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ArbitrLikvidat [17]
3 years ago
9

Determine the angles made by the vector v = (67)i + (-15)j with the positive x- and y-axes. write the unit vector n in the direc

tion of v

Mathematics
1 answer:
vivado [14]3 years ago
6 0
Consider the picture attached.

From right triangle trigonometry: 

tan(α)=(opposite side)/(adjacent side)=15/67=0.2239

using a scientific calculator we find that arctan(0.2239)=12.62°

thus α=12.62°, is the angle that the vector makes with the positive x-axis.

The angle made with the + y-axis is 12.62°+90°=102.62°.



The length of the vector v can be determined using the Pythagorean theorem:

|v|= \sqrt{ 67^{2} + 15^{2} }= \sqrt{4489+225}= \sqrt{4714}=68.8


Thus, to make v a unit vector, without changing its direction, we need to divide v by |v|=68.8. 

This means that the x and y components will also be divided by 68.8, by proportionality.

So, the unit vector in the direction of v is:

<span>(67/68.8)i + (-15/68.8)j=0.97 i + (- 0.22)j
</span>

Answer: 12.62°;  102.62°;  0.97 i + (- 0.22)j

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daser333 [38]
To complete the identity, we need these fundamental identities:

1)\displaystyle{sec(x)=\frac{1}{cos(x)}

2) cos(x-y)=cos(x)cos(y)+sin(x)sin(y)

\displaystyle{csc(x)= \frac{1}{sin(x)}


Thus, by identity 1 we have:

\displaystyle{ sec( \frac{ \pi }{2}-\theta )= \frac{1}{cos(\frac{ \pi }{2}-\theta)}

by identity :

\displaystyle{cos(\frac{ \pi }{2}-\theta)=cos(\frac{ \pi }{2})cos(\theta)+sin(\frac{ \pi }{2})sin(\theta)

recall the values :

\displaystyle{ sin(\frac{ \pi }{2})^R=sin(90^o)=1\\\\

\displaystyle{ cos(\frac{ \pi }{2})^R=cos(90^o)=0, 


so: 

cos(\frac{ \pi }{2})cos(\theta)+sin(\frac{ \pi }{2})sin(\theta)=0+sin(\theta)=sin(\theta)


Putting all these together, we have:


\displaystyle{ sec( \frac{ \pi }{2}-\theta )= \frac{1}{cos(\frac{ \pi }{2}-\theta)}= \frac{1}{cos(\frac{ \pi }{2})cos(\theta)+sin(\frac{ \pi }{2})sin(\theta)}= \frac{1}{sin(\theta)}}

which is equal to csc(\theta), by identity 3


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