Find a vector along the vector < 3,6 > of length 7

Question

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

Answer 1

Can you compute the length of $\lt 3,6\gt$? Now just multiply by a scalar that makes the length $7$

Answer 2

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 $$ 7\frac{<3,6>}{\|<3,6>\|} $$

Answer 3

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 $$\frac{7}{\|\langle 3, 6 \rangle\|} = \frac{7}{\sqrt{3^2 + 6^2}} = \frac{7}{3\sqrt{5}}.$$