \( \textbf{u}= 3\textbf{i} + 2\textbf{j}+\textbf{k}\) and \( \textbf{v}= 2\textbf{i} - 3\textbf{j}\).
Solution Because \( \textbf{u}= \langle 3,2,1 \rangle \) and \( \textbf{v}= \langle 2,-3,0 \rangle \).

$$ \cos \theta = \frac{0}{\sqrt{14}\sqrt{13}} = 0 $$

(a) 0 radians
(b) 0° degrees

Exercise 11.3.42 Finding the Projection for \( \textbf{u} \) onto \( \textbf{v} \)

For (a) find the projection for \( \textbf{u} \) onto \( \textbf{v} \), and (b) find the vector component for \( \textbf{u} \) orthogonal to \( \textbf{v} \) where, \( \textbf{u}= \textbf{i} + 4\textbf{k}\) and \( \textbf{v}= 3\textbf{i} + 2\textbf{k}\).
Solution

$$\textbf{(a) } \frac{11}{13}(3\textbf{i} + 2\textbf{k})= \frac{33}{13}\textbf{i} + \frac{22}{13}\textbf{k}$$

$$\textbf{(b) } \textbf{w}_{2}=\textbf{u}-\textbf{w}_{1}= (\textbf{i} + 4\textbf{k}) - \left (\frac{33}{13}\textbf{i} + \frac{22}{13}\textbf{k} \right ) =\frac{20}{13}\textbf{i}+\frac{30}{13}\textbf{k} $$

Exercise 11.3.57 Finding an Angle

Cube rendered along the positive axises. Figure 11.3.12

Find the angle between a cube’s diagonal and one of its edges. Solution Assume length on each side is 1. The vector \( \textbf{d}=\langle 1,1,1 \rangle\) is the chosen diagonal vector \(\textbf{d}\) as shown in Figure 11.3.12. The three sides are described by the vectors \( \textbf{i}=\langle 1,0,0 \rangle\), \( \textbf{j}=\langle 0,1,0 \rangle\), and \( \textbf{k}=\langle 0,0,1 \rangle\). Each with the same angle. $$\cos \theta = \frac{ \textbf{v} \cdot \textbf{i}}{ \| \textbf{v}\| \| \textbf{i} \|} = \frac{1}{\sqrt{3}} $$ $$\theta = \cos^{-1} \left ( \frac{1}{\sqrt{3}} \right ) \approx 55^{\circ} $$

Exercise 11.3.62 Work

Figure 11.3.13

A toy wagon is pulled by exerting a force of 25 pounds on a handle that makes a 20° angle with the horizontal, as shown in Figure 11.3.13. Find the work done in pulling the wagon 50 feet. Solution \( W= \| \text{proj}_{\vec{PQ}}\textbf{F} \| \| \vec{PQ} \| = \cos (20^{\circ}) \| \textbf{F} \| \| \text{proj}_{\vec{PQ}} \| = (0.93969)(25)(50) = 1,174.6 \text{ foot-pounds }\).

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Parent Article: Calculus III Advanced (Course)