Find a vector along the vector < 3,6 > of length 7
I know I'm supposed to use the dot product, but I have no idea how to set it up, considering our professor didn't cover THIS type of problem. I decided to draw it out to start off.
What formula should I be using in this situation? Thanks so much
$\endgroup$ 23 Answers
$\begingroup$Well, the vector $<3,6>$ is in the direction $\theta=\tan^{-1}(\frac{6}{3})$. So you can define a vector in that direction with magnitude 7. If you want the components of that new vector, it could be $$ <7\cos(\theta),7\sin(\theta)> $$Or if you'd like, the unit vector in the direction $<3,6>$ is given by $$ \hat{v}=\frac{<3,6>}{\|<3,6>\|} $$so you could define your vector as $$ 7\frac{<3,6>}{\|<3,6>\|} $$
$\endgroup$ $\begingroup$Can you compute the length of $\lt 3,6\gt$? Now just multiply by a scalar that makes the length $7$
$\endgroup$ 3 $\begingroup$The question is asking you to find a vector, pointing in the same direction as $\langle 3, 6\rangle$, but has length $7$. The picture you made doesn't accurately represent the situation, as both vectors need to be parallel.
To find this, you simply need to scale $\langle 3, 6 \rangle$ by a positive scalar that makes the length $7$. Such a scalar is simply$$\frac{7}{\|\langle 3, 6 \rangle\|} = \frac{7}{\sqrt{3^2 + 6^2}} = \frac{7}{3\sqrt{5}}.$$
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