Answer to Question #105782 in Analytic Geometry for Henry Finebone

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

Expert's answer

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.

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

"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;"

"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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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)

Answer to Question #105782 in Analytic Geometry for Henry Finebone

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