Homework Help: Parallel vectors and scalar product rule

1. Sep 11, 2006

hey,
ive been given a problem where vector a = 2i + 3j and vector b = $$\lambda$$i + 12j and also told that these vectors are parallel of each other. i understand since the vectors are parallel of each other, the angle between them would be equal to zero, thus i could apply the scalar product rule to help solve this problem, which in this case i did.

My working:

$$\begin{array}{c} {\bf{a}} = 2{\bf{i}} + 3{\bf{j}} \\ {\bf{b}} = \lambda {\bf{i}} + 12{\bf{j}} \\ {\bf{a}}\parallel {\bf{b}} \\ \theta = 0 \\ \cos \theta = 1 \\ {\bf{a}} \cdot {\bf{b}} = \left| {\bf{a}} \right|\left| {\bf{b}} \right|\cos \theta \\ \left( {2{\bf{i}} + 3{\bf{j}}} \right) \cdot \left( {\lambda {\bf{i}} + 12{\bf{j}}} \right) = \left( {\sqrt {13} } \right)\left( {\sqrt {144 + \lambda ^2 } } \right)\left( 1 \right) \\ 2\lambda + 36 = \left( {\sqrt {13} } \right)\left( {\sqrt {144 + \lambda ^2 } } \right) \\ \left( {2\lambda + 36} \right)^2 = 13\left( {144 + \lambda ^2 } \right) \\ 4\lambda ^2 + 144\lambda + 1296 = 1872 + 13\lambda ^2 \\ - 9\lambda ^2 + 144\lambda - 576 = 0 \\ x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\ \lambda = \frac{{ - 144 \pm \sqrt {144^2 - 4\left( { - 9} \right)\left( { - 576} \right)} }}{{2\left( { - 576} \right)}} \\ \lambda = \frac{1}{8} \\ \end{array}$$

this is the wrong answer i know, as the correct answer would result in $$\lambda$$ being equal to eight, however i do not know how to obtain this answer, any help is greatly appriciated, thanks. sorry for the many post lately but i have a test comeing up which i realy wish to do well in to prove the teacher that i can do her subject. thanks once again,

2. Sep 11, 2006

Tomsk

Check your working, it should be OK. The second last equation you quoted is right, but you didn't put the numbers in correctly.

3. Sep 11, 2006

okay thanks for your help ill try that right now

4. Sep 11, 2006

hey that worked thanks~! i got the answer of 8 and -8. can i ask you why it didn't work when i used the quadratic formula? thanks

5. Sep 11, 2006

Astronuc

Staff Emeritus
$$\lambda = \frac{{ - 144 \pm \sqrt {144^2 - 4\left( { - 9} \right)\left( { - 576} \right)} }}{{2\left( { - 9} \right)}}$$
One divided by 2c instead of 2a.

As Tomsk suggested, dividing through by -9 would have simplied the solution.

$$\lambda ^2 - 16\lambda + 64 = 0$$

Two vectors are parallel if one is a multiple of the other, i.e.

a i + b j = $\lambda$a i + $\lambda$b j

Last edited: Sep 11, 2006
6. Sep 11, 2006

oh i see thanks, i feel so stupid now
thanks once again,

7. Sep 11, 2006

0rthodontist

It would have been a lot simpler to use the fact that parallel vectors are multiples of each other from the start. You have:
2xi + 3xj = yi + 12j
3x = 12
x = 4
2x = y
y = 8

8. Sep 12, 2006