Question

Suppose the position vector of x and y are (1,2,4) and (2,3,5) ratio 2:3

Answer 1

The required position vector of point  of bisection = $\frac{1}{7} (7 i + 12 j +22 k)$

Step-by-step explanation:

In the given question, two points of a line are given .

It is said that there is a point (let Z), which is incident on the line such that,

It cuts the line in the ratio of 2: 3.

So,

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

$Z = \frac{bx_1 + ax_2}{a + b} i+ \frac{by_1 + by_2}{a+b}j + \frac{bz_1 + az_2}{a+b}k$            (vector form)         (equation 1)

Putting all the values  in equation 1:

$x_1 = 1\\x_2 = 2\\y_1 = 2\\y_2 = 3\\z_1 = 4\\z_2 = 5\\a = 2\\b = 3$

 $Z = \frac{3\times 1 + 2\times 2}{2+3 } i+ \frac{3\times 2+ 2\times 3}{2+3}j + \frac{3\times 4 + 2\times 5}{2+3}k$

$\textbf{\Large So, the required position vector Z = } \frac{\textbf{\Large 1}}{\textbf{\Large 7}} \textbf{\Large (7 i + 12j + 22 k)}$