1.Find the distance between P(7, -4) and the line with equation x - 3y + 5 = 0 round to nearest tenth
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.
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) =
24/sqrt(10)
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)
= 24/√10
1. Finding the distance between a point and a line can be done using the formula:
Distance = |Ax + By + C| / sqrt(A^2 + B^2)
In this case, the equation of the line is x - 3y + 5 = 0. To find the values of A, B, and C, we can convert the equation into the standard form Ax + By + C = 0.
First, isolate y in terms of x:
x - 3y = -5
-3y = -x - 5
y = (x + 5) / 3
Comparing this to the standard form, we have A = 1, B = -3, and C = 0. Now, plug in the coordinates of the point P(7, -4) into the distance formula:
Distance = |1 * 7 - 3 * (-4) + 0| / sqrt(1^2 + (-3)^2)
= |7 + 12| / sqrt(1 + 9)
= |19| / sqrt(10)
= 19 / sqrt(10)
≈ 6.02 (rounded to the nearest tenth)
So, the distance between P(7, -4) and the line x - 3y + 5 = 0 is approximately 6.02 units.
2. To write XY with a line over it as the sum of unit vectors, first, calculate the vector from X(8, 2, -9) to Y(-12, -1, 10). This can be done by subtracting the coordinates of X from the corresponding coordinates of Y:
XY = (-12 - 8, -1 - 2, 10 - (-9))
= (-20, -3, 19)
Next, calculate the magnitude of XY using the formula:
Magnitude = sqrt((-20)^2 + (-3)^2 + (19)^2)
= sqrt(400 + 9 + 361)
= sqrt(770)
≈ 27.8
Now, to express XY as the sum of unit vectors, divide each component of XY by its magnitude:
XY = ((-20) / 27.8, (-3) / 27.8, 19 / 27.8)
= (-0.719, -0.108, 0.684)
So, XY with a line over it can be written as the sum of unit vectors (-0.719)i + (-0.108)j + (0.684)k.