17Calculus - Dot Product Application - Direction Cosines and Direction Angles

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Direction angles are the angles between a given vector $$\vec{v}$$ and each coordinate axis (usually in three dimensions, so there are three of them). Basically, we use the equation for the angle between vectors to get the direction cosine equations and the direction angles. For example, to find the direction cosine and the direction angle between a vector $$\vec{v}$$ and the x-axis, we have
$$\displaystyle{ \cos(\alpha) = \frac{\vec{v} \cdot \hat{i}}{\norm{\vec{v}} \norm{\hat{i}}} }$$
Let's label the components of $$\vec{v}$$ as $$\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$$
Since $$\hat{i}$$ is the unit vector in the direction of the x-axis, we can write $$\hat{i} = 1\hat{i} + 0\hat{j} + 0\hat{k}$$ and $$\|\hat{i}\| = 1$$.
$$\displaystyle{ \cos(\alpha) = \frac{\vec{v} \cdot \hat{i}}{\|\vec{v}\| \|\hat{i}\|} = \frac{v_1}{\|\vec{v}\|} }$$
Similar results can be obtained for the other two angles. Most textbooks and mathematicians use special greek letters for these angles as listed below.

Angle Description Direction Cosine Direction Angle $$\alpha$$ is the angle between $$\vec{v}$$ and $$\hat{i}$$ $$\displaystyle{ \cos(\alpha) = \frac{v_1}{\|\vec{v}\|} }$$ $$\displaystyle{ \alpha = \arccos\left(\frac{v_1}{\|\vec{v}\|} \right) }$$ $$\beta$$ is the angle between $$\vec{v}$$ and $$\hat{j}$$ $$\displaystyle{ \cos(\beta) = \frac{v_2}{\|\vec{v}\|} }$$ $$\displaystyle{ \beta = \arccos\left(\frac{v_2}{\|\vec{v}\|} \right) }$$ $$\gamma$$ is the angle between $$\vec{v}$$ and $$\hat{k}$$ $$\displaystyle{ \cos(\gamma) = \frac{v_3}{\|\vec{v}\|} }$$ $$\displaystyle{ \gamma = \arccos\left(\frac{v_3}{\|\vec{v}\|} \right) }$$

Study Hint: Since you already need to know the equation for the angle between two vectors, just remember what the direction cosines and direction angles are. You can derive the equations in the above table from that information. Additionally, if you just memorize the equations, you may not remember what they represent or where they come from. What are you going to do when your instructor asks you to define they mean and where they come from?

Practice

Find the direction angle of $$3 \vhat{i} - 4\vhat{j}$$.

Problem Statement

Find the direction angle of $$3 \vhat{i} - 4\vhat{j}$$.

Solution

1809 video

video by Larson Calculus

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 basics of vectors dot product

Wikipedia - Dot Product

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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