Determine wether the given points are collinear.?

Question

P (1,2,3) Q (0,3,7) R(3,5,11)

Answer 1

No, they are not.

Look at Q (0,3,7) to P (1,2,3). As x increases by 1 (1-0=1), y decreases by 1 (2-3=-1), and z decreases by 4 (3-7=-4).

From P (1,2,3) to R (3,5,11), x increases by twice as much (3-1=2), so y should decrease by twice as much (2*-1 = -2) and z should decrease by twice as much (2*-4 = -8).

The point colinear with PQ and having x=3 (like R does) would be P (1,2,3) plus (2,-2,-8):
(1+2, 2-2, 3-8) = (3, 0, -5). Since that is not the same as the y and z values of R, R is not on the PQ line -- and the three points are not colinear.

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The slightly longer version is to derive the equations for the lines. For PQ, you'll have two equations (3-d equations of the first degree define a plane, so you have to have two equations for the intersection of two planes to define a line).

You can derive them as if they were a linear equation in x&y, then an independent one in x&z, just is if it were two 2-d problems:

y = mx + b
y = (y2-y1)/(x2-x1) * x + b
y = (2-3)/(1-0) * x + b
y = -x + b

when x=0, y=3 (point Q):

y = -x + b
3 = -0 + b
b = 3

Thus:

y = -x + 3

Then the second equation:

z = mx + b
z = (z2-z1)/(x2-x1) * x + b
z = (3-7)/(1-0) * x + b
z = -4x + b

When x=0, z=7 (point Q)

z = -4x + b
7 = -4*0 + b
7 = b

Thus:

z = -4x + 7

So...

Your two equations for line PQ are:

y = -x + 3
z = -4x + 7

You can plug the values for R into BOTH equations and see if they satisfy them:

y =? -x + 3
5 =? -3 + 3
5 =? 0 (doesn't work)

z =? -4x + 7
11 =? -4*3 + 7
11 =? -5 (doesn't work)

If (and only if) the values for R satisfy both of the equations for PQ, then R is colinear with PQ.

Answer 2

Ignore the z-value and find the slope between P and Q then Q and R. If those values are not the same then the three points are not collinear.

Answer 3

its quiet obivious not