FP4 Vectors question

In general a line defined by two points, $A$ and $B$, where $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$, has starting point $A$ and direction $AB$.

If we let $\vec{AB} = \vec{AO} + \vec{OB} = \vec{b} - \vec{a} = \vec{c}$, the equation of the line can be written as:

$\begin{aligned} \vec{r} & = \vec{OA} + \lambda \left( \vec{AB} \right) \\ & = \vec{a} + \lambda \left( \vec{b} - \vec{a} \right) \\ & = \vec{a} + \lambda \vec{c} \end{aligned}$

After a bit of rearranging and taking the vector product of the equality with $\vec{c}$ gives:



$\because \vec{c} \times \vec{c} = 0, \ \left( \vec{r} - \vec{a} \right) \times \vec{c} = 0$

Precisely!

No problem! I'm glad you get it!

The vector product of two vectors $\vec{a}$ and $\vec{b}$, $\vec{a} \times \vec{b}$, gives another vector that is perpendicular to both $\vec{a}$ and $\vec{b}$. Since you've found that $\vec{a} \cdot \vec{b} = 0$ and both $\vec{a}$ and $\vec{b}$ are non-zero, this leaves us with the fact that the angle, $\theta$, between $\vec{a}$ and $\vec{b}$ is 90 degrees so they are perpendicular to each other. This means that they form a right handed set like the Cartesian (positive $x$, $y$ and $z$) axes. I think...

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That's good! Vectors can be pretty darn tricky when you're trying to visualise them

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Which exam board are you on?

I thought that the vectors in C4 were primarily the scalar product and it's applications.

I'm guessing you got 93 in C3 then?

Remember that the scalar product of two vectors $\vec{a}$ and $\vec{b}$ is the projection of $\vec{a}$ onto $\vec{b}$.

It's all a bit like resolving a vector into its components

That's a nice score! Are you applying to university this year?

Nice! I'm on my gap year at the moment. Takin' it easy