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ive been given a problem where vectora= 2i+ 3jand vectorb= [tex]\lambda[/tex]i+ 12jand 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:

[tex]

\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}

[/tex]

this is the wrong answer i know, as the correct answer would result in [tex]\lambda[/tex] 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,

Pavadrin

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# Homework Help: Parallel vectors and scalar product rule

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