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November 08, 2025, 04:29:36 pm

Author Topic: perpendicular vector  (Read 1690 times)  Share 

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enpassant

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perpendicular vector
« on: April 02, 2008, 09:58:29 pm »
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find a unit vector perpendicular to both 2i - j and i + 3j.

Collin Li

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Re: perpendicular vector
« Reply #1 on: April 02, 2008, 10:35:03 pm »
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There are none in two dimensional space. If we assign them 0k, we can find some in three dimensional space:

Let some vector ai + bj + ck be perpendicular to 2i - j and i + 3j.

and

Solving these sets of simultaneous equations yields

This means the vector perpendicular to both those vectors must have while is a 'free' variable (can choose anything).

However, for a unit vector, .

Therefore, are unit vectors perpendicular to both 2i - j and i + 3j.
« Last Edit: April 02, 2008, 10:37:41 pm by coblin »

evaporade

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Re: perpendicular vector
« Reply #2 on: April 03, 2008, 07:16:12 am »
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Obviously +/-k is a unit vector perpendicular to 2i - j and i + 3j because both are in the x-y plane (i-j plane). So is it necessary to do the working as suggested?

Collin Li

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Re: perpendicular vector
« Reply #3 on: April 03, 2008, 08:09:25 am »
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Obviously +/-k is a unit vector perpendicular to 2i - j and i + 3j because both are in the x-y plane (i-j plane). So is it necessary to do the working as suggested?

Perhaps in Specialist Maths, yes, since geometry is not taught properly.

enpassant

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Re: perpendicular vector
« Reply #4 on: April 03, 2008, 03:44:20 pm »
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find a unit vector perpendicular to i - 2j + 3k

Mao

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Re: perpendicular vector
« Reply #5 on: April 03, 2008, 04:23:49 pm »
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there's an infinite numbers of unit vectors perpendicular to this one (imagine it as an axis)

but to find "a" unit vector:

let a be a vector perpendicular to





one such set can be:



then

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