- #1
pavadrin
- 154
- 0
hey,
ive been given a problem where vector a = 2i + 3j and vector b = \lambdai + 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:
<br /> \begin{array}{c}<br /> {\bf{a}} = 2{\bf{i}} + 3{\bf{j}} \\ <br /> {\bf{b}} = \lambda {\bf{i}} + 12{\bf{j}} \\ <br /> {\bf{a}}\parallel {\bf{b}} \\ <br /> \theta = 0 \\ <br /> \cos \theta = 1 \\ <br /> {\bf{a}} \cdot {\bf{b}} = \left| {\bf{a}} \right|\left| {\bf{b}} \right|\cos \theta \\ <br /> \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) \\ <br /> 2\lambda + 36 = \left( {\sqrt {13} } \right)\left( {\sqrt {144 + \lambda ^2 } } \right) \\ <br /> \left( {2\lambda + 36} \right)^2 = 13\left( {144 + \lambda ^2 } \right) \\ <br /> 4\lambda ^2 + 144\lambda + 1296 = 1872 + 13\lambda ^2 \\ <br /> - 9\lambda ^2 + 144\lambda - 576 = 0 \\ <br /> x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\ <br /> \lambda = \frac{{ - 144 \pm \sqrt {144^2 - 4\left( { - 9} \right)\left( { - 576} \right)} }}{{2\left( { - 576} \right)}} \\ <br /> \lambda = \frac{1}{8} \\ <br /> \end{array}<br />
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 really wish to do well into prove the teacher that i can do her subject. thanks once again,
Pavadrin
ive been given a problem where vector a = 2i + 3j and vector b = \lambdai + 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:
<br /> \begin{array}{c}<br /> {\bf{a}} = 2{\bf{i}} + 3{\bf{j}} \\ <br /> {\bf{b}} = \lambda {\bf{i}} + 12{\bf{j}} \\ <br /> {\bf{a}}\parallel {\bf{b}} \\ <br /> \theta = 0 \\ <br /> \cos \theta = 1 \\ <br /> {\bf{a}} \cdot {\bf{b}} = \left| {\bf{a}} \right|\left| {\bf{b}} \right|\cos \theta \\ <br /> \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) \\ <br /> 2\lambda + 36 = \left( {\sqrt {13} } \right)\left( {\sqrt {144 + \lambda ^2 } } \right) \\ <br /> \left( {2\lambda + 36} \right)^2 = 13\left( {144 + \lambda ^2 } \right) \\ <br /> 4\lambda ^2 + 144\lambda + 1296 = 1872 + 13\lambda ^2 \\ <br /> - 9\lambda ^2 + 144\lambda - 576 = 0 \\ <br /> x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\ <br /> \lambda = \frac{{ - 144 \pm \sqrt {144^2 - 4\left( { - 9} \right)\left( { - 576} \right)} }}{{2\left( { - 576} \right)}} \\ <br /> \lambda = \frac{1}{8} \\ <br /> \end{array}<br />
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 really wish to do well into prove the teacher that i can do her subject. thanks once again,
Pavadrin
