How To Find Direction Angle Of A 3d Vector

Direction Angles of Vectors

Effigy 1 shows a unit vector u that makes an angle θ with the positive x-axis. The angle θ is called the directional bending of vector u.

The terminal bespeak of vector u lies on a unit circumvolve and thus u can be denoted by:

$u=\langle \mathrm{10},y\rangle =\langle \cos \theta ,\sin \theta \rangle =(\cos \theta )i+(\sin \theta )j$

Any vector that makes an angle θ with the positive x-axis tin be written as the unit vector times the magnitude of the vector.

$\mathrm{five}=\parallel 5\parallel (\cos \theta )i+\parallel v\parallel (\sin \theta )j$ $v=ai+bj$

Therefore the direction angle of θ of whatsoever vector tin exist calculated as follows:

DIRECTIONAL ANGLE:

$\text{tan θ}=\frac{\mathrm{southward}i\mathrm{due\; north}\theta }{\cos \theta }=\frac{\parallel v\parallel sin\theta }{\parallel v\parallel \cos \theta }=\frac{b}{a}$

Let's look at some examples.

To piece of work these examples requires the use of various vector rules. If you are not familiar with a rule go to the associated topic for a review.

Instance 1:Observe the management angle of w = -2i + 9j.

Step ane: Identify the values for a and b and calculate θ. $ta\mathrm{north}\theta =\frac{b}{a}$	a = -2, b = 9 $\tan \theta =\frac{b}{a}=\frac{9}{-2}$ $\theta ={\tan }^{-1}\left|\frac{\mathrm{nine}}{-2}\right|$ $\theta \approx 78°$
Step ii: Make up one's mind the Quadrant the vector lies in.	Because the vector terminus is (-2, nine), it will autumn in quadrant II and and so will θ.
Step 3: Brand any necessary adjustments to find the directional angle θ from the positive x-axis.	Since the reference angle is 78°, the directional angle from the positive ten-axis is 180° - 78° = 102°.

Instance two:Detect the management angle of $5=\mathrm{three}\left(\cos \mathrm{sixty}°i+\sin 60°j\right)$.

Step 1: Simplify vector v using scalar multiplication. $\mathrm{1000}\mathrm{five}=\mathrm{thou}{v}_{\mathrm{i}},{\mathrm{five}}_2=k{v}_{\mathrm{one}},k{5}_{\mathrm{two}}\to ScalarGultiplicatio\mathrm{north}$	$v=\mathrm{iii}\left(\cos \mathrm{sixty}°i+\sin \mathrm{sixty}°j\right)$ $v=3·\cos \mathrm{sixty}°i+3·\sin 60°j$ $\mathrm{five}=3·\frac{1}{2}i+3·\frac{\sqrt{\mathrm{three}}}{2}j$ $v=\frac{3}{2}i+\frac{3\sqrt{\mathrm{iii}}}{\mathrm{two}}$
Step 2: Identify the values for a and b and calculate θ.	$a=\frac{3}{2},b=\frac{3\sqrt{\mathrm{iii}}}{2}$ $\tan \theta =\frac{b}{a}=\frac{\frac{3\sqrt{\mathrm{three}}}{\mathrm{two}}}{\frac{\mathrm{three}}{\mathrm{ii}}}=\frac{3\sqrt{\mathrm{iii}}}{2}·\frac{2}{3}=\sqrt{3}$ $\theta ={\tan }^{-1}\left|\sqrt{3}\right|$ $\theta =60°$
Stride 3: Determine the Quadrant of the vector lies in.	Because the vector terminus is $\left(\frac{3}{\mathrm{ii}},\frac{\mathrm{three}\sqrt{3}}{2}\right)=\left(1.5,2.6\right)$ and both components are positive the vector will fall in quadrant I and so volition θ.
Step 4: Make whatever necessary adjustments to discover the directional angle θ from the positive x-axis.	Since the reference bending is 60°, the directional angle from the positive x-axis is 60° - 0° = 60°.

Source: https://www.softschools.com/math/pre_calculus/direction_angles_of_vectors/

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