Question

Suppose the position vector of X and Y are (1,2,4) and (2,3,5), find the position vector of a point Z that bisect XY in the ratio 2:3.

a.\\(7i+12j+22k\\)
b.\\(7i-12j+22k\\)
c.\\(\\frac{1}{7} (7i+12j+22k)\\)
d.\\(\\frac{1}{17} (7i-12j+22k)\\)

Answer 1

Coordinates of the point Z:

$x=\frac{x_1+\frac{2}{3}x_2}{1+\frac{2}{3}}=\frac{1+\frac{2}{3}*2}{1+\frac{2}{3}}=\frac{7}{5}.$

$y=\frac{y_1+\frac{2}{3}y_2}{1+\frac{2}{3}}=\frac{2+\frac{2}{3}*3}{1+\frac{2}{3}}=\frac{12}{5}.$

$z=\frac{z_1+\frac{2}{3}z_2}{1+\frac{2}{3}}=\frac{4+\frac{2}{3}*5}{1+\frac{2}{3}}=\frac{22}{5}.$

So, the position vector of a point Z is:

$\frac{1}{5}(7i+12j+22k).$