Vectors: Find all the possible coordinates of a pt P that is twice as far from R as it is from Q?

Question

A line contains the pts Q=(5,5,3) & R=(1,-3,8).

I know there is 2 answer. I found one of the pts which is (11/3,7/3,14/3), but i don't know how to find the 2nd pt. I know the 2nd pt is (9,13,-2), but i have no clue on how to solve it

Answer 1

if you have 1 point, then the other point should be the reflection of the point on the other side of the line.

but I think its an equation, cuz i doubt its only 2 points.

let (x,y,z) be the coordinates of P.

the distance from P to Q is given by:

sqrt [(x-5)^2 + (y-5)^2 + (z-3)^2] and the distance from P to R:
sqrt [(x-1)^2 + (y+3)^2 + (z-8)^2]

PR = 2 PQ since distance from P to R is twice as much as distance from P to Q

sqrt [(x-1)^2+(y+3)^2+(z-8)^2]=2*sqrt [(x-5)^2+(y-5)^2+(z-3)^2]
square both sides
[(x-1)^2+(y+3)^2+(z-8)^2]=4 * [(x-5)^2+(y-5)^2+(z-3)^2]

expand brackets

[x^2-2x+1+y^2+6y+9+z^2-16z+64]=4[x^2-10x+25+y^2-10y+25+z^2-6z+9]

multiply RHS by 4

[x^2-2x+1+y^2+6y+9+z^2-16z+64]=[4x^2-40x+100+4y^2-40y+100+4z^2-24z+36]

collect like terms
x^2-2x+y^2+6y+z^2-16z+74 = 4x^2-40x+4y^2-40y+4z^2-24z+236

-3x^2 + 38x - 3y^2 + 46y - 3z^2 + 8z - 162 = 0
3x^2 - 38x + 3y^2 - 46y + 3z^2 - 8z + 162 = 0

There is the equation, I think you can most likely simplify it
3(x^2 - 38/3x + 361/9)+3(y^2 - 46/3y +529/9) + 3(z^2 - 8/3z +16/9) + 1458/9 - 361/9 - 529/9 - 16/9 = 0

In this step, I just completed the square of each of the x,y,z terms. The term I added to each of them is half the coefficient of the x squared. so for the first 38/3 term, I added (19/3)^2 and then subtracted it at the end. the 1458/9 is same as 162.

3(x - 19/3)^2 + 3(y - 23/3)^2 + 3(z - 4/3)^2 = 552/9

(x - 19/3)^2 + (y - 23/3)^2 + (z - 4/3)^2 = 552/27

(x - 19/3)^2 + (y - 23/3)^2 + (z - 4/3)^2 = (4.52155)^2

which is the equation of a sphere, with centre (19/3,23/3,4/3).

Hope that helped.

Answer 2

The question isn't really clear. Is P supposed to be on the same line as Q and R??

If so, it's quite simple. Let P = R + k(RQ), where RQ = Q-R = (4, 8, -5). We want PR = ±2PQ. We have PR = -k(RQ) and PQ = RQ - k(RQ) = (-k+1) (RQ). So we want -k(RQ) = ±2 (-k+1) RQ, i.e. -k = ±2 (-k+1).
-k = 2(-k+1) => k = 2
-k = -2(-k+1) => k = 2/3
So P = (1, -3, 8) + 2(4, 8, -5) = (9, 13, -2)
or P = (1, -3, 8) + 2/3(4, 8, -5) = (11/3, 7/3, 14/3).

Now, if P is not restricted to be on the line it gets harder.

Let P have coordinates (x, y, z).

||P-Q||^2 = (x-5)^2 + (y-5)^2 + (z-3)^2
||P-R||^2 = (x-1)^2 + (y+3)^2 + (z-8)^2.
If ||P-R|| = 2||P-Q||, then we get
(x-1)^2 + (y+3)^2 + (z-8)^2 = 4(x-5)^2 + 4(y-5)^2 + 4(z-3)^2.
<=> x^2 - 2x + 1 + y^2 + 6y + 9 + z^2 - 16z + 64 = 4x^2 - 40x + 100 + 4y^2 - 40y + 100 + 4z^2 - 24z + 36
<=> 3x^2 - 38x + 3y^2 - 46y + 3z^2 - 8z + 162 = 0
<=> 3(x - 38/6)^2 + 3(y-46/6)^2 + 3(z-8/6)^2 - 3(38/6)^2 - 3(46/6)^2 - 3(8/6)^2 + 162 = 0
<=> 3[(x - 19/3)^2 + (y-23/3)^2 + (z-4/3)^2] - 140 = 0.
So the locus of possible points P is the surface of the sphere with centre (19/3, 23/3, 4/3) and radius √140 = 2√35.

Answer 3

Given the points Q=(5,5,3) and R=(1,-3,8) find all the points that are twice as far from R as from Q.

If you are talking about the line thru Q and R there are two points. However, if you want all the points in 3space that are twice as far from R as from Q, there are a infinite number of them in the form of a sphere.

I assume you are only talking about the points on the line thru Q and R.

The first point is between Q and R and can be found by taking a weighted average of the two points. Q gets double the weight of R because it is twice as close.

P = (2Q + R)/3 = [2(5,5,3) +(1,-3,8)] / 3

P = (2*5+1, 2*5-3, 2*3+8) / 3 = (11/3, 7/3, 14/3).

The other point is not between Q and R but rather beyond Q. Q is in the middle between P and R.

Let P = (x,y,z)

PR = PQ + QR = 2PQ
QR = PQ

Q = (P + R) / 2
(5,5,3) = [(x,y,z) + (1,-3,8)] / 2

5 = (x + 1)/2
10 = x + 1
x = 9

5 = (y - 3)/2
10 = y - 3
y = 13

3 = (z + 8)/2
6 = z + 8
z = -2

So P = (9, 13, -2).

Answer 4

Case 1 P lies on QR between Q and R
Difficult to show vectors so will have to assume that RP and RQ and OQ are vectors with O the origin.
OP = OR + (2/3)RQ
OP = (1 -3 8) + (2/3).(4 8 -5)
OP = (11/3 7/3 14/3) as required
Case 2 P lies outwith line QR on same side as Q
RP = 2RQ
RP = 2 x (4 8 5) = (8 16 -10)
OP = OR + RP (vectors)
OP = (1 -3 8) + (8 16 -10)
OP = (9 13 -2)
P is point (9,13,-2)

Answer 5

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