Question

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

Answer 1

Position vector of point Z = $\textbf{\Large 1i + }\frac{\textbf{\Large 12}}{\textbf{\Large 7} }\textbf{\Large j} + \frac{\textbf{\Large 22}}{\textbf{\Large 7} }\textbf{\Large k}$

Step-by-step explanation:

The position vector of X = (1,2,4) = $(x_1, y_1, z_1)$

The position vector of Y = (2,3,5) = $(x_2, y_2, z_2)$

A point Z bisects XY in the ratio 2: 3  = a:b     (given )

The formula to find the points of Z where the line is bisected in the ratio 2:3

$\textbf{\large Z}= \frac{bx_1 + ax_2}{a + b} i + \frac{by_1 + ay_2}{a + b}j + \frac{bz_1 + az_2}{a + b}k$           (vector form)         (equation 1)

$\textbf{\large x}_3= \frac{bx_1 + ax_2}{a + b} \\\textbf{\large y}_3= \frac{by_1 + ay_2}{a + b}\\\textbf{\large z}_3= \frac{bz_1 + az_2}{a + b}$                                              (coordinate form)     (equation 2)

putting all the values in equation 1 and 2.

$\textbf{\Large So, The solution is :}$

$\textbf{\large Z}= \frac{(3\times 1)+ (2\times 2)}{2 + 3} i + \frac{(3\times 2) + (2\times 3)}{2 + 3}j + \frac{(3\times4) + (2\times 5)}{2 + 3}k$

$\textbf{\large Z}= \frac{1}{7}(7i + 12j +22k)$

$\\\textbf{\Large x}_3 =\textbf{\Large 1}\\ \\\textbf{\Large y}_3 = \frac{\textbf{\Large12}}{\textbf{\Large7}}\\\textbf{\Large z}_3 =\frac{\textbf{\Large 22}}{\textbf{\Large 7}}$                  

Answer 2

Supose the position of X and Y are (1,2,4) and (2,3,5). The position vector of the point Z that bisect XY in the ratio 2:3 is 1i + 12/7j + 22/7k.

Stepwise explanation is given below:

- The position vector of X = (1,2,4) = (x1,y1,z1)

The position vector of Y = (2,3,5) = (x2,y2,z2)

- The position vector of Z = (x3,y3,z3)

A point Z bisects XY in the ratio 2: 3  = a:b     (given )

- The formula to find the points of Z where the line is bisected in the ratio 2:3.

Z=[(bx1+ax2)/(a+b)]i + [(by1+ay2)/(a+b)]i + [(bz1+az2)/(a+b)]i

(vector form) (equation 1)

x3= [(bx1+ax2)/(a+b)]i

y3= [(by1+ay2)/(a+b)]i

z3= [(bz1+az2)/(a+b)]i

(coordinate form) (equation 2)

- By putting all the values in equation 1 and 2.

- So, the solution is-

Z= [(3*1)+(2*2)/(2+3)]i + [(3*2)+(2*3)/(2+3)]j + [(3*4)+(2*5)/(2+3)]k

Z= (1/7)(7i+12j+22k)

x3= 7/7=1

y3= 12/7

z3= 22/7