Find the value of p for which the points A(-1, 3, 2), B(-4, 2, -2), and C(5, 5, p) are collinear.
We have to show that the three points are colinear , i.e. they all lie on the same line,
If we define a line which is having a parallel line to AB and the points A and B lie on it, as the points are colinear so C must satisfy the line,
Given A(-1, 3, 2) and B(-4, 2, -2), AB = -3i – j -4k
The points on the line AB with A on the line can be written as,
R = (-1, 3, 2) +a(-3, -1, -4)
Let C = (-1-3a, 3-1a, 2-4a)
ð (5, 5, p) = (-1-3a, 3-1a, 2-4a)
ð As L.H.S = R.H.S, thus
ð 5 = -1 – 3a, a = -2
Substituting a = -2 we get, p = 2-4(-2) = 10
Hence, p = 10.