Vectors

It's 3am, long past the time for thinking straight, but we might as well write down what we know:

We want a vector x that is parallel to a, so let

x = \alpha (1 \ -2 \ 4)

.
We want a vector y that is perpendicular to a. So

y.a = 0

. In other words,

y_1 - 2y_2 + 4y_3 = 0

, and let us call this equation

(\dagger )

We want

x + y = (2 \ -5 \ 4)

, so this means:

\alpha + y_1 = 2

-2 \alpha + y_2 = -5

4\alpha + y_3 = 4

.

Using these three equations and

\dagger

, you should be able to finish the problem off.

Reply 2

DFranklin

Writing capital letters for vectors, suppose we want to write R as a vector parallel to T and a vector perpendicular to T.

We can write the vector perpendicular to T as tT for some scalar t.

Then we must have R = tT + (R-tT), with (R-tT) perpendicular to T.

So T.(R-tT) = 0. So t = T.R/T^2

Reply 3

bubbles50

I can't do those vectors yet, it looks like you've got the answers now anyway but just had to say: you're CALLED vector... wtf? lol

Reply 4

vector

Haha, yeah, named after the fountain pen :p:

Thanks for the help :smile:

Reply 5

Kolya

In the light of day, DFranklin takes a more mature approach to the question. :smile:

Reply 6

DFranklin

Kolya

In the light of day, DFranklin takes a more mature approach to the question. :smile:

People sometimes ask "do you ever use any of this stuff in real life?". In fact, calculations like this come up all the time in computer graphics. Which is why I know this particular one like the back of my hand.

Reply 7

Kolya

DFranklin

People sometimes ask "do you ever use any of this stuff in real life?". In fact, calculations like this come up all the time in computer graphics.

In what way?

(As the OP seems to have had their question answered, I think it's safe to derail slightly.)

Reply 8

DFranklin

This particular one comes up when you have a line A+tB and a point C and you want to find the point on the line that is closest to C. Which is equivalent to finding the value of t s.t. (A+tB-C).B = 0.

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