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Athanasios
Level 3
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Message 1 of 6
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Vector calculus on HP

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HP Prime

Hi All,

 

Can someone please show me how to solve these questions using the HP Prime?

 

2017-10-27_9-14-08.jpg

2017-10-27_9-16-14.jpg

2017-10-27_9-16-33.jpg

I am able to complete the calculations by hand however given that these questions are worth so few marks and that this is a calclator I would like to know how to use the HP Prime to solve these questions.

 

 

Thanks,

Athanasios

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Joe_Horn
Level 8
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Message 2 of 6
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a. A unit vector can be found by dividing any vector by its own absolute value (Frobenius norm, AKA Euclidean norm):

 

sign2

 

Important: Exact results are only obtained in CAS (not Home), and ABS must be spelled using uppercase letters (or the |x| symbol may be used instead), and the CAS Setting for Simplify must be set to Maximum, to achieve the above results.  If Textbook Display mode is turned on (Home Settings, page 2), the above will look like this:

 

sign

 

I'll leave the other problems for others to work on.

 

Disclaimer: I don't work for HP. I'm just another happy HP calculator user.

 

-Joe-
Jan_D
Level 6
305 297 32 76
Message 3 of 6
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I will try to solve question c.

 

When you know the formula which describes the relation between the dot product and cos(phi) it is not so difficult.

In fact, we only need the dot product and the length of a vector.

 

The dot product can be found by Toolbox - Math - Matrix - Vector - Dot Product.

The length of a vector is on Key M.

 

This is my solution:

 

2017-10-27 22.57.55.jpg

 

 

Of course, I did not type the argument of solve manually, I just tapped twice the previous result.

So it is clear that the solution is the second element of the list.

 

When you still have questions feel free to ask them.

Jan_D
Level 6
305 297 32 76
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Question b I would solve this way:

 

a=k*b+x, where k is real and x is a vector. I want to solve for k and x, and I know that DOT(x,b)=0.

So x=a-k*b

 

With CAS settings set to: Simplify=Maximum, I get:

 

2017-10-28 17.58.57.jpg

Jan_D
Level 6
305 297 32 76
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I just found out that it is also possible to solve for a vector.

For this purpose we first need to define the vector variable explicitly as such a variable by assigning eg:

x:=[x1,x2,x3].

For this reason in this particular case this is more of theoretical interest, because it asks more effort.

Good to know nevertheless that this is possible. 

 

2017-10-28 23.14.26.jpg

Jan_D
Level 6
305 297 32 76
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additional remark:

 

For real vectors we can simply write

a*b instead of DOT(a,b).

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