Show that the points A(3,2,−4),B(5,4,−6) and C(9,8,−10) are collinear, find the ratio in which B divides .

Engineering

Mathematics

Equation of Straight Line in 3D

Section Formulae and Centres of a Triangle

Question

Show that the points A(3,2,−4),B(5,4,−6) and C(9,8,−10) are collinear, find the ratio in which B divides $\vec{AC}$ .

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Given that

Point A(3,2,–4), B(5,4,–6) & C(9,8,–10) are collinear

B must divide line segment AC in some ratio externally and internally

We know that

Co-ordinate of point A(x,y,z) that divides line segment joining (x₁,y₁,z₁) and (x₂,y₂,z₂) in ratio m:n is $(x, y, z) = (\frac{{mx}_{2} + {nx}_{1}}{m + n}, \frac{{my}_{2} + {ny}_{1}}{m + n})$

Let point B(5,4,−6) divide line segment A(3, 2, −4) and C(9, 8, −10) in the ratio k : 1

B (5, 4, - 6) = (\frac{9 k + 3}{k + 1}, \frac{8 k + 2}{k + 1}, \frac{- 10 k - 4}{k + 1})

Comparing x-coordinate of B

5 = \frac{9 k + 3}{k + 1}

5k + 5 = 9k + 3

9k − 5k = 5 − 3

4k = 2

k = 12

So, k : 1 = 1 : 2

Thus, point B divides AC in the ratio 1:2