Question

Using section formula, show that the points A(2, -3, 4), B (-1, 2, 1) and $C(0, \frac{1}{3}, 2)$ are collinear.

Answer 1

Please see the attachment

Answer 2

The points A(2, -3, 4), B (-1, 2, 1) and  C (0,$\frac{1}{3}$,2) are collinear.

• Given :

Points A(2, -3, 4), B (-1, 2, 1) and C (0,$\frac{1}{3}$,2) .

• The points will be collinear if any one point divides the other two points in some ratio internally or externally.
• We assume that the point A divides the points B and C in some ratio k:1.
• Therefore , By section formula

(2 , -3 , 4) = ($\frac{k(-1) + 1(0)}{k+1}$ , $\frac{k(2) + 1(1/3)}{k+1}$ , $\frac{k(1) + 1(2)}{k+1}$)

• Now,

Comparing both sides we get

2 = $\frac{k(-1) + 1(0)}{k+1}$

2k + 2 = -k

k = -2/3

-3 = $\frac{k(2) + 1(1/3)}{k+1}$

-9k -9 = 6k + 1

-15k = 10

k = -2/3

4 = $\frac{k(1) + 1(2)}{k+1}$

4k + 4 = k + 2

k = -2/3

• Therefore, as the value of k is same for all the coordinates, the point A divides the points B and C in the ratio -2/3 : 1.

• That is the points A lies outside the segment B and C.

• Therefore, the points are collinear.