Question #105782
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
1
Expert's answer
2020-03-18T18:03:21-0400

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+2/3x2)/(1+2/3)=(1+4/3)/(5/3)=7/33/5=1.4;X_z = (x_1 + 2/3 * x_2) / (1 + 2/3)\\ = (1+ 4/3) / (5/3) =7/3 * 3/5 = 1.4;

Yz=(y1+2/3y2)/(1+2/3)=(2+2/33)/(1+2/3)=43/5=12/5=2.4;Y_z = (y_1 + 2/3 * y_2) / (1 + 2/3) \\= (2 + 2/3*3) / (1 + 2/3) = 4 * 3/5 = 12/5 = 2.4;

Zz=(z1+2/3z2)/(1+2/3)=(4+2/35)/(1+2/3)=22/33/5=4.4;Z_z = (z_1 + 2/3 * z_2) / (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).



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Comments

Assignment Expert
25.03.20, 15:31

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.

Victoria NOUN
25.03.20, 01:46

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)

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