Friday

April 25, 2014

April 25, 2014

Posted by **Jon** on Saturday, April 25, 2009 at 3:43pm.

3Y=x+5

y=(x+5)/3

-1/m

= -3

y=-3x+b

sub in P(7,-4)

-4=-3(7)+b

b=21-4

b=17

y=-3x+17

set the two equations equal.

-3x+17=(x+5)/3

-9x+51=x+5

46=10x

x=4.6

sub back in:

y= -3(4.6)+17= -13.8+17=3.2

distance=sqrt((7-4.6)^2+(-4-3.2)^2)

=sqrt((2.4)^2+(7.2)^2)

=sqrt(5.76+51.84)

=sqrt(57.6)=7.59

=7.6

Is there a shorter way to do this?

2.Write XY(with line over it) as the sum of unit vectors for X(8,2,-9) and Y(-12,-1,10).

There's an example just like this in my book but I really don't understand it.

- Math -
**Count Iblis**, Saturday, April 25, 2009 at 6:27pmThere is indeed a shorter way. What you do is you shift the origin of the coordinate system so that the line moves to the origin.

To do that just find a random point that lies on the line, say, the point

(-5,0). Then if we translate the entire coordinate system so that this point moves to the orgin, the point P will have coordinates

(7, -4) - (-5,0) = (12, -4)

Now consider the unit vector e1 that points in the direction along the line and the unit vector e2 that points orthogonal to the line.

If you express the point P = (12,-4) in terms of e1 and e2, like:

P = r e1 + s e2

then you can interpret this as moving from the origin to P as moving along the line over a distance r and orthogonal to the line over a distance s.

So, clearly all we need to do is expand P in terms of e1 and e2 and then the coefficient of e2 is the answer.

e2 is, of course, proportional to

(1,-3). You have to normalize it:

e2 = 1/sqrt(10) (1,-3)

An then s follows from the general expansion formula of vectors in terms of unit vectors:

P = (P dot e1) e1 + (P dot e2) e2

s = P dot e2 =

(12,-4) dot (1,-3)/sqrt(10) =

24/sqrt(10)

- Math -
**Reiny**, Sunday, April 26, 2009 at 11:49amOr an even shorter way is to use the formula for the distance from a point (p,q) to the line Ax + By + C = 0

= │Ap + Bq + C│/√(A^2 + B^2)

= │7 +12 +5│/(1+9)

= 24/√10

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