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OpenStudy (anonymous):

Find the vector and parametric equations of the line in R^2 that passes through the origin and is orthogonal to v, where v = (-2,3).

OpenStudy (anonymous):

@LizzyMcWire

myininaya (myininaya):

So say that u is the vector we are looking for. So we know u dot v=0

myininaya (myininaya):

We also know u=<a,b> since u is a vector in R^2

myininaya (myininaya):

so we have -2a+3b=0

myininaya (myininaya):

But do you know how to take care of the part where it passes through the origin?

OpenStudy (anonymous):

at the origin, a=0 and b=0

myininaya (myininaya):

yeah but that seems too easy...

OpenStudy (anonymous):

how does v becomes (3,2)?

myininaya (myininaya):

that actually is a vector that also satisfies the -2a+3b=0 thing but one sec

myininaya (myininaya):

we need it to pass through the origin .... so we also need to choose the vector so that the terminal points terminates on one side of 0 and the initial point starts on the other side of 0 and then we would have a vector passing through the origin so |dw:1412816843148:dw| The slope of that is (d-q)/(c-p) y=mx (this line goes through (0,0)) Now we need to figure out what points those are... and what if we chose the starting point (the intial point as (0,0) then we have m=d/c y=dx/c |dw:1412817105124:dw|

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