Posted by Jon on Saturday, April 25, 2009 at 3:43pm.
1.Find the distance between P(7, -4) and the line with equation x - 3y + 5 = 0 round to nearest tenth
sub in P(7,-4)
set the two equations equal.
sub back in:
y= -3(4.6)+17= -13.8+17=3.2
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 6:27pm
There 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) =
Math - Reiny, Sunday, April 26, 2009 at 11:49am
Or 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)
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