Help me with these vectors, please
 

Problem:

u=i-3j
v=ai+j
Vectors are parallel.

Find a.

What I have so far:

We know that:

Code:

---

              a₁a₂+ b₁b₂
 cos Θ =    ---------------
                  |u||v|

cos Θ = 1 

            (a)(1) + (-3)(1)
        1 = ------------------
                  |u||v|


|u| = √(1² +(-3)²)
|v| = √(a² + 1²)

---

So what you get when you put them in is this:

Code:

---

                  a - 3
            ------------------    = 0
            √(10)  √(a² + 1)

---

How on earth do you solve for a?

The answer is: -1/3


The other way to attack this is to say:

u=kv where k is a constant scalar since the two vectors are parallel, right? How can this be used?

Seems you don't need an equation, just draw the two vectors on a X-Y chart.

i is the unit vector on the x-axis, i.e. the vector from (0,0) to (1,0)
j is the unit vector on the y-axis, i.e. the vector from (0,0) to (0,1)
unit vectors in red below

vector u = i - 3 j is the vector from (0,0) to (1,-3). In blue below. It takes 1 step right for every three steps down.

vector v = a i + j. Vector v is "parallel" to vector u. In green below. I guess that must mean it is parallel as in lies on the same line, but it must face the other way since the j coefficient is positive. It is the vector from (0,0) to (a,1). It takes 1 step left for every 3 steps up. But since it only takes 1 step up, it can only take 1/3 step left. Thus a = -1/3.

Make sense? Maybe I didn't understand the problem or the word "parallel".

http://forums.pelicanparts.com/uploa...1240960513.jpg

Check this --> http://www.onlinemathlearning.com/parallel-vectors.html

If you need to answer it using a computation, see here

http://www.onlinemathlearning.com/parallel-vectors.html

This also gives you a = -1/3

(1, -3) = k (a, 1)

1 = k a => k = 1/a

-3 = k 1 = 1/a 1 => a = -1/3

Edit: Oops, funny/sorry, jordi pointed to the same link.

Yeah was already there. Can't seem to wrap around how to apply it to the question. Drawing a picture gets me to the answer, but it doesn't do anything for the application of the material in the section of the book though.

So when given two vectors that are parallel (or anti-parallel I guess), using u=kv is the best way to attack it?

My question that remains is this:

This equation:
should be able to be used whether or not the angle is 20°, 160°, or pi/4 or 0°. Correct?

Jeepers you guys are smart. It's Greek to me...

87 blk coupe

Do these Vectors help?

http://www.bligblog.com/media/03-13-...ctor_5_450.jpg

http://files.conceptcarz.com/img/vector/vwx3-2.jpg

http://blogs.venturacountystar.com/m...ector%20W8.jpg

Randy

No. I'm already frustrated.

Ahhhhh, a classic case of Vector Envy! ;)

Randy

In the equation in the initial post, square both sides. Left side is still 1. Right side is a polynomial numerator divided by a polynomial denominator, and will have gotten rid of the SQRT. Move denominator to numerator of left side, rewrite equation in polynomial = 0 form, and solution will be clear.

I am SO glad there we guys like you to cheat off in HS. Now give me your lunch money.

I thought you'd been banned for starting the Miss Australia thread.

The solution is not clear though on future problems.

u=2i+3j
v=ai-j
angle between vectors: 2pi/3

This ends up with some incredibly obnoxious polynomial equal to zero, which doesn't have any obvious solutions. Solving for it via a calculator comes out with two answers but the book only lists one.

The dot product works doesn't it? The cos(0) = 1 so from your equation above

Sqrt(10)*sqrt(a^2+1) = a-3
Square both sides
10*a^2+10=a^2-6*a+9
9*a^2 + 6 * a + 1 = 0
a^2 + 2/3 * a + 1/9 = 0

a=-1/3

if they are parallel then the coefficients must have the same relations

1/a=-3/1 a=-1/3