Dot Product: Unit Vector & Projections

**lilaziz1** · September 12th 2010, 05:32 PM

Question 1.

Find a unit vector that is orthogonal to both $i^\rightharpoonup + j^\rightharpoonup$ and $i^\rightharpoonup + k^\rightharpoonup$

My attempt:

$a^\rightharpoonup = <1, 1, 0>$

$b^\rightharpoonup = <1, 0, 1>$

$a^\rightharpoonup * x^\rightharpoonup = 0$

$<1, 1, 0> * <x_1, x_2, x_3> = 0$

$x_1 + x_2 = 0$

$b^\rightharpoonup * x^\rightharpoonup = 0$

$<1, 0, 1> * <x_1, x_2, x_3> = 0$

$x_1 + x_3 = 0$

Now I have 3 variables and 2 equations. What do I do?

Question 2.

If $a^\rightharpoonup = <3, 0, -1>$ , find a vector $b^\rightharpoonup$ such that $comp_{a^\rightharpoonup}b^\rightharpoonup = 2$

My attempt:

$|a^\rightharpoonup| = \sqrt{3^2 + 0^2 + (-1)^2} = \sqrt{10}$

$|b^\rightharpoonup|cos\theta = \frac{a^\rightharpoonup * b^\rightharpoonup}{|a^\rightharpoonup|} = 2$

$2 = \frac{<3, 0, -1> * <b_1, b_2, b_3>}{\sqrt{10}}$

$2 = \frac{3b_1 + (-1)b_3}{\sqrt{10}}$

Once again, I havmore variables than the # of equations. What should I doo?

Thanks in advance!

**sa-ri-ga-ma** · September 12th 2010, 09:04 PM

If A and B are the vectors, then AXB will be a vector orthogonal to A and B.

| i j k |
|11 0|
|10 1|

Solve this determinant to get AXB.

**HallsofIvy** · September 13th 2010, 05:27 AM

By the way, the LaTex \vec{v}, $\vec{v}$ , gives, in my opinion, a better looking vector than v^\rightharpoonup, $v^\rightharpoonup$ , and is simpler.

**lilaziz1** · September 13th 2010, 10:45 AM

I can't use the cross product to solve this because we have not covered it yet. Is there an alternative method? Thanks! (especially for the LaTex statement).

**HallsofIvy** · September 13th 2010, 11:47 AM

Originally Posted by lilaziz1

Question 1.

Find a unit vector that is orthogonal to both $i^\rightharpoonup + j^\rightharpoonup$ and $i^\rightharpoonup + k^\rightharpoonup$

My attempt:

$a^\rightharpoonup = <1, 1, 0>$

$b^\rightharpoonup = <1, 0, 1>$

$a^\rightharpoonup * x^\rightharpoonup = 0$

$<1, 1, 0> * <x_1, x_2, x_3> = 0$

$x_1 + x_2 = 0$

$b^\rightharpoonup * x^\rightharpoonup = 0$

$<1, 0, 1> * <x_1, x_2, x_3> = 0$

$x_1 + x_3 = 0$

Now I have 3 variables and 2 equations. What do I do?

Yes, because the "orthogonal to" determines only about the direction, not the length. You have the third equation $\sqrt{x_1^2+ y_1^2+ z_1^2}= 1$ . Alternatively, take any one of the variables, say x, equal to 1 and solve for the other two. Then normalize to get a unit vector.

Question 2.

If $a^\rightharpoonup = <3, 0, -1>$ , find a vector $b^\rightharpoonup$ such that $comp_{a^\rightharpoonup}b^\rightharpoonup = 2$

My attempt:

$|a^\rightharpoonup| = \sqrt{3^2 + 0^2 + (-1)^2} = \sqrt{10}$

$|b^\rightharpoonup|cos\theta = \frac{a^\rightharpoonup * b^\rightharpoonup}{|a^\rightharpoonup|} = 2$

$2 = \frac{<3, 0, -1> * <b_1, b_2, b_3>}{\sqrt{10}}$

$2 = \frac{3b_1 + (-1)b_3}{\sqrt{10}}$

Once again, I havmore variables than the # of equations. What should I doo?

Thanks in advance!

**matheagle** · September 13th 2010, 01:38 PM

this is from stewart, isn't it.
the next section has cross products
Just let X1=1 and the the other two are -1,
<1,-1,-1> works or it's negative
NOW divide by the magnitude and you are done.

$\pm{1\over \sqrt{3}}<1,-1,-1>$

**lilaziz1** · September 13th 2010, 02:55 PM

Wait how can you just plug any number you want in the equation? Doesn't that break all the rules to solving an equation?

And yea this is from Stewart. Btw, what about Question #2? Can we just plug in w/e the hell we want? If so, whyy?

**Plato** · September 13th 2010, 03:12 PM

Originally Posted by lilaziz1

Wait how can you just plug any number you want in the equation? Doesn't that break all the rules to solving an equation?

It breaks no rules.
Consider this: solve $<1,1,0>\cdot <a,b,c>=0~\&~<1,0,1>\cdot <a,b,c>=0$ .

It is easy to see that if $c\ne 0$ then $<c,-c,-c>$ solves that system.

So make a unit vector out of $<c,-c,-c>$ .

**lilaziz1** · September 13th 2010, 04:22 PM

Um let's step back in your example.

$<1,1,0>\cdot <a,b,c>=0~\&~<1,0,1>\cdot <a,b,c>=0$

a + b = 0
a + c = 0

How do you know which variable to substitute with any number that doesn't equal to 0? If I let b = 1 in the 2 equations above, I get
<-1, 1, 1>
and if I let a = 1 I get
<1, -1, -1>
and if I let c = 1 I get
<-1, 1, 1>

Thus, they are not the same. They have the same magnitude, but not the same direction.

Am I having a brain fart or am I missing something?

Thanks a ton btw.

**Plato** · September 13th 2010, 05:16 PM

Originally Posted by lilaziz1

$<1,1,0>\cdot <a,b,c>=0~\&~<1,0,1>\cdot <a,b,c>=0$
a + b = 0
a + c = 0
How do you know which variable to substitute with any number that doesn't equal to 0?

The point is very simple, it does not matter what non-zero value of $c$ we choose.
$<c,-c,-c>$ works as long as we make it a unit vector.
So why not choose $c=1?$
It makes thing relatively easy.

**matheagle** · September 13th 2010, 05:37 PM

THERE are TWO answers,
there are 2 unit vectors orthogonal to both
(I assign this problem each year.)

**lilaziz1** · September 13th 2010, 05:41 PM

how are you getting <c, -c, -c> ? If you let c = 1 in what equations?

a + b = 0
a + c = 0

If c = 1,
a + 1 = 0
a = -1

(-1) + b = 0
b = 1

which equals:
<-1, 1, 1>

and you have <c, -c, -c> :\

Thanks!

EDIT: Just read matheagle's post

Okay that makes sense. So it doesn't matter if you let a = 1, b = 1, or c = 1. Right?

but how is Plato getting <c, -c, -c>

**lilaziz1** · September 15th 2010, 12:14 PM

So I was looking through the back of the book and the answer I saw was: <1, -1, -1> * 3^(-1/2).

I am still confused why we can just plug in any # we want for one of the variables. I mean it's a vector so it doesn't have infinite # of solutions like a line has.

**Plato** · September 15th 2010, 12:38 PM

If $A\cdot B=0$ then $A\cdot (-B)=0$ .
So both $<c,-c,-c>~\&~<-c,c,c>$ are perpendicular to $<1,0,1>~\&~<1,1,0>$ .

BTW: $<c,-c,-c>=-<-c,c,c>$

**matheagle** · September 15th 2010, 12:55 PM

Originally Posted by lilaziz1

So I was looking through the back of the book and the answer I saw was: <1, -1, -1> * 3^(-1/2).

I am still confused why we can just plug in any # we want for one of the variables. I mean it's a vector so it doesn't have infinite # of solutions like a line has.

THERE are an infinite number of vector orthogonal to those two
BUT there are only two unit vectors that are orthogonal to those two

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