# Determine scale of vector from base vector

I'm fairly new to vectors and using them in algebra, so please forgive me if this is a really basic question!

Say I have a point $P_2$ where $P_2 = P_1 + \vec{v}t$. I know $P_2$, $P_1$, $\vec{v}$. I know $P_2$ satisfies the equation due to prior calculations and want to solve for $t$. My confusion comes on the algebraic manipulation:

$$P_2 = P_1 + \vec{v}t$$ $$P_2 - P_1 = \vec{v}t$$ $$\frac{P_2 - P_1}{\vec{v}} = t$$

The last step is wrong, isn't it? One can't divide by a vector AFAIK. Then I thought, if I was to solve this just using an arbitrary element ($P_2-P_1$) divided by the corresponding element of $\vec{v}$, then I risk dividing by zero.

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The part after your "Then I thought..." is quite acceptable. – J. M. Oct 2 '11 at 21:06

Assuming $\vec{v} \not=0$.

If $P_2-P_1=\vec{v}t$ then $P_2-P_1$ and $\vec{v}$ are parallel. This means each of their components are off by the same scalar multiple (namely $t$). Suppose a non-zero component of $\vec{v}$ is component number $j$, then $t$ is the $j^{\mbox{th}}$-component of $P_2-P_1$ divided by the $j^{\mbox{th}}$-component of $\vec{v}$.

Alternatively, you could use dot products. $(P_2-P_1)\cdot \vec{v} = t \vec{v} \cdot \vec{v}$ and so $t = ((P_2-P_1)\cdot \vec{v} )/|\vec{v}|^2$. [This should remind you of "projections".]

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You cannot divide vectors. But if you have a norm on your vector space, you could do the following:

$$|| P_2 -P_1|| = ||\overrightarrow{v}t|| = ||\overrightarrow{v}|| \cdot |t| \ .$$

And now you can divide numbers:

$$|t| = \frac{|| P_2 -P_1||}{||\overrightarrow{v}||} \ .$$

And, if for some reason, you know that your $t$ is a positive number, you'll have:

$$t = \frac{|| P_2 -P_1||}{||\overrightarrow{v}||} \ .$$

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