Vectors

[cazza90]

How do I go about solving this?
V is a vector of length 3 and u is a vector of length 2. The angle between u and v is pi/3. find the exact value of the length of u-2v?

 

[Deleted member 4993]

How do I go about solving this?
V is a vector of length 3 and u is a vector of length 2. The angle between u and v is pi/3. find the exact value of the length of u-2v?

Have you covered graphical method?

In graphical method, draw (-2v) as x-axis going from origin to the negetive x direction.

Draw vector u, making an angle ?/3 with the positive x-axis.

Now use law of parallelogrm and add (-2v) and (u)

 

[soroban]

Hello, cazza90!

This is a slight variation of Subhotosh's solution.

\(\displaystyle \vec v\text{ is a vector of length 3, and }\vec u\text{ is a vector of length 2.}\)

\(\displaystyle \text{The angle between }\vec u\text{ and }\vec v\text{ is }\frac{\pi}{3}\)

\(\displaystyle \text{Find the exact value of the length of }\vec u-2\vec v\)

We have:

            *
           /
          /
   |v|=3 /
        /
       / 60d 
      * - - - * 
       |u|=2

\(\displaystyle \text{Then we have: }\:\vec u - 2\vec v\)

              2
        P o - - - o Q
             60d /
                /
         *     /
              /
             /
        *   * 6
           /
          /
       * /
        /
       /
    R o

\(\displaystyle \text{We have: }\;\vec u \,=\,\overrightarrow{PQ},\;\;-2\vec v \:=\:\overrightarrow{QR},\;\;\angle PQR = 60^o\)

. . \(\displaystyle \text{Then: }\:|\vec u - 2\vec v| \;=\;|\overrightarrow{PR}|\)


\(\displaystyle \text{Law of Cosines: }\;|\vec u - 2\vec v|^2 \;=\;|\overrightarrow{PR}|^2 \;=\;2^2 + 6^2 - 2(2)(6)\cos60^o \;=\;28\)


\(\displaystyle \text{Therefore: }\:|\vec u - 2\vec v| \;=\;2\sqrt{7}\)

 

[cazza90]

Thanks guys, that really helped. Except that I don't know what the parallogram law is?