Determine if two vectors are orthogonal

**hemster83** · September 26th 2012, 03:07 PM

Hey everyone! I have a question about the orthogonality of two vectors

The question is: It is given: llmll = 4; llnll = square root of 3; (m^,n)= 150 degrees

(i) find the norm of the vector m + 2n
(ii) Determine if the vectors (m+2n) and (-m+n) are orthogonal

I got an answer of 2 for the norm but I don't know what to do for the second part. Do I just substitute the values of m and n into the vectors and if they both equal 0 when multiplied, they are orthogonal? Any help would be greatly appreciated

**Plato** · September 26th 2012, 03:12 PM

Originally Posted by hemster83

Hey everyone! I have a question about the orthogonality of two vectors

The question is: It is given: llmll = 4; llnll = square root of 3; (m^,n)= 150 degrees

(i) find the norm of the vector m + 2n
(ii) Determine if the vectors (m+2n) and (-m+n) are orthogonal

What does the notation (m^,n)= 150 degrees mean?
I have never seem that used.

**hemster83** · September 26th 2012, 03:14 PM

Originally Posted by Plato

What does the notation (m^,n)= 150 degrees mean?
I have never seem that used.

It means the angle between m and n equals 150 degrees. I think my prof made it up himself

**Plato** · September 26th 2012, 03:54 PM

Originally Posted by hemster83

It means the angle between m and n equals 150 degrees. I think my prof made it up himself

How very odd is that?

Originally Posted by hemster83

The question is: It is given: llmll = 4; llnll = square root of 3; (m^,n)= 150 degrees
(i) find the norm of the vector m + 2n
(ii) Determine if the vectors (m+2n) and (-m+n) are orthogonal

Well you know that $\frac{{m \cdot n}}{{\left\| m \right\|\left\| n \right\|}} = \cos \left( {\frac{{5\pi }}{6}} \right) = \frac{{ - \sqrt 3 }}{2}$ .

From that you can find $m\cdot n~.$

Now $\|m+2n\|^2=(m+2n)\cdot(m+2n)=m\cdot m+4(n\cdot m)+4(n\cdot n)~.$

**Kristoffermk3** · January 25th 2013, 02:39 PM

I have this same exact question.

Well you know that $\frac{{m \cdot n}}{{\left\| m \right\|\left\| n \right\|}} = \cos \left( {\frac{{5\pi }}{6}} \right) = \frac{{ - \sqrt 3 }}{2}$ .

From that you can find $m\cdot n~.$

Now $\|m+2n\|^2=(m+2n)\cdot(m+2n)=m\cdot m+4(n\cdot m)+4(n\cdot n)~.$

Now this is the method I used to find the norm, or length of the vector (m+2n). However, I am still at a loss as to how to test the orthogonality of the two vectors (m+2n) and (-m+n).

Thanks in advance.

**Plato** · January 25th 2013, 02:50 PM

Originally Posted by Kristoffermk3

I have this same exact question.
Now this is the method I used to find the norm, or length of the vector (m+2n). However, I am still at a loss as to how to test the orthogonality of the two vectors (m+2n) and (-m+n).

What is $(m+2n)\cdot(-m+n)=~?$

**Kristoffermk3** · January 25th 2013, 02:52 PM

Yes, that's what i'm looking for.

I suppose the part that is confusing me is simply that its not asking something as simple as (m dot n). Which is very simple to figure out, but I am confused on how to approach this when they are combinations of vectors such as (m+2n) and (-m+n)...

**Plato** · January 25th 2013, 03:06 PM

Originally Posted by Kristoffermk3

Yes, that's what i'm looking for.

I suppose the part that is confusing me is simply that its not asking something as simple as (m dot n). Which is very simple to figure out, but I am confused on how to approach this when they are combinations of vectors such as (m+2n) and (-m+n)...

$(m+2n)\cdot(-m+n)=-m\cdot m-2 m\cdot n+m\cdot n+2n\cdot n$

**Kristoffermk3** · January 25th 2013, 03:58 PM

Oh my, that seems so self-evident now.

Thank you very much for your clarification.

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