Let`s count out the coordinates of point Z using the formula, where x1, y1, z1 are coordinates of point X, x2, y2, z2 are coordinates of point Y and 2/3 is ratio.
Xz=(x1+2/3∗x2)/(1+2/3)=(1+4/3)/(5/3)=7/3∗3/5=1.4;
Yz=(y1+2/3∗y2)/(1+2/3)=(2+2/3∗3)/(1+2/3)=4∗3/5=12/5=2.4;
Zz=(z1+2/3∗z2)/(1+2/3)=(4+2/3∗5)/(1+2/3)=22/3∗3/5=4.4;
So, point Z has coordinates (1.4, 2.4, 4.4).
Comments
The question was correctly solved. If X(1,2,4), Y(2,3,5), then Z(7/5,12/5,22/5) bisects XY in the ratio 2:3 because |XZ|=2sqrt(3)/5, |ZY|=3sqrt(3)/5, |XY|=sqrt(3), hence |XZ|:|ZY|=2/3.
u:v of X and Y Given (vX + uY)/u+v (3(1,2,4)+2(2,3,5))/(2+3) (3(i+2j+4k)+2(2i+3j+5k))/5 ((3i+6j+12k)+(4i+6j+10k))/5 (3i+4i+6j+6j+12k+10k)/5 Answer = 1/5 (7i+12j+22k) If you a NOUN student with this option 1/7 (7i+12j+22k)